RS Aggarwal Class 10 Mathematics Solutions Chapter 14 Height and Distances

Exercise 14(A)

Question 1. A tower stands vertically on the ground. From a point O on the ground 20 m away from the base of the tower, the angle of elevation to the top is 60°, and the angle of elevation to a point halfway up the tower is 90°. Find the height of the tower.
Answer: Let AB represent the tower standing upright on the ground, with O being the observer's location. We are given that \( OA = 20 \text{ m} \), \( \angle OAB = 90° \), and \( \angle AOB = 60° \).

Let \( AB = h \text{ m} \).

In the right triangle OAB:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AB}{OA} = \sqrt{3} \]
\[ \frac{h}{20} = \sqrt{3} \]
\[ h = 20\sqrt{3} = (20 \times 1.732) = 34.64 \text{ m} \]

Therefore, the tower's height is 34.64 m.
In simple words: When you measure the angle up to the top of the tower from a point 20 m away, it makes a 60° angle. Using the trigonometric ratio for tangent, you can calculate the height as approximately 34.64 metres.

Exam Tip: Always identify the right angle in the triangle first, then apply the correct trigonometric ratio (sine, cosine, or tangent) based on which sides are known and which is unknown.

Question 2. A kite is flying with a string of length 75 m. The string makes an angle of 60° with the horizontal ground. Find the height of the kite and the horizontal distance from the observer.
Answer: Let OX be the horizontal ground and A be the position of the kite. O is the observer's location and OA represents the thread.

We are given: \( \angle BOA = 60° \), \( OA = 75 \text{ m} \), and \( \angle OBA = 90° \).

Height of the kite from the ground: \( AB = 75 \text{ m} \)
Length of the string: \( OA = x \text{ m} \)

In the right triangle OBA:
\[ \sin 60° = \frac{\sqrt{3}}{2} \]
\[ \frac{AB}{OA} = \sin 60° = \frac{\sqrt{3}}{2} \]
\[ \frac{75}{x} = \frac{\sqrt{3}}{2} \]
\[ x = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{1.732} = 86.6 \text{ m} \]

Therefore, the length of the string is 86.6 m.
In simple words: The kite hangs 75 m above the ground. By measuring the angle the string makes with the ground and knowing basic trigonometry, you can figure out that the string is roughly 86.6 metres long.

Exam Tip: When a problem involves a string or rope at an angle, draw a perpendicular from the object to the ground to form a right triangle, then use sine or cosine appropriately.

Question 3. From a point on the ground 30 m away from the base of a chimney, the angle of elevation to the top is 60°. The observer's eyes are 1.5 m above the ground. Find the height of the chimney.
Answer: Let CE and AD denote the heights of the observer and the chimney, respectively. We have: \( BD = CE = 1.5 \text{ m} \), \( BC = DE = 30 \text{ m} \), and \( \angle ACB = 60° \).

In triangle ABC:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AB}{BC} = \sqrt{3} \]
\[ \frac{AB}{30} = \sqrt{3} \]
\[ AB = 30\sqrt{3} \]

Since \( AD = AB + BD \):
\[ AD = 30\sqrt{3} + 1.5 \]
\[ AD = 30 \times 1.732 + 1.5 \]
\[ AD = 51.96 + 1.5 \]
\[ AD = 53.46 \text{ m} \]

Therefore, the chimney's height is approximately 53.46 m.
In simple words: An observer standing 30 m away looks up at a 60° angle. After doing the trigonometric calculation and adding the observer's eye height (1.5 m), the total chimney height comes to about 53.46 metres.

Exam Tip: When the observer is not at ground level, subtract or add their height to the calculated height to get the total height of the object.

Question 4. A tower stands on a ground level. From point C, which is 5 m away from the base, the angle of elevation is \( \theta \). From point D, which is 20 m away from the base, the angle of elevation is \( 90° - \theta \). Find the height of the tower.
Answer: Let AB be the tower's height. We have: \( AC = 5 \text{ m} \) and \( AD = 20 \text{ m} \).

From point C, let the angle of elevation be \( \theta \). Then from point D, the angle is \( (90° - \theta) \).

In triangle ABC:
\[ \tan \theta = \frac{AB}{AC} = \frac{AB}{5} \quad \cdots (i) \]

In triangle ABD:
\[ \cot(90° - \theta) = \frac{AD}{AB} \]
\[ \tan \theta = \frac{20}{AB} \quad \cdots (ii) \]

From equations (i) and (ii):
\[ \frac{AB}{5} = \frac{20}{AB} \]
\[ AB^2 = 100 \]
\[ AB = \sqrt{100} = 10 \text{ m} \]

Therefore, the tower's height is 10 m.
In simple words: When angles at two different positions are complementary (add up to 90°), you can set up two equations and solve them to find the height. In this case, the height turns out to be exactly 10 metres.

Exam Tip: Complementary angles (those that add to 90°) often lead to elegant algebraic solutions - use them to eliminate variables and solve faster.

Question 5. A tower of height BC stands on level ground. A flagstaff CD is mounted on top of it. From a point A on the ground 120 m away, the angle of elevation to the base of the flagstaff is 45°, and to the top is 60°. Find the heights of the tower and the flagstaff.
Answer: Let BC represent the tower and CD represent the flagstaff. We have: \( AB = 120 \text{ m} \), \( \angle BAC = 45° \), and \( \angle BAD = 60° \).

Let \( CD = x \text{ m} \).

In triangle ABC:
\[ \tan 45° = 1 \]
\[ \frac{BC}{AB} = 1 \]
\[ BC = 120 \text{ m} \]

In triangle ABD:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{BD}{AB} = \sqrt{3} \]
\[ \frac{BC + CD}{120} = \sqrt{3} \]
\[ 120 + x = 120\sqrt{3} \]
\[ x = 120\sqrt{3} - 120 \]
\[ x = 120(\sqrt{3} - 1) \]
\[ x = 120(1.732 - 1) \]
\[ x = 120 \times 0.732 \]
\[ x = 87.8 \text{ m} \]

Therefore, the height of the tower is 120 m and the height of the flagstaff is 87.8 m.
In simple words: From a point 120 m away, the angle to the flagstaff's base is 45° and to its top is 60°. Using these two angles, you can find both the tower height and the flagstaff height separately.

Exam Tip: When two angles of elevation are given for different levels on the same vertical structure, set up separate equations and subtract to isolate the smaller component.

Question 6. A tower BC and a water tank beneath it are located on level ground. From point A, 40 m away, the angle of elevation to the top of the tank is 45°, and to the bottom of the tank (which is also the top of the tower) is 30°. Find the height of the tower and the depth of the tank.
Answer: Let BC be the tower and CD be the water tank. We have: \( AB = 40 \text{ m} \), \( \angle BAC = 30° \), and \( \angle BAD = 45° \).

In triangle ABD:
\[ \tan 45° = 1 \]
\[ \frac{BD}{AB} = 1 \]
\[ BD = 40 \text{ m} \]

In triangle ABC:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{BC}{AB} = \frac{1}{\sqrt{3}} \]
\[ BC = \frac{40}{\sqrt{3}} = \frac{40\sqrt{3}}{3} \text{ m} \]

(i) Height of the tower:
\[ BC = \frac{40\sqrt{3}}{3} = \frac{40 \times 1.73}{3} = 23.1 \text{ m} \]

(ii) Depth of the tank:
\[ CD = BD - BC = 40 - 23.1 = 16.9 \text{ m} \]

Therefore, the tower's height is 23.1 m and the tank's depth is 16.9 m.
In simple words: From 40 m away, a 45° angle tells you the total depth to the tank bottom, while a 30° angle tells you where the tank starts. The difference between these two heights gives the tank's depth.

Exam Tip: When calculating depth or distance between two levels, always find both heights separately using their respective angles, then subtract to find the difference.

Question 7. A tower AB and a flagstaff BC of height 6 m are mounted together. From point O, the angle of elevation to the top of the tower is 30°, and to the top of the flagstaff is 60°. Find the height of the tower.
Answer: Let AB be the tower and BC be the flagstaff. We have: \( BC = 6 \text{ m} \), \( \angle AOB = 30° \), and \( \angle AOC = 60° \).

Let \( AB = h \text{ m} \).

In triangle AOB:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{AB}{OA} = \frac{1}{\sqrt{3}} \]
\[ OA = h\sqrt{3} \quad \cdots (i) \]

In triangle AOC:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AC}{OA} = \sqrt{3} \]
\[ \frac{AB + BC}{h\sqrt{3}} = \sqrt{3} \quad \text{[Using (i)]} \]
\[ 3h = h + 6 \]
\[ 3h - h = 6 \]
\[ 2h = 6 \]
\[ h = 3 \text{ m} \]

Therefore, the tower's height is 3 m.
In simple words: The tower and a 6 m flagstaff sit one above the other. From the observer's spot, the angles to their tops are 30° and 60° respectively. Working through the trigonometry shows the tower is 3 metres tall.

Exam Tip: When two structures are stacked vertically, express the upper angle in terms of the lower angle and the height difference to set up a solvable equation.

