ML Aggarwal Class 10 Maths Solutions Chapter 20 Heights and Distances

Question 1. An electric pole is 10 metres high. If its shadow is 10√3 metres in length, find the elevation of the sun.
Answer: Let θ represent the angle of elevation. Consider triangle ABC where AB is the pole's height and BC is the shadow length. Since the pole stands perpendicular to the ground, angle ABC is 90 degrees. Working from the triangle, we find:
\( \Rightarrow \tan \theta = \frac{AB}{BC} \)
\( \Rightarrow \tan \theta = \frac{10}{10\sqrt{3}} \)
\( \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} \)
\( \Rightarrow \tan \theta = \tan 30° \)
\( \therefore \theta = 30° \)
So the sun's angle of elevation is 30 degrees.
In simple words: When an object and its shadow are the same length, the sun is at 45 degrees. When the shadow is longer than the object, the angle is smaller, like 30 degrees here.

Exam Tip: Always set up the trigonometric ratio carefully - identify which side is opposite, adjacent, and the hypotenuse relative to the angle of elevation, then match to the appropriate trig function.

Question 2. The angle of elevation of the top of a tower, from a point on the ground and at a distance of 150 m from its foot, is 30°. Find the height of the tower correct to one place of decimal.
Answer: Let the tower be represented as MP with height h metres, and O as a point 150 m away on level ground from the tower's base. The angle of elevation at this point is 30 degrees. In triangle OMP, the angle at M is 90 degrees. Using trigonometry:
\( \Rightarrow \tan 30° = \frac{MP}{OM} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{150} \)
\( \Rightarrow h = \frac{150}{\sqrt{3}} \)
\( \Rightarrow h = 86.6 \) metres
Therefore, the tower stands 86.6 metres tall.
In simple words: When you stand far from a tall structure and look up at an angle of 30 degrees, you can use that angle and your distance to calculate how high it is.

Exam Tip: Always check your final answer is rounded to the requested decimal place - the question asks for one decimal place, so 86.6 is correct, not 86.60.

Question 3. A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 metres away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall.
Answer: Let MP denote the wall of height h metres and O denote a point on level ground 1.5 m from the wall's base. The angle of elevation is 60 degrees. In right triangle OMP with the right angle at M, we apply trigonometry:
\( \Rightarrow \tan 60° = \frac{MP}{OM} \)
\( \Rightarrow \sqrt{3} = \frac{h}{1.5} \)
\( \Rightarrow h = 1.5 \times \sqrt{3} \)
\( \Rightarrow h = 2.598 \approx 2.6 \) metres
The wall's height is 2.6 metres.
In simple words: The ladder, the ground distance, and the wall form a right triangle. The steeper the angle, the taller the wall must be compared to the ground distance.

Exam Tip: Recognise that tan relates the opposite side (wall height) to the adjacent side (ground distance) - this is faster than using sin or cos when both horizontal and vertical distances are involved.

Question 4. What is the angle of elevation of sun when the length of shadow of a vertical pole is equal to its height?
Answer: Let MP be the pole's height and OM be its shadow length. We are given that both are equal, each measuring h metres. Let θ be the angle of elevation. In right triangle OMP with angle 90 degrees at M:
\( \Rightarrow \tan \theta = \frac{MP}{OM} \)
\( \Rightarrow \tan \theta = \frac{h}{h} \)
\( \Rightarrow \tan \theta = 1 \)
\( \Rightarrow \tan \theta = \tan 45° \)
\( \therefore \theta = 45° \)
The sun's angle of elevation is 45 degrees.
In simple words: Whenever an object casts a shadow equal to its own height, the sun is always at exactly 45 degrees above the horizon.

Exam Tip: This is a standard result worth memorising - when height equals shadow length, the angle of elevation is always 45 degrees, regardless of the actual height value.

Question 5. From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100m high, how far is P from the foot of the tower ?
Answer: Let MO represent the tower with O at its base. The angle of elevation from point P to the top M is 30 degrees, and the tower height is 100 metres. In right triangle POM with angle 90 degrees at O:
\( \Rightarrow \tan 30° = \frac{OM}{OP} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{OP} \)
\( \Rightarrow OP = 100 \times \sqrt{3} \)
\( \Rightarrow OP = 173.2 \) metres
Point P is located 173.2 metres from the tower's base.
In simple words: The smaller the angle of elevation, the farther away you must be standing from the base of the structure.