Question 8. A statue of height 1.46 m stands on a pedestal. From a point on the ground, the angle of elevation to the top of the pedestal is 45°, and to the top of the statue is 60°. Find the height of the pedestal.
Answer: Let AC be the pedestal and BC be the statue, where \( BC = 1.46 \text{ m} \). Let \( AC = h \text{ m} \) and \( AD = x \text{ m} \).

In the right triangle ADC:
\[ \tan 45° = 1 \]
\[ \frac{AC}{AD} = 1 \]
\[ \frac{h}{x} = 1 \]
\[ h = x \]

In the right triangle ADB:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AB}{AD} = \sqrt{3} \]
\[ \frac{h + 1.46}{x} = \sqrt{3} \]

Substituting \( x = h \):
\[ \frac{h + 1.46}{h} = \sqrt{3} \]
\[ h + 1.46 = h\sqrt{3} \]
\[ h(\sqrt{3} - 1) = 1.46 \]
\[ h = \frac{1.46}{(\sqrt{3} - 1)} = \frac{1.46}{0.73} = 2 \text{ m} \]

Therefore, the pedestal's height is 2 m.
In simple words: A 1.46 m statue sits atop a pedestal. From ground level, looking at the pedestal's top gives a 45° angle, while looking at the statue's top gives 60°. The math shows the pedestal is 2 metres high.

Exam Tip: Complementary angle problems (like 45° and 60°) often simplify when you express one height in terms of another and substitute to eliminate variables.

Question 9. An unfinished tower AB has a raised portion AC. From point O, 75 m away, the angle of elevation to the top of the unfinished section is 30°, and to the raised portion is 60°. Find the height gained by raising the tower.
Answer: Let AB be the unfinished tower and AC be the raised tower such that \( BC = (H - h) \text{ m} \). We have: \( OA = 75 \text{ m} \), \( \angle AOB = 30° \), and \( \angle AOC = 60° \).

Let \( AC = H \text{ m} \) such that \( BC = (H - h) \text{ m} \).

In triangle AOB:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{AB}{OA} = \frac{1}{\sqrt{3}} \]
\[ \frac{h}{75} = \frac{1}{\sqrt{3}} \]
\[ h = \frac{75}{\sqrt{3}} = \frac{75\sqrt{3}}{3} = 25\sqrt{3} \text{ m} \]

In triangle AOC:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AC}{OA} = \sqrt{3} \]
\[ \frac{H}{75} = \sqrt{3} \]
\[ H = 75\sqrt{3} \text{ m} \]

Required height gain:
\[ H - h = (75\sqrt{3} - 25\sqrt{3}) = 50\sqrt{3} = 86.6 \text{ m} \]

Therefore, the height gained by raising the tower is 86.6 m.
In simple words: Two angles of elevation (30° and 60°) from 75 m away show the before and after heights of a tower project. Subtracting these heights reveals that roughly 86.6 metres of height was added.

Exam Tip: When comparing two elevations, calculate each height separately, then find their difference to get the change in height.

Question 10. A tower AD with a flagstaff CD is located on horizontal ground. From point B, 9 m away, the angle of elevation to the base of the flagstaff is 30°, and to its top is 60°. Find the heights of the tower and the flagstaff.
Answer: Let OX be the horizontal ground, AD be the tower, and CD be the vertical flagstaff. We have: \( AB = 9 \text{ m} \), \( \angle DBA = 30° \), and \( \angle CBA = 60° \).

Let \( AD = h \text{ m} \) and \( CD = x \text{ m} \).

In the right triangle ABD:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{AD}{AB} = \frac{1}{\sqrt{3}} \]
\[ \frac{h}{9} = \frac{1}{\sqrt{3}} \]
\[ h = \frac{9}{\sqrt{3}} = 5.19 \text{ m} \]

In the right triangle ABC:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AC}{BA} = \sqrt{3} \]
\[ \frac{h + x}{9} = \sqrt{3} \]
\[ h + x = 9\sqrt{3} \]

By substituting \( h = \frac{9}{\sqrt{3}} \):
\[ \frac{9}{\sqrt{3}} + x = 9\sqrt{3} \]
\[ x = 9\sqrt{3} - \frac{9}{\sqrt{3}} \]
\[ x = \frac{27 - 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18}{1.73} = 10.4 \text{ m} \]

Therefore, the height of the tower is 5.19 m and the height of the flagstaff is 10.4 m.
In simple words: From 9 m away, a 30° angle points to where the flagstaff begins, and 60° points to its top. These two angles let you calculate the tower height and flagstaff height separately.

Exam Tip: Always set up separate equations for each given angle, then solve them in sequence - first finding the lower height, then subtracting to get the upper section's height.

Question 11. Two equal poles AB and CD stand on opposite sides of a road 80 m wide. From the midpoint O on the road, the angle of elevation to the top of each pole is 60° on one side and 30° on the other. Find the height of each pole and the distances from O to each pole.
Answer: Let AB and CD be the equal poles, and BD be the width of the road. We have: \( \angle AOB = 60° \) and \( \angle COD = 60° \).

In triangle AOB:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{AB}{BO} = \sqrt{3} \]
\[ BO = \frac{AB}{\sqrt{3}} \]

In triangle COD:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{CD}{DO} = \frac{1}{\sqrt{3}} \]
\[ DO = \sqrt{3} \cdot CD \]

Since \( BD = 80 \):
\[ BO + DO = 80 \]
\[ \frac{AB}{\sqrt{3}} + \sqrt{3} \cdot CD = 80 \]

Given \( AB = CD \):
\[ \frac{AB}{\sqrt{3}} + \sqrt{3} \cdot AB = 80 \]
\[ AB \left( \frac{1}{\sqrt{3}} + \sqrt{3} \right) = 80 \]
\[ AB \left( \frac{1 + 3}{\sqrt{3}} \right) = 80 \]
\[ AB \left( \frac{4}{\sqrt{3}} \right) = 80 \]
\[ AB = 20\sqrt{3} \text{ m} \]

Also:
\[ BO = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \text{ m} \]
\[ DO = 80 - 20 = 60 \text{ m} \]

Therefore, the height of each pole is \( 20\sqrt{3} \) m, point P is 20 m from the left pole and 60 m from the right pole.
In simple words: Two identical poles face each other across an 80 m road. Different angles from the midpoint reveal how far each pole is from the centre and their exact height.

Exam Tip: When poles or structures are equal in height, use this fact to create a single equation that relates the distances and eliminate one variable.

Question 12. A tower CD stands between two men at positions A and B on the same straight line. The angles of elevation from A and B to the top are 30° and 45° respectively. The tower is 50 m tall. Find the distance between the two men.
Answer: Let CD be the tower and A and B be the positions of the two men standing on opposite sides. We have: \( \angle DAC = 30° \), \( \angle DBC = 45° \), and \( CD = 50 \text{ m} \).

Let \( AB = x \text{ m} \) and \( BC = y \text{ m} \) such that \( AC = (x - y) \text{ m} \).

In the right triangle DBC:
\[ \tan 45° = 1 \]
\[ \frac{CD}{BC} = 1 \]
\[ \frac{50}{y} = 1 \]
\[ y = 50 \text{ m} \]

In the right triangle ACD:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{CD}{AC} = \frac{1}{\sqrt{3}} \]
\[ \frac{50}{(x - y)} = \frac{1}{\sqrt{3}} \]
\[ x - y = 50\sqrt{3} \]

By substituting \( y = 50 \):
\[ x - 50 = 50\sqrt{3} \]
\[ x = 50 + 50\sqrt{3} = 50(\sqrt{3} + 1) \]
\[ x = 50(1.732 + 1) = 50 \times 2.732 = 136.6 \text{ m} \]

Therefore, the distance between the two men is 136.6 m.
In simple words: Two observers stand on opposite sides of a 50 m tower. Their angles of elevation (30° and 45°) tell you how far each one is from the tower base. Adding these distances gives the total separation.

Exam Tip: When two observers are on opposite sides of an object, add their individual distances from the object to get the total separation between them.

Question 13. A tower PQ of height 100 m stands vertically. Two cars travel on a straight road and pass points A and B. From A, the angle of elevation is 30°, and from B, it is 45°. Find the distance between the two cars.
Answer: Let PQ be the tower. We have: \( PQ = 100 \text{ m} \), \( \angle PQR = 30° \), and \( \angle PBQ = 45° \).

In triangle APQ:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{PQ}{AP} = \frac{1}{\sqrt{3}} \]
\[ \frac{100}{AP} = \frac{1}{\sqrt{3}} \]
\[ AP = 100\sqrt{3} \text{ m} \]

In triangle BPQ:
\[ \tan 45° = 1 \]
\[ \frac{PQ}{BP} = 1 \]
\[ \frac{100}{BP} = 1 \]
\[ BP = 100 \text{ m} \]

Now:
\[ AB = AP + BP \]
\[ AB = 100\sqrt{3} + 100 \]
\[ AB = 100(\sqrt{3} + 1) \]
\[ AB = 100(1.73 + 1) \]
\[ AB = 100 \times 2.73 \]
\[ AB = 273 \text{ m} \]

Therefore, the distance between the cars is 273 m.
In simple words: A 100 m tower creates two different angles (30° and 45°) when viewed from two cars' locations. Adding the distances from each car to the tower base gives the separation between them: 273 metres.

Exam Tip: When two objects are on the same side of a structure, the distance between them is the difference in their distances from the structure's base.

Question 14. A moving car passes a tower. At one moment, the angle of elevation from the car to the tower top is 30°. Six seconds later, when the car is nearer, the angle is 60°. Find the time taken for the car to reach the foot of the tower.
Answer: Let PQ be the tower. We have: \( \angle PBO = 60° \) and \( \angle PAQ = 30° \).

Let \( PQ = h \), \( AB = x \), and \( BQ = y \).

In triangle APQ:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{PQ}{AQ} = \frac{1}{\sqrt{3}} \]
\[ \frac{h}{x + y} = \frac{1}{\sqrt{3}} \]
\[ x + y = h\sqrt{3} \quad \cdots (i) \]

In triangle BPQ:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{PQ}{BQ} = \sqrt{3} \]
\[ \frac{h}{y} = \sqrt{3} \]
\[ h = y\sqrt{3} \quad \cdots (ii) \]

Substituting \( h = y\sqrt{3} \) in (i):
\[ x + y = \sqrt{3}(y\sqrt{3}) \]
\[ x + y = 3y \]
\[ 3y - y = x \]
\[ 2y = x \]
\[ y = \frac{x}{2} \]

Speed of the car from A to B = \( \frac{AB}{6} = \frac{x}{6} \) units/sec.

Time to reach the foot of the tower (from B):
\[ \text{Time} = \frac{BQ}{\text{speed}} = \frac{y}{\frac{x}{6}} = \frac{y \times 6}{x} = \frac{\frac{x}{2} \times 6}{x} = \frac{6}{2} = 3 \text{ sec} \]

Therefore, the time taken to reach the foot of the tower is 3 seconds.
In simple words: A car travels toward a tower. At one point, the angle is 30°; after 6 seconds of driving, it becomes 60°. By calculating distances and the car's speed, you can find it takes 3 more seconds to arrive at the tower base.

Exam Tip: When a moving object changes its angle of elevation to a fixed point, use the time difference and angle change to determine speed, then calculate remaining time.

Question 15. A TV tower stands on the bank of a canal. From point A on the opposite bank, 20 m away from point B (the foot of the tower), the angle of elevation to the tower top is 30°. From point B, the angle is 60°. Find the height of the tower and the width of the canal.
Answer: Let \( PQ = h \text{ m} \) be the height of the TV tower and \( BQ = x \text{ m} \) be the width of the canal. We have: \( AB = 20 \text{ m} \), \( \angle PAQ = 30° \), and \( \angle BQ = x \).

In triangle PBQ:
\[ \tan 60° = \sqrt{3} \]
\[ \frac{PQ}{BQ} = \sqrt{3} \]
\[ \frac{h}{x} = \sqrt{3} \]
\[ h = x\sqrt{3} \quad \cdots (i) \]

In triangle APQ:
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
\[ \frac{PQ}{AQ} = \frac{1}{\sqrt{3}} \]
\[ \frac{h}{20 + x} = \frac{1}{\sqrt{3}} \]
\[ h\sqrt{3} = 20 + x \]
\[ x\sqrt{3} \times \sqrt{3} = 20 + x \quad \text{[Using (i)]} \]
\[ 3x = 20 + x \]
\[ 2x = 20 \]
\[ x = 10 \text{ m} \]

From (i):
\[ h = 10\sqrt{3} \approx 17.3 \text{ m} \]

Therefore, the height of the tower is approximately 17.3 m and the width of the canal is 10 m.
In simple words: A tower stands across a canal from an observer. Two different angles of elevation (30° and 60°) from points 20 m apart help determine both the canal width (10 m) and the tower height (about 17.3 metres).

Exam Tip: When two observation points are on opposite sides of water or a gap, express heights in terms of the unknown distance, then use both equations to solve for it.

Question 15. A TV tower stands on a canal bank. From a point on the opposite bank directly across from the tower, the angle of elevation to the top is 30°. From another point 20 m away (measured along the bank), the angle of elevation is 60°. Find the height of the TV tower and the width of the canal.
Answer: Let PQ represent the TV tower with height h metres. Let the width of the canal be x metres. When observing from the point directly opposite the tower, the angle of elevation is 30°. This gives us \( \tan 30° = \frac{h}{x} \), which simplifies to \( \frac{1}{\sqrt{3}} = \frac{h}{x} \). Therefore, \( h = \frac{x}{\sqrt{3}} \). From the second observation point, which is 20 metres further along the bank, the angle of elevation becomes 60°. At this location, the distance from the tower's base is (x + 20) metres. This produces the equation \( \tan 60° = \frac{h}{x + 20} \), giving us \( \sqrt{3} = \frac{h}{x + 20} \). Substituting the earlier relationship \( h = \frac{x}{\sqrt{3}} \) into this equation: \( \sqrt{3} = \frac{x/\sqrt{3}}{x + 20} \). Multiplying both sides by (x + 20): \( \sqrt{3}(x + 20) = \frac{x}{\sqrt{3}} \). Cross-multiplying yields \( 3(x + 20) = x \), which expands to \( 3x + 60 = x \). Solving: \( 2x = -60 \) appears incorrect. Let me reconsider: from \( \sqrt{3} = \frac{x/\sqrt{3}}{x + 20} \), we get \( \sqrt{3}(x + 20) = \frac{x}{\sqrt{3}} \). Rearranging: \( 3(x + 20) = x \) gives \( 3x + 60 = x \), so \( 2x = -60 \). Actually, correcting the setup: \( \tan 30° = \frac{h}{x} \) means from point A directly opposite, and \( \tan 60° = \frac{h}{20} \) means 20 metres away. From the first: \( h = \frac{x}{\sqrt{3}} = \frac{x\sqrt{3}}{3} \). From the second: \( h = 20\sqrt{3} \). Therefore: \( \frac{x\sqrt{3}}{3} = 20\sqrt{3} \), giving x = 60. Wait, let me use the correct interpretation from the source. Actually: \( \tan 30° = \frac{h}{AB + BQ} \) where BQ = x and AB = 20. So \( \frac{1}{\sqrt{3}} = \frac{h}{20 + x} \). And \( \tan 60° = \frac{h}{x} \) gives \( \sqrt{3} = \frac{h}{x} \), so \( h = x\sqrt{3} \). Substituting: \( \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{20 + x} \). Cross-multiplying: \( 20 + x = 3x \), giving \( 20 = 2x \), so \( x = 10 \) metres. Therefore: \( h = 10\sqrt{3} \approx 17.32 \) metres. The height of the TV tower is \( 10\sqrt{3} \) metres (approximately 17.32 metres) and the width of the canal is 10 metres.
In simple words: The tower is about 17.32 metres tall and the canal is 10 metres wide. Using two different viewing positions with their angles, we can set up equations that let us find both distances.

Exam Tip: Always set up two separate angle equations from the two observation points, then substitute one into the other to eliminate a variable and solve systematically.

Question 16. A 60 m tall tower is observed from point A. The angle of elevation from point B to the top of the building AB is 30°, and the angle of elevation from point B to the top of the tower is 60°. Find the height of the building AB.
Answer: Let AB denote the building with unknown height, and let PQ denote the tower with height 60 metres. Point B is where both angles are measured. From point B, the angle of elevation to the building's top is 30°, and to the tower's top is 60°. In triangle ABP, we apply the tangent ratio: \( \tan 60° = \frac{PQ}{AP} \). This gives \( \sqrt{3} = \frac{60}{AP} \). Solving for AP: \( AP = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \) metres. Now, in triangle ABP (where we find the height of the building), we use: \( \tan 30° = \frac{AB}{AP} \). This simplifies to \( \frac{1}{\sqrt{3}} = \frac{AB}{20\sqrt{3}} \). Multiplying both sides by \( 20\sqrt{3} \): \( AB = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \) metres. The height of the building is 20 metres.
In simple words: Using the angle of 60° to the tower top, we find the horizontal distance. Then using the angle of 30° to the building top with that same distance, we calculate the building's height to be 20 metres.

Exam Tip: When two objects are observed from the same point with different angles of elevation, always compute the horizontal distance first from one angle, then use it with the other angle to find the unknown height.

Question 17. Two towers DE and AB stand such that DE = 55.36 m and AB = 90 m. A horizontal distance CE = 60 m separates them. An observer at point B sees point E on the first tower at an angle of depression of 30°. Find the height of the first tower DE.
Answer: Let DE represent the first tower with unknown height h metres, and let AB represent the second tower with height 90 metres. Point C marks the top of the first tower, and E is positioned such that the horizontal distance CE equals 60 metres. Point B is the top of the second tower from which the angle of depression to point E is 30°. In the right triangle BCE, the angle of depression of 30° means the angle of elevation from E upward to B is also 30°. Therefore: \( \tan 30° = \frac{BC}{CE} \), which gives \( \frac{1}{\sqrt{3}} = \frac{BC}{60} \). Solving: \( BC = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \) metres. The vertical distance BC represents the difference in heights: \( BC = AB - AC = AB - DE \). So: \( 20\sqrt{3} = 90 - DE \). Rearranging: \( DE = 90 - 20\sqrt{3} = 90 - 20(1.732) = 90 - 34.64 = 55.36 \) metres. The height of the first tower is 55.36 metres.
In simple words: The angle of depression from the taller tower to a point on the shorter tower creates a right triangle. Using that angle and the known horizontal distance, we find how much taller the second tower is, then subtract to get the first tower's height.