Exam Tip: Notice this question gives the tower height and asks for the ground distance (opposite of Question 2) - use tan with the given height in the numerator, not the denominator.

Question 6. From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.
Answer: Let MP represent the cliff and O represent the buoy. From the diagram, the angle of depression from the cliff top is 20 degrees. By the property of alternate angles, the angle at O in the triangle is also 20 degrees. In right triangle POM with angle 90 degrees at M:
\( \Rightarrow \tan 20° = \frac{PM}{OM} \)
\( \Rightarrow 0.3640 = \frac{92}{OM} \)
\( \Rightarrow OM = \frac{92}{0.3640} \)
\( \Rightarrow OM = 252.74 \approx 253 \) metres
The buoy is 253 metres from the cliff's base.
In simple words: Depression angles work the same way as elevation angles, but you measure downward instead of upward. The alternate angle rule makes the working identical.

Exam Tip: For angle of depression problems, draw a horizontal line from the observer and remember that the angle of depression equals the angle of elevation at the target point (alternate angles with a horizontal line are equal).

Question 7. A boy is flying a kite with a string of length 100 m. If the string is tight and the angle of elevation of the kite is 26° 32', find the height of the kite correct to one decimal place (ignore the height of the boy).
Answer: Let O represent the boy's position and P represent the kite. In right triangle POM, angle PMO is 90 degrees, angle POM is 26 degrees 32 minutes, and the hypotenuse OP is 100 metres. The height MP is what we seek:
\( \Rightarrow \sin 26° 32' = \frac{MP}{OP} \)
\( \Rightarrow 0.4467 = \frac{MP}{100} \)
\( \Rightarrow MP = 0.4467 \times 100 \)
\( \Rightarrow MP = 44.67 \approx 44.7 \) metres
The kite reaches a height of 44.7 metres.
In simple words: When you know the string length (hypotenuse) and the angle, use sine to find the vertical height, since sine involves the opposite side and the hypotenuse.

Exam Tip: When the hypotenuse is given (not the horizontal or vertical distance), use sine or cosine, not tangent. Here, sine is correct because we need the opposite side (height).

Question 8. An electric pole is 10 m high. A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the wire makes an angle of 45° with the horizontal through the foot of the pole, find the length of the wire.
Answer: Let MP denote the pole and the wire extend from P to point O on the ground. In right triangle POM with angle PMO equal to 90 degrees, angle POM equal to 45 degrees, and MP equal to 10 metres:
\( \Rightarrow \sin 45° = \frac{MP}{OP} \)
\( \Rightarrow \frac{1}{\sqrt{2}} = \frac{10}{OP} \)
\( \Rightarrow OP = 10 \times \sqrt{2} \)
\( \Rightarrow OP = 10 \times 1.414 \)
\( \Rightarrow OP = 14.14 \) metres
The wire's length is 14.14 metres.
In simple words: At a 45-degree angle, the hypotenuse (wire) is always √2 times the height of the pole - this is a useful fact to remember.

Exam Tip: For 45-degree angles, √2 ≈ 1.414 always appears - recognising this can save calculation time and help verify your answer.

Question 9. A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ?
Answer: Let θ be the angle of elevation. We know that cos θ = 0.53, which corresponds to cos 58°, so θ = 58 degrees. Let MP represent the tower and O the man's position. In right triangle POM with angle PMO equal to 90 degrees, angle POM equal to 58 degrees, and MP equal to 20 metres:
\( \Rightarrow \tan 58° = \frac{MP}{OM} \)
\( \Rightarrow 1.6003 = \frac{20}{OM} \)
\( \Rightarrow OM = \frac{20}{1.6003} \)
\( \Rightarrow OM = 12.49 \approx 12.5 \) metres
The man stands 12.5 metres from the tower's base.
In simple words: First convert the cosine value to find the actual angle, then use standard trigonometry with that angle to solve for the unknown distance.

Exam Tip: When given a trig ratio value (like cos = 0.53), always use trigonometric tables or a calculator to find the corresponding angle before applying it to the triangle problem.

Question 10. The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38° 30' at a point 6 m from the foot of the tree. Calculate :
(i) the height at which the tree is broken.
(ii) the original height of the tree correct to two decimal places.
Answer:
(i) Let ACB represent the tree. When the wind breaks it at point C, the upper part A falls so that it touches the ground at a point where angle CAB is 38 degrees 30 minutes and the distance AB is 6 metres. Using the right triangle ABC:
\( \Rightarrow \tan 38° 30' = \frac{BC}{AB} \)
\( \Rightarrow 0.7954 = \frac{BC}{6} \)
\( \Rightarrow BC = 0.7954 \times 6 \)
\( \Rightarrow BC = 4.77 \) metres
The tree broke at a height of 4.77 metres.