Exam Tip: Angles of depression from an upper point equal the corresponding angles of elevation from the lower point - use this relationship to set up your tangent equation correctly.

Question 18. A chimney PQ stands 120 m tall. From point P on the ground near a tower AB, the angle of elevation to the top of the tower is 30°. From the same point, the angle of elevation to the top of the chimney is 60°. The tower AB is 40 m high. Verify that the chimney's height meets pollution control standards (height must exceed 100 m).
Answer: Let AB denote the tower with height 40 metres, and let PQ represent the chimney with unknown height. Point P is the observation location. From point P, the angle of elevation to the tower's top is 30°. Using the tangent ratio in triangle ABP: \( \tan 30° = \frac{AB}{AP} \), which gives \( \frac{1}{\sqrt{3}} = \frac{40}{AP} \). Solving: \( AP = 40\sqrt{3} \) metres. Now, from the same point P, the angle of elevation to the chimney's top is 60°. Applying the tangent ratio in triangle PQA (with the same horizontal distance AP): \( \tan 60° = \frac{PQ}{AP} \), which gives \( \sqrt{3} = \frac{PQ}{40\sqrt{3}} \). Solving: \( PQ = 40\sqrt{3} \times \sqrt{3} = 40 \times 3 = 120 \) metres. The height of the chimney is 120 metres, which exceeds the 100-metre standard, so it satisfies environmental regulations.
In simple words: From the same point on the ground, steeper angles mean taller objects. The 60° angle (steeper than 30°) corresponds to the taller chimney of 120 metres, which is well above the 100-metre pollution limit.

Exam Tip: When two objects of different heights are viewed from the same point, larger angles of elevation correspond to taller structures. Always verify your final answer against any stated constraints mentioned in the problem.

Question 19. A 7 m high building AB stands near a cable tower CD. From point A, the angle of elevation to the base D of the tower is 45°, and the angle of elevation to the top C of the tower is 60°. Find the height of the tower CD.
Answer: Let AB denote the 7-metre-high building, and let CD represent the cable tower with unknown height. The base of the tower, D, is on the same level as the top of the building, so DE = AB = 7 metres. In right triangle ABD (where E is at the height of point A), the angle of elevation is 45°. Using: \( \tan 45° = \frac{AB}{BD} \), we get \( 1 = \frac{7}{BD} \), so \( BD = 7 \) metres. This means the horizontal distance AE = BD = 7 metres. In right triangle ACE (measuring from A upward to C), the angle of elevation is 60°. Using: \( \tan 60° = \frac{CE}{AE} \), we get \( \sqrt{3} = \frac{CE}{7} \), so \( CE = 7\sqrt{3} \) metres. The total height of the tower is the sum: \( CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \) metres. Substituting \( \sqrt{3} \approx 1.732 \): \( CD = 7(1.732 + 1) = 7(2.732) = 19.124 \approx 19.12 \) metres. The height of the tower is approximately 19.12 metres.
In simple words: The building is 7 metres tall. From its top, the tower extends further up. Using two different angles (45° to the tower's base level and 60° to its top), we can calculate how much higher the tower reaches, which turns out to be about 19.12 metres total.

Exam Tip: When an upper observation point already has height above ground level, clearly identify your reference points and work with segments above and below that level separately.

Question 20. A tower PQ is observed from two points A and B on level ground. From point A, the angle of elevation to the top of the tower is 30°. From point B, which is 20 m closer to the tower, the angle of elevation is 60°. Find the height of the tower and its horizontal distance from point A.
Answer: Let PQ denote the tower with height h metres. Let BQ = x metres denote the distance from point B to the tower's base. Then AQ = 20 + x metres, since point A is 20 metres further away. From point B, the angle of elevation is 60°. In triangle PBQ: \( \tan 60° = \frac{PQ}{BQ} \), giving \( \sqrt{3} = \frac{h}{x} \). Therefore: \( h = x\sqrt{3} \) ... (i) From point A, the angle of elevation is 30°. In triangle PAQ: \( \tan 30° = \frac{PQ}{AQ} \), giving \( \frac{1}{\sqrt{3}} = \frac{h}{20 + x} \). Substituting equation (i): \( \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{20 + x} \). Cross-multiplying: \( 20 + x = 3x \). Solving: \( 20 = 2x \), so \( x = 10 \) metres. From equation (i): \( h = 10\sqrt{3} = 10(1.732) = 17.32 \) metres. The distance from A to the tower's base: \( AQ = 20 + 10 = 30 \) metres. The height of the tower is 17.32 metres, and its horizontal distance from point A is 30 metres.
In simple words: Two observers see the same tower from different distances. The closer observer (B) sees it at a steeper angle (60°) while the farther observer (A) sees it at a gentler angle (30°). Using these two angles and the 20-metre separation, we can work out both the tower's height and the distances.

Exam Tip: Always express distances from both observation points in terms of the same unknown, ensuring your equations reflect the actual geometric layout before solving.

Question 21. From a point M above a building of height 10 m, a tower PQ is observed. The angle of elevation from M to the top of the tower is 30°. From a point N directly below M at the building's base, the angle of elevation to the top of the tower is 60°. Find the height of the tower PQ.
Answer: Let AB represent the building with height 10 metres, and let PQ denote the tower with unknown height h metres. Point M is at the top of the building, and point N is at the base. The tower stands at a horizontal distance from the building. Let BQ = x metres denote this horizontal distance. From point N (ground level), the angle of elevation to the tower's top is 60°. In triangle NBQ: \( \tan 60° = \frac{PQ}{BQ} \), giving \( \sqrt{3} = \frac{h}{x} \). Therefore: \( x = \frac{h}{\sqrt{3}} \) ... (i) From point M (at height 10 metres), the angle of elevation to the tower's top is 30°. The vertical distance from M to P is (h - 10) metres, and the horizontal distance remains x. In triangle PQM: \( \tan 30° = \frac{h - 10}{x} \), giving \( \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \). Substituting equation (i): \( \frac{1}{\sqrt{3}} = \frac{h - 10}{h/\sqrt{3}} \). Simplifying: \( \frac{1}{\sqrt{3}} = \frac{(h - 10)\sqrt{3}}{h} \). Cross-multiplying: \( h = 3(h - 10) \), which gives \( h = 3h - 30 \). Solving: \( 30 = 2h \), so \( h = 15 \) metres. The height of the tower PQ is 15 metres.
In simple words: An observer at ground level sees the entire 15-metre tower at a 60° angle. An observer 10 metres higher up (on the building) sees only the top 5 metres of the tower at a 30° angle. These two sightings fit together perfectly when the tower is 15 metres tall.

Exam Tip: When the observation point is elevated, carefully track the vertical segment you are measuring (the full height vs. the portion above the observer's eye level) in your tangent ratio.

Question 22. A tower AD stands on level ground. From a point C at a horizontal distance of 60 m, the angle of depression to the base of the tower is 60°, and the angle of elevation to the top is 45°. Find the height of the tower AD.
Answer: Let AD represent the tower with unknown height h metres. Let BC denote the cliff or elevated position with height 60\(\sqrt{3}\) metres. Point C is at the observation location. From point C, the angle of depression to the tower's base D is 45°. The angle of depression equals the angle of elevation from D looking back to C. This means in triangle CDE (where E is at the tower's base level relative to C): \( \tan 45° = \frac{CE}{DE} \). Given that CE = 60\(\sqrt{3}\) - h (the vertical drop from C), and letting DE be the horizontal distance, we have: \( 1 = \frac{60\sqrt{3} - h}{DE} \), so \( DE = 60\sqrt{3} - h \) ... (i) From point C, the angle of elevation to the tower's top A is 60°. The vertical rise from C to A is h - 60\(\sqrt{3}\). Wait, this needs reconsideration based on the diagram description. Let me recalculate using the correct setup. Actually, if BC = 60\(\sqrt{3}\) is the cliff height, BE = h is the tower height, then CE = BC - BE = 60\(\sqrt{3}\) - h. In triangle CDE with angle of depression 45° (at C): \( \tan 45° = \frac{CE}{DE} = \frac{60\sqrt{3} - h}{DE} \), giving \( DE = 60\sqrt{3} - h \). Also, AB = DE, so AB = 60\(\sqrt{3}\) - h ... (i) In triangle ABC with angle of elevation 60° (at C): \( \tan 60° = \frac{BC}{AB} = \frac{60\sqrt{3}}{60\sqrt{3} - h} \). This gives \( \sqrt{3} = \frac{60\sqrt{3}}{60\sqrt{3} - h} \). Cross-multiplying: \( \sqrt{3}(60\sqrt{3} - h) = 60\sqrt{3} \), which expands to \( 180 - h\sqrt{3} = 60\sqrt{3} \). Solving: \( h\sqrt{3} = 180 - 60\sqrt{3} \), so \( h = \frac{180 - 60\sqrt{3}}{\sqrt{3}} = \frac{180}{\sqrt{3}} - 60 = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1) = 60(1.732 - 1) = 60(0.732) = 43.92 \) metres. The height of the tower is 43.92 metres.
In simple words: From an elevated position, an observer sees the tower's base below at a steep 45° angle, and the tower's top above at a steeper 60° angle. The different angles tell us how high the observer is relative to both the base and top of the tower, letting us calculate the tower's exact height.

Exam Tip: Angles of depression and elevation from the same point are measured in opposite directions (down vs. up). Set up separate equations for each, using the vertical distances and horizontal distance appropriately.

Question 23. A ship's deck is 16 m above water level. From the deck, an observer sees the top of a cliff at an angle of elevation of 60°, and the base of the cliff at an angle of depression of 30°. Find the height of the cliff and the horizontal distance from the ship.
Answer: Let AB represent the ship's deck at 16 metres above water. Let DE denote the cliff with unknown height h metres, where D is at water level. The observer at point B on the deck measures angles to both the cliff's top C and base D. The angle of depression of 30° to the cliff's base D means: \( \tan 30° = \frac{AB}{AD} \), giving \( \frac{1}{\sqrt{3}} = \frac{16}{AD} \). Solving: \( AD = 16\sqrt{3} \approx 27.68 \) metres (horizontal distance). The angle of elevation of 60° to the cliff's top C means the vertical rise from B to C must be measured. If DE = h is the cliff's full height and AB = 16 is the deck height, then the rise from B to C is (h - 16) metres. Thus: \( \tan 60° = \frac{h - 16}{AD} \), giving \( \sqrt{3} = \frac{h - 16}{16\sqrt{3}} \). Solving: \( h - 16 = 16\sqrt{3} \times \sqrt{3} = 16 \times 3 = 48 \). Therefore: \( h = 48 + 16 = 64 \) metres. The height of the cliff is 64 metres, and the horizontal distance from the ship is approximately 27.68 metres.
In simple words: The deck is 16 metres high. Looking down at 30° finds the cliff base 27.68 metres away. Looking up at 60° finds the cliff top, which must be 64 metres above water to create that steeper upward angle.

Exam Tip: When angles of elevation and depression are measured from the same point, ensure you correctly identify the vertical segments - one measures down from your position, the other measures up from your position.

Question 24. An observer at point Y, which is 40 m above ground, views a cliff. The angle of depression to the cliff's base X is 60°, and the angle of elevation to the cliff's top Z is 45°. Find the height of the cliff PQ.
Answer: Let XY denote the vertical drop from the observer's position Y to ground level (40 metres). The observer measures an angle of depression of 60° downward to point X at ground level, and an angle of elevation of 45° upward to point Z at the cliff's top. From the angle of depression of 60° to the base: \( \tan 60° = \frac{40}{MY} \), where MY is the horizontal distance. This gives \( \sqrt{3} = \frac{40}{MY} \), so \( MY = \frac{40}{\sqrt{3}} = \frac{40\sqrt{3}}{3} \) metres ... (i) From the angle of elevation of 45° to the cliff's top: \( \tan 45° = \frac{MQ}{MY} \), where MQ is the vertical height from Y to Z. This gives \( 1 = \frac{MQ}{MY} \), so \( MQ = MY = \frac{40\sqrt{3}}{3} \) metres. The cliff's total height is: \( PQ = MQ + YP = \frac{40\sqrt{3}}{3} + 40 = \frac{40\sqrt{3} + 120}{3} \). Factoring: \( PQ = \frac{40(\sqrt{3} + 3)}{3} \) metres. Numerically: \( \sqrt{3} + 3 \approx 1.732 + 3 = 4.732 \). So: \( PQ = \frac{40 \times 4.732}{3} \approx \frac{189.28}{3} \approx 63.09 \) metres. Wait, let me recalculate using the source's approach. From \( \tan 60° = \frac{MQ}{PX} \) where PX is the horizontal distance, we get... Actually, reviewing the source setup: \( MY = h - 40 \) where h is the tower height. \( \tan 45° = \frac{MQ}{MY} \) gives \( 1 = \frac{h - 40}{MY} \), so \( MY = h - 40 \). Then \( \tan 60° = \frac{PQ}{PX} = \frac{h}{h - 40} \), giving \( \sqrt{3} = \frac{h}{h - 40} \). Cross-multiplying: \( h\sqrt{3} - 40\sqrt{3} = h \). Solving: \( h(\sqrt{3} - 1) = 40\sqrt{3} \), so \( h = \frac{40\sqrt{3}}{\sqrt{3} - 1} \). Rationalizing: \( h = \frac{40\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{40\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{40\sqrt{3}(\sqrt{3} + 1)}{2} = 20\sqrt{3}(\sqrt{3} + 1) = 20(3 + \sqrt{3}) = 60 + 20\sqrt{3} \approx 60 + 34.6 = 94.6 \) metres. The height of the cliff is approximately 94.6 metres.
In simple words: Standing 40 metres above ground, the observer sees both down to the cliff base and up to the cliff top. The two angles (60° down and 45° up) along with the 40-metre height of the observer create a solvable system that reveals the cliff reaches about 94.6 metres high.

Exam Tip: When working with combined angles of depression and elevation from an elevated observation point, set up your variables carefully - the horizontal distance is shared between both sight lines.

Question 25. An aeroplane flies at constant height 2500 m. From point A on the ground, the angle of elevation to the aeroplane at position P is 45°. After some time, the angle of elevation to position Q is 30°. Assuming the aeroplane travels horizontally, find its speed if it takes 15 seconds to move from P to Q. Express the answer in m/s and km/h.
Answer: Let PQ denote the flight path at constant height 2500 metres. Point A is the ground observation point. When the aeroplane is at position P, the angle of elevation is 45°. In the right triangle APQ (where Q is directly below P): \( \tan 45° = \frac{PQ}{AQ} \), giving \( 1 = \frac{2500}{AQ} \). Therefore: \( AQ = 2500 \) metres (horizontal distance to P). When the aeroplane is at position Q, the angle of elevation is 30°. In the right triangle AQ'Q (where Q' is directly below Q): \( \tan 30° = \frac{PQ}{AQ'} \), giving \( \frac{1}{\sqrt{3}} = \frac{2500}{AQ'} \). Therefore: \( AQ' = 2500\sqrt{3} \) metres (horizontal distance to Q). The horizontal distance travelled by the aeroplane is: \( QC = AQ' - AQ = 2500\sqrt{3} - 2500 = 2500(\sqrt{3} - 1) \) metres. Calculating: \( QC = 2500(1.732 - 1) = 2500(0.732) = 1830 \) metres. The speed of the aeroplane is: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1830}{15} = 122 \) m/s. Converting to km/h: \( 122 \text{ m/s} = 122 \times \frac{3600}{1000} = 122 \times 3.6 = 439.2 \) km/h. The speed of the aeroplane is 122 m/s or 439.2 km/h.
In simple words: The plane stays at 2500 metres high. From different observation angles (45° then 30°), we can find how far it travelled horizontally. Dividing that distance by the travel time gives us the speed - about 122 metres per second or roughly 439 kilometres per hour.

Exam Tip: When an object maintains constant height, changes in elevation angle directly reveal changes in horizontal distance. Always compute the horizontal distances first, then find the path distance the object travelled.

Question 26. A tower AB stands between two observation points C and D. From point C, the angle of elevation to the tower's top is 30°, and from point D, the angle of elevation is 60°. The distance CD between the points is 150 m. Find the height of the tower.
Answer: Let AB denote the tower with unknown height h metres. Points C and D are on the ground 150 metres apart, with D closer to the tower than C. From point C, the angle of elevation to the tower's top is 30°. If the horizontal distance from C to the tower's base is CB, then: \( \tan 30° = \frac{h}{CB} \), giving \( \frac{1}{\sqrt{3}} = \frac{h}{CB} \). Therefore: \( CB = h\sqrt{3} \) ... (i) From point D, the angle of elevation to the tower's top is 60°. If the horizontal distance from D to the tower's base is DB, then: \( \tan 60° = \frac{h}{DB} \), giving \( \sqrt{3} = \frac{h}{DB} \). Therefore: \( DB = \frac{h}{\sqrt{3}} = \frac{h\sqrt{3}}{3} \) ... (ii) The distance between C and D is: \( CD = CB - DB = 150 \) metres. Substituting equations (i) and (ii): \( h\sqrt{3} - \frac{h\sqrt{3}}{3} = 150 \). Simplifying: \( h\sqrt{3}\left(1 - \frac{1}{3}\right) = 150 \), so \( h\sqrt{3} \times \frac{2}{3} = 150 \). Therefore: \( h\sqrt{3} = 225 \), giving \( h = \frac{225}{\sqrt{3}} = \frac{225\sqrt{3}}{3} = 75\sqrt{3} \approx 75 \times 1.732 = 129.9 \) metres. The height of the tower is approximately 129.9 metres.
In simple words: Two observers at different distances see the same tower at different angles. The steeper 60° angle is from the closer point, and the gentler 30° angle is from the farther point. Their 150-metre separation and these angles uniquely determine the tower's height to be about 129.9 metres.

Exam Tip: Always set up distance equations showing how the observation points relate to the object (whether the object is between them, or they are on the same side). This prevents sign errors in your algebra.

Question 27. A lighthouse OA stands 100 m tall on a coast. A ship at position C observes the lighthouse top at an angle of elevation of 60°, and later at position B observes it at an angle of 30°. The ship moves from B to C (moving away from the lighthouse). Find the distance the ship travels.
Answer: Let OA denote the lighthouse with height 100 metres. The ship is initially at position B, then moves to position C, both on the water at sea level. From position C (closer to the lighthouse), the angle of elevation to the lighthouse top is 60°. In the right triangle OAC: \( \tan 60° = \frac{OA}{OC} \), giving \( \sqrt{3} = \frac{100}{OC} \). Therefore: \( OC = \frac{100}{\sqrt{3}} = \frac{100\sqrt{3}}{3} \) metres. From position B (farther from the lighthouse), the angle of elevation is 30°. In the right triangle OAB: \( \tan 30° = \frac{OA}{OB} \), giving \( \frac{1}{\sqrt{3}} = \frac{100}{OB} \). Therefore: \( OB = 100\sqrt{3} \) metres. The distance the ship travels is: \( BC = OB - OC = 100\sqrt{3} - \frac{100\sqrt{3}}{3} = 100\sqrt{3}\left(1 - \frac{1}{3}\right) = 100\sqrt{3} \times \frac{2}{3} = \frac{200\sqrt{3}}{3} \) metres. Calculating: \( BC = \frac{200 \times 1.732}{3} \approx \frac{346.4}{3} \approx 115.47 \) metres. The distance the ship travels is approximately 115.47 metres.
In simple words: As the ship sails away from the lighthouse, the angle at which it sees the lighthouse top decreases from 60° to 30°. These two angles, combined with the 100-metre height, tell us the ship moved roughly 115 metres further out to sea.

Exam Tip: When an object moves away from a fixed reference point, larger angles of elevation occur at closer positions. Always subtract the farther distance from the closer distance to get the distance travelled in the correct direction.

Question 28. A bridge spans a river at height 2.5 m above the water. From point P on the bridge, two points A and B on opposite banks are observed at angles of depression 30° and 45° respectively. Find the width of the river.
Answer: Let P represent a point on the bridge 2.5 metres above the water surface. Points A and B are on opposite banks of the river. The angle of depression to point A is 30°, and to point B is 45°. From the angle of depression of 30° to point A: \( \tan 30° = \frac{DP}{AD} \), where DP = 2.5 and AD is the horizontal distance to bank A. This gives \( \frac{1}{\sqrt{3}} = \frac{2.5}{AD} \). Solving: \( AD = 2.5\sqrt{3} \) metres. From the angle of depression of 45° to point B: \( \tan 45° = \frac{DP}{BD} \), where DP = 2.5 and BD is the horizontal distance to bank B. This gives \( 1 = \frac{2.5}{BD} \). Solving: \( BD = 2.5 \) metres. The width of the river is the sum of these distances (if the banks are on opposite sides of the bridge): \( AB = AD + BD = 2.5\sqrt{3} + 2.5 = 2.5(\sqrt{3} + 1) = 2.5(1.732 + 1) = 2.5(2.732) = 6.83 \) metres. The width of the river is approximately 6.83 metres.
In simple words: From a point 2.5 metres high on the bridge, we look down at steeper and gentler angles to reach opposite banks. The steeper 45° angle finds the closer bank 2.5 metres away, while the gentler 30° angle finds the far bank 4.33 metres away, for a total width of about 6.83 metres.

Exam Tip: Angles of depression are measured downward from the horizontal line through your observation point. Use complementary geometry to set up your tangent ratios correctly when working with depression angles.

Question 29. A tower AB has unknown height. From point C at a distance of 4 m, the angle of elevation is \(\theta\). From point D at a distance of 9 m, the angle of elevation is (90° - \(\theta\)). Find the height of the tower.
Answer: Let AB represent the tower with height h metres. Point C is 4 metres from the tower's base, and point D is 9 metres away. From point C, the angle of elevation is \(\theta\). This gives: \( \tan \theta = \frac{h}{4} \) ... (1) From point D, the angle of elevation is (90° - \(\theta\)). This angle satisfies the complementary relationship: \( \tan(90° - \theta) = \cot \theta = \frac{1}{\tan \theta} \). Therefore: \( \cot \theta = \frac{h}{9} \), which means \( \frac{1}{\tan \theta} = \frac{h}{9} \) ... (2) Multiplying equations (1) and (2): \( \tan \theta \times \frac{1}{\tan \theta} = \frac{h}{4} \times \frac{h}{9} \). This simplifies to: \( 1 = \frac{h^2}{36} \). Solving: \( h^2 = 36 \), so \( h = 6 \) metres. The height of the tower is 6 metres.
In simple words: Two observers at different distances see a tower at complementary angles (they add up to 90°). The observer closer to the tower sees a steeper angle, while the farther observer sees a gentler angle. This complementary relationship, combined with the distances 4 and 9 metres, uniquely determines the tower is 6 metres tall.

Exam Tip: Complementary angles in elevation problems often lead to elegant algebraic solutions. When angles are complementary, their tangent and cotangent values are reciprocals, which creates powerful relationships for solving.

Question 30. Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at the point O on the ground. We have, AO = CO = 6 m, ∠AOB = 60° and ∠COD = 45°. In △ABO, find BO. Also, in △CDO, find DO. Now, calculate the distance between two walls of the room = BD.
Answer: In triangle ABO, using the cosine ratio:
\( \cos 60° = \frac{BO}{AO} \)
\( \frac{1}{2} = \frac{BO}{6} \)
\( BO = 3 \text{ m} \)

In triangle CDO, using the cosine ratio:
\( \cos 45° = \frac{DO}{CO} \)
\( \frac{1}{\sqrt{2}} = \frac{DO}{6} \)
\( DO = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \text{ m} \)

The horizontal distance between the two walls equals:
\( BD = BO + DO = 3 + 3\sqrt{2} = 3(1 + \sqrt{2}) = 3(1 + 1.414) = 3(2.414) = 7.242 \approx 7.24 \text{ m} \)
In simple words: Use the cosine function to get the distances from each wall to the ladder's base. Add them together to find how far apart the walls are.

Exam Tip: Always identify which trigonometric ratio to use by checking which sides are given relative to the angle; for this problem, the cosine ratio connects the adjacent side and the hypotenuse directly.

Question 31. Let OP be the tower and points A and B be the positions of the cars. We have, AB = 100 m, ∠OAP = 60° and ∠OBP = 45°. Let OP = h. Find the height of the tower.
Answer: In triangle AOP, using the tangent ratio:
\( \tan 60° = \frac{OP}{OA} \)
\( \sqrt{3} = \frac{h}{OA} \)
\( OA = \frac{h}{\sqrt{3}} \)

In triangle BOP, using the tangent ratio:
\( \tan 45° = \frac{OP}{OB} \)
\( 1 = \frac{h}{OB} \)
\( OB = h \)

Since the two cars are 100 m apart on the same line:
\( OB - OA = 100 \)
\( h - \frac{h}{\sqrt{3}} = 100 \)
\( \frac{h\sqrt{3} - h}{\sqrt{3}} = 100 \)
\( \frac{h(\sqrt{3} - 1)}{\sqrt{3}} = 100 \)
\( h = \frac{100\sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \)
\( h = \frac{100\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{100\sqrt{3}(\sqrt{3} + 1)}{2} \)
\( h = \frac{100(3 + \sqrt{3})}{2} = 50(3 + 1.732) = 50(4.732) = 236.6 \text{ m} \)
In simple words: Set up two equations using tangent for each angle. Use the fact that the two points are 100 m apart to find the tower height.

Exam Tip: When two observation points lie on the same line, always write the distance between them as the difference of the two calculated distances from the base; rationalize the denominator when needed to simplify the final expression.

Question 32. Let AC be the pole and BD be the ladder. We have, AC = 4 m, AB = 1 m and ∠BDC = 60°. Also, BC = AC - AB = 4 - 1 = 3 m. In △BDC, find the length of the ladder BD.
Answer: In triangle BDC, using the sine ratio:
\( \sin 60° = \frac{BC}{BD} \)
\( \frac{\sqrt{3}}{2} = \frac{3}{BD} \)
\( BD = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} \)
\( BD = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \)
\( BD = 2 \times 1.73 = 3.46 \text{ m} \)
In simple words: The ladder makes a 60° angle with the ground. Use the sine function with the height difference between the ladder's top and bottom to find the ladder's length.

Exam Tip: When dealing with a ladder leaning against a pole, identify the angle formed at the base and use the sine ratio since you know the opposite side (vertical distance) and need the hypotenuse (ladder length).

Question 33. We have, AB = 60 m, ∠ACE = 30° and ∠ADB = 60°. Let BD = CE = x and CD = BE = y. So, AE = AB - BE = 60 - y. Find (i) the horizontal distance between AB and CD = BD = x, (ii) the height of the lamp post CD = y, and (iii) the difference between the heights of the building and the lamp post = AB - CD.
Answer: In triangle ACE, using the tangent ratio:
\( \tan 30° = \frac{AE}{CE} \)
\( \frac{1}{\sqrt{3}} = \frac{60 - y}{x} \)
\( x = (60 - y)\sqrt{3} \)
\( x = 60\sqrt{3} - y\sqrt{3} \) ... (i)

In triangle ABD, using the tangent ratio:
\( \tan 60° = \frac{AB}{BD} \)
\( \sqrt{3} = \frac{60}{x} \)
\( x = \frac{60}{\sqrt{3}} = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \)

Substituting \( x = 20\sqrt{3} \) in equation (i):
\( 20\sqrt{3} = 60\sqrt{3} - y\sqrt{3} \)
\( y\sqrt{3} = 60\sqrt{3} - 20\sqrt{3} = 40\sqrt{3} \)
\( y = 40 \text{ m} \)

(i) The horizontal distance between AB and CD = \( BD = x = 20\sqrt{3} = 20 \times 1.732 = 34.64 \text{ m} \)
(ii) The height of the lamp post = \( CD = y = 40 \text{ m} \)
(iii) The difference between the heights of the building and the lamp post = \( AB - CD = 60 - 40 = 20 \text{ m} \)
In simple words: Use the two angles of elevation to set up a system of equations. Solve for the unknown distances using tangent ratios, and then calculate each requested distance.

Exam Tip: When multiple heights and distances are unknown, express them in terms of one variable and use the two given angles to create independent equations; always verify your solution by substituting back into both original equations.

Question 1. If the height of the pole AB equals the length of the shadow BC, and θ is the angle of elevation of the sun, what is the measure of θ?
(a) 30°
(b) 45°
(c) 60°
(d) None of the options
Answer: (c) 45°
In simple words: When the pole height and shadow length are equal, the angle of elevation is 45° because tan(45°) = 1, meaning the opposite side equals the adjacent side.

Exam Tip: For angle of elevation problems, always set up the tangent ratio using opposite over adjacent; if these two sides are equal, the angle must be 45°.

Question 2. A pole AO has a shadow BO of length x. The height of the pole is \( x\sqrt{3} \), and θ is the angle of elevation of the sun. What is the measure of θ?
(a) 30°
(b) 45°
(c) 60°
(d) None of the options
Answer: (c) 60°
In simple words: When the pole height is √3 times the shadow length, the tangent of the angle is √3, which corresponds to 60°.

Exam Tip: Memorize the standard tangent values: tan(30°) = 1/√3, tan(45°) = 1, and tan(60°) = √3; recognizing these helps solve angle of elevation questions quickly.

Question 3. A pole AB has height h and casts a shadow BC such that BC = \( \sqrt{3}h \). If θ is the angle of elevation of the sun, what is θ?
(a) 60°
(b) 30°
(c) 45°
(d) None of the options
Answer: (b) 30°
In simple words: When the shadow length is √3 times the pole height, the tangent of the angle is 1/√3, which equals tan(30°).

Exam Tip: Always take the reciprocal of the tangent ratio when needed; if the shadow is longer than the pole, the angle of elevation will be less than 45°.

Question 4. A pole AB has height 12 m and casts a shadow BC of length \( 4\sqrt{3} \) m. If θ is the angle of elevation of the sun, what is θ?
(a) 60°
(b) 45°
(c) 30°
(d) None of the options
Answer: (a) 60°
In simple words: Calculate the tangent as height divided by shadow length: 12 ÷ (4√3) = √3, which gives an angle of 60°.

Exam Tip: Always simplify the tangent ratio completely before matching it to a known angle value; rationalizing the denominator helps reveal the standard form.

Question 5. A stick AB has height 5 m and casts a shadow BC of 2 m. A tree PQ has height 12.5 m. If both are viewed from the same angle of elevation of the sun, how long is the tree's shadow QR?
(a) 2 m
(b) 3 m
(c) 4 m
(d) 5 m
Answer: (d) 5 m
In simple words: Since the sun's angle is the same for both objects, their height-to-shadow ratios must be equal. Set up a proportion: 5/2 = 12.5/QR and solve to get QR = 5 m.

Exam Tip: When two objects cast shadows at the same angle of elevation, use the property of similar triangles: the ratio of height to shadow length is constant for all objects in that situation.

Question 6. A ladder AC leans against a wall AB. The horizontal distance BC from the wall base to the ladder base is 2 m, and the angle between the ladder and the wall is 60°. What is the length of the ladder?
(a) 2 m
(b) 3 m
(c) 3.5 m
(d) 4 m
Answer: (d) 4 m
In simple words: Use the cosine ratio: cos(60°) = BC/AC = 2/AC. Since cos(60°) = 1/2, solve to get AC = 4 m.

Exam Tip: When a ladder makes an angle with a wall, use cosine if you know the adjacent side (distance from wall) and the hypotenuse (ladder length); the angle given is measured from the wall, not the ground.

Question 7. A ladder AC of length 15 m leans against a wall AB. The angle ∠BAC at the wall is 60°. What is the height AB that the ladder reaches up the wall?
(a) 10 m
(b) 8 m
(c) 7.5 m
(d) 6 m
Answer: (c) 7.5 m
In simple words: Use the cosine ratio: cos(60°) = AB/AC = AB/15. Since cos(60°) = 1/2, we get AB = 15/2 = 7.5 m.

Exam Tip: The angle at the wall determines which trigonometric ratio to use; here, cosine connects the wall height (adjacent to the angle) with the ladder length (hypotenuse).

Question 8. A tower AB stands vertically on level ground. From a point C on the ground 30 m from the base B, the angle of elevation to the top A is 30°. What is the height of the tower?
(a) \( 5\sqrt{3} \) m
(b) \( 10\sqrt{3} \) m
(c) 15 m
(d) 20 m
Answer: (b) \( 10\sqrt{3} \) m
In simple words: Use the tangent ratio: tan(30°) = AB/BC = AB/30. Since tan(30°) = 1/√3, solve to get AB = 30/√3 = 10√3 m.

Exam Tip: For towers viewed from the ground, use tangent of the angle of elevation to relate the height (opposite) to the horizontal distance (adjacent).

Question 9. A tower AB stands vertically. From point C on the ground, the angle of elevation to the top is 30°. The tower is 150 m tall. What is the horizontal distance BC from the tower base?
(a) 100√3 m
(b) 150√3 m
(c) 200 m
(d) 300 m
Answer: (b) 150√3 m
In simple words: Use the tangent ratio: tan(30°) = AB/BC = 150/BC. Since tan(30°) = 1/√3, solve to get BC = 150√3 m.

Exam Tip: When the height is given and distance is unknown, rearrange the tangent formula to isolate the distance term; always rationalize if needed for the final answer.

Question 10. A kite is at position A, and its string AC has length 60 m. The vertical height of the kite above the ground is AB = 30 m. If θ = ∠ACB, what is θ?
(a) 45°
(b) 30°
(c) 60°
(d) None of the options
Answer: (b) 30°
In simple words: Use the sine ratio: sin(θ) = AB/AC = 30/60 = 1/2. Since sin(30°) = 1/2, the angle is 30°.

Exam Tip: When the hypotenuse and opposite side are known, use sine to find the angle; recognize standard sine values such as sin(30°) = 1/2, sin(45°) = 1/√2, and sin(60°) = √3/2.

Question 11. A cliff AB has height 20 m. A tower CD is located beyond the cliff such that CE = AB = 20 m. From point A on the cliff, the angles ∠ACB = ∠CAE = ∠DAE = θ. Find the height of the tower CD.
(a) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Answer: (b) 40 m
In simple words: From the cliff, use tangent to find the relationship between the angle and distances. In the first triangle, tan(θ) = 20/BC, which gives BC = 20/tan(θ). In the second triangle, DE = 20·tan(θ). Then CD = CE + DE = 20 + 20·tan(θ). Solving these equations yields CD = 40 m.

Exam Tip: For problems with two vertical structures and equal angles of elevation from the same observation point, express distances in terms of the angle, then use the constraint that the angles are equal to solve the system.

Question 12. A lamp post AB stands vertically. A girl CD of height 1.5 m stands at horizontal distance AD = 3 m from the post. Her shadow DE extends 4.5 m beyond her position. If the angle of elevation of the sun to the top of the girl's head is θ, find the height of the lamp post AB.
(a) 2 m
(b) 2.2 m
(c) 2.5 m
(d) 3 m
Answer: (c) 2.5 m
In simple words: From the girl's position, find tan(θ) = 1.5/4.5 = 1/3 using her height and shadow length. Then apply this same ratio to the lamp post from point E: tan(θ) = AB/(AD + DE) = AB/7.5. Since tan(θ) = 1/3, we get AB = 2.5 m.

Exam Tip: When a person casts a shadow under the sun, their angle of elevation is consistent throughout that location; use the known measurements from a smaller object to determine the trigonometric ratio, then apply it to find the larger object's height.

Question 13. An object D is viewed from point A at an angle of elevation 30°, and from point B at an angle of elevation 45°. Points A and B are on the ground 2x apart, with A closer to the base. The height of object D above point C (which is between A and B on the ground) is h. Find the height h of the object.
(a) \( x(\sqrt{3} - 1) \)
(b) \( x(\sqrt{3} + 1) \)
(c) \( x\sqrt{3} \)
(d) 2x
Answer: (a) \( x(\sqrt{3} - 1) \) (or another equivalent form depending on interpretation)
In simple words: Set up equations using tan(30°) and tan(45°) from each observation point. From point A at distance AC from the base, tan(30°) = h/AC. From point B at distance BC from the base, tan(45°) = h/BC. Use the constraint that A and B are 2x apart to solve for h in terms of x.

Exam Tip: When two angles of elevation are given from different points on a horizontal line, express the height using both angles, then use the distance constraint to eliminate one variable and solve for the height.

Question 13. Let CD = h be the height of the tower. We have AB = 2x, ∠DAC = 30° and ∠DBC = 45°. In △BCD, tan 45° = CD/BC ⟹ 1 = h/BC ⟹ BC = h. Now, in △ACD, tan 30° = CD/AC ⟹ 1/√3 = h/(AB + BC) ⟹ 1/√3 = h/(2x + h) ⟹ 2x + h = h√3 ⟹ h√3 - h = 2x ⟹ h(√3 - 1) = 2x ⟹ h = 2x/(√3 - 1) × (√3 + 1)/(√3 + 1) ⟹ h = 2x(√3 + 1)/(3 - 1) ⟹ h = 2x(√3 + 1)/2 ⟹ h = x(√3 + 1)m. Hence, the correct answer is option (d).

Exam Tip: When dealing with two angles of elevation from points on the same line, set up separate trigonometric equations and eliminate the unknown distance to find the height directly.

Question 14. Let AB be the rod and BC be its shadow; and θ be the angle of elevation of the sun. We have AB : BC = 1 : √3. Let AB = x. Then, BC = x√3. In △ABC, tan θ = AB/BC ⟹ tan θ = x/(x√3) ⟹ tan θ = 1/√3 ⟹ tan θ = tan 30° ⟹ θ = 30°. Hence, the correct answer is option (a).

Exam Tip: The angle of elevation of the sun is determined by the ratio of the object's height to its shadow length - remember that tan 30° = 1/√3.

Question 15. Let AB be the pole and BC be its shadow. We have BC = 2√3 m and ∠ACB = 60°. In △ABC, tan 60° = AB/BC ⟹ √3 = AB/(2√3) ⟹ AB = 6 m. Hence, the correct answer is option (b).

Exam Tip: In shadow problems, the angle at the top of the shadow is the angle of elevation; use tan directly with the height over the shadow length.

Question 16. Let the sun's altitude be θ. We have AB = 20 m and BC = 20√3 m. In △ABC, tan θ = AB/BC ⟹ tan θ = 20/(20√3) ⟹ tan θ = 1/√3 ⟹ tan θ = tan 30° ⟹ θ = 30°. Hence, the correct answer is option (a).

Exam Tip: Altitude of the sun refers to the angle of elevation; when the shadow is √3 times the height, the sun's angle is always 30°.

Question 17. Let AB and CD be the two towers such that AB = x and CD = y. We have ∠AEB = 30°, ∠CED = 60° and BE = DE. In △ABE, tan 30° = AB/BE ⟹ 1/√3 = x/BE ⟹ BE = x√3. Also, in △CDE, tan 60° = CD/DE ⟹ √3 = y/DE ⟹ DE = y/√3. As, BE = DE ⟹ x√3 = y/√3 ⟹ x/y = 1/(3) ⟹ x : y = 1 : 3. Hence, the correct answer is option (c).

Exam Tip: When two towers are observed from a common point with equal horizontal distances, the ratio of their heights equals the ratio of the tangents of their angles of elevation.

Question 18. Let AB be the tower and O be the point of observation. Also, ∠AOB = 30° and OB = 30 m. Let AB = h m. In △AOB, we have: tan 30° = AB/OB ⟹ 1/√3 = h/30 ⟹ h = 30/√3 × √3/√3 ⟹ h = 30√3/3 ⟹ h = 10√3 m. Hence, the height of the tower is 10√3 m.

Exam Tip: Always rationalize the denominator when you get a square root in the denominator - multiply both numerator and denominator by the square root.

Question 19. Let AB be the string of the kite and AX be the horizontal line. If BC ⊥ AX, then AB = 100 m and ∠BAC = 60°. Let: BC = h m. In the right △ACE, we have: sin 60° = BC/AB ⟹ √3/2 = h/100 ⟹ h = 100√3/2 ⟹ h = 50√3 m. Hence, the height of the kite is 50√3 m.

Exam Tip: In kite problems, the angle given is typically measured from the horizontal; use sine when finding the vertical height from the string length.

Question 20. Let AB be the tower and C and D be the points of observation on AC. ∠ACB = θ, ∠ADB = 90 - θ and AB = h m. Thus, we have: AC = a, AD = b and CD = a - b. Now, in the right △ABC, we have: tan θ = AB/AC ⟹ h/a = tan θ .........(i). In the right △ABD, we have: tan(90 - θ) = AB/AD ⟹ cot θ = h/b .......(ii). On multiplying (i) and (ii), we have: tan θ × cot θ = (h/a) × (h/b) ⟹ (h/a) × (h/b) = 1 ⟹ h² = ab ⟹ h = √(ab) m. Hence, the height of the tower is √(ab) m.

Exam Tip: When angles are complementary (add to 90°), their tangent and cotangent are reciprocals - this relationship helps simplify products of trigonometric ratios.

Question 21. Let AB be the tower and C and D be the points of observation such that ∠BCD = 30°, ∠BDA = 60°, CD = 20 m and AD = x m. Now, in △ADB, we have: AB/AD = tan 60° = √3 ⟹ AB/x = √3 ⟹ AB = √3 x. In △ACB, we have: AB/AC = tan 30° = 1/√3 ⟹ AB/(20 + x) = 1/√3 ⟹ AB = (20 + x)/√3. ∴ √3 x = (20 + x)/√3 ⟹ 3x = 20 + x ⟹ 2x = 20 ⟹ x = 10. ∴ Height of the tower AB = √3 × 10 = 10√3 m.

Exam Tip: When you have two observation points with different angles, set up two equations using the same height and solve for the distance first, then find the height.

Question 22. Let ABCD be the rectangle in which ∠BAC = 30° and AC = 8 cm. In △BAC, we have: AB/AC = cos 30° = √3/2 ⟹ AB/8 = √3/2 ⟹ AB = 8 × √3/2 = 4√3 m. Again, BC/AC = sin 30° = 1/2 ⟹ BC/8 = 1/2 ⟹ BC = 4 m. ∴ Area of the rectangle = (AB × BC) = (4√3 × 4) = 16√3 cm².

Exam Tip: When a diagonal of a rectangle is given with an angle at one corner, use cosine for the side adjacent to the angle and sine for the opposite side.

Question 23. Let AB be the hill making angles of depression at points C and D such that ∠ADB = 45°, ∠ACB = 30° and CD = 1 km. Let: AB = h km and AD = x km. In △ADB, we have: tan 45° = 1 ⟹ AB/AD ⟹ h/x = 1 ⟹ h = x .........(i). In △ACB, we have: tan 30° = 1/√3 ⟹ AB/AC ⟹ h/(1 + x) = 1/√3 .........(ii). On putting the value of h taken from (i) in (ii), we get: h/(h + 1) = 1/√3 ⟹ √3 h = h + 1 ⟹ √3 h - h = 1 ⟹ h(√3 - 1) = 1 ⟹ h = 1/(√3 - 1). On multiplying the numerator and denominator by (√3 + 1), we get: h = [1 × (√3 + 1)]/[(√3 - 1)(√3 + 1)] ⟹ h = (√3 + 1)/(3 - 1) ⟹ h = (√3 + 1)/2 km. Hence, the height of the hill is [(√3 + 1)/2] km.

Exam Tip: Always rationalize denominators containing surds by multiplying by the conjugate - this gives a cleaner final answer.

Question 24. Let AB be the pole and AC and AD be its shadows. We have: ∠ACB = 30°, ∠ADB = 60° and AB = 15 m. In △ACB, we have cot 30° = AC/AB ⟹ √3 = AC/15 ⟹ AC = 15√3 m. Now, in △ADB, we have: cot 60° = AD/AB ⟹ 1/√3 = AD/15 ⟹ AD = 15/√3 = 15√3/3 = 5√3 m. Difference between the lengths of the shadows = AC - AD = 15√3 - 5√3 = 10√3 m.

Exam Tip: When finding shadow lengths for different sun angles, use cotangent (the reciprocal of tangent) - cot = adjacent/opposite from the angle's perspective.

Question 25. Let AB be the observer and CD be the tower. Draw BE ⊥ CD, let CD = h meters. Then, AB = 1.5 m, BE = AC = 28.5 m and ∠EBD = 45°. DE = (CD - EC) = (CD - AB) = (h - 1.5) m. In right △BED, we have: DE/BE = tan 45° = 1 ⟹ (h - 1.5)/28.5 = 1 ⟹ h - 1.5 = 28.5 ⟹ h = 28.5 + 1.5 = 30 m. Hence, the height of the tower is 30 m.

Exam Tip: In observer-tower problems, account for the observer's eye level by subtracting it from the final tower height - the angle of elevation is measured from the eye level, not the ground.