(ii) Next, find the length of the broken upper part:
\( \Rightarrow \cos 38° 30' = \frac{AB}{AC} \)
\( \Rightarrow 0.7826 = \frac{6}{AC} \)
\( \Rightarrow AC = \frac{6}{0.7826} \)
\( \Rightarrow AC = 7.67 \) metres
The original tree height equals the break point height plus the fallen length: 4.77 + 7.67 = 12.44 metres. The tree was originally 12.44 metres tall.
In simple words: When a tree breaks and falls, you can find both the break height and the original height by using two different trig ratios on the same fallen triangle.

Exam Tip: In broken tree problems, identify which side you can find from the given information, then use other ratios to find remaining sides - typically you'll use tan for one part and cos or sin for another part of the same triangle.

Question 11. An observer 1.5 m tall is 20.5 meters away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Answer: Let CD represent the observer standing 20.5 metres away from tower AB. The observer's eye level is at height 1.5 metres (point E on the diagram), and the tower rises to 22 metres. The relevant height difference from eye level to the tower top is 22 - 1.5 = 20.5 metres. In the right triangle formed, the horizontal distance is 20.5 metres and the vertical height above eye level is 20.5 metres. Therefore:
\( \Rightarrow \tan \theta = \frac{20.5}{20.5} \)
\( \Rightarrow \tan \theta = 1 \)
\( \Rightarrow \theta = 45° \)
The angle of elevation from the observer's eye is 45 degrees.
In simple words: When the height difference equals the horizontal distance, the angle is always 45 degrees, no matter what the actual measurements are.

Exam Tip: Remember to subtract the observer's height from the tower height - the angle is measured from the observer's eye level, not from ground level. This is a common mistake.

Question 12. In the adjoining figure, the angle of elevation from a point P of the top of a tower QR, 50 m high is 60° and that of the tower PT from a point Q is 30°. Find the height of the tower PT, correct to the nearest metre.
Answer: Let the height of tower PT be h metres. From the right-angled triangle PQR, we get \( \tan 60° = \frac{QR}{PQ} \). Substituting the known value, \( \sqrt{3} = \frac{50}{PQ} \), which gives us \( PQ = \frac{50}{\sqrt{3}} \). Now, looking at the right-angled triangle PQT, we apply \( \tan 30° = \frac{h}{PQ} \). This becomes \( \frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}} \), which simplifies to \( h = \frac{50}{\sqrt{3} \times \sqrt{3}} = \frac{50}{3} = 16.7 \). When rounded to the nearest metre, the height of tower PT is 17 m.
In simple words: Use the first triangle to find how far P is from Q. Then use the second triangle with that distance to find the height of PT.

Exam Tip: Always identify which triangle to use for each step and ensure you substitute the correct values. Watch out for the order of substitution when working with two linked triangles.

Question 13. From a point P on the ground, the angle of elevation of the top of a 10 m tall building and a helicopter, hovering over the top of the building are 30° and 60° respectively. Find the height of the helicopter above the ground.
Answer: Let QR represent the tall building, which is 10 m high, and let S be the location of the helicopter. Using the right-angled triangle PQR, we calculate \( \tan 30° = \frac{QR}{PQ} \). Substituting, \( \frac{1}{\sqrt{3}} = \frac{10}{PQ} \), which gives \( PQ = 10\sqrt{3} \) m. Next, from the right-angled triangle PQS, we apply \( \tan 60° = \frac{QS}{PQ} \). This becomes \( \sqrt{3} = \frac{QS}{10\sqrt{3}} \), so \( QS = 10\sqrt{3} \times \sqrt{3} = 30 \) m. Since QS represents the height of the helicopter above the ground, the answer is 30 m.
In simple words: First find the ground distance using the building's angle. Then use that distance with the helicopter's angle to find how high the helicopter is.

Exam Tip: The height QS is measured from ground level, not from the top of the building. Make sure your final answer represents the total height from the ground.

Question 14. An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at the instant.
Answer: Let R be the position of the lower aeroplane at 3125 m height, and S be the position of the upper plane. Using the right-angled triangle PQR, where the angle of elevation is 30°, we write \( \tan 30° = \frac{QR}{PQ} \). Substituting the height, \( \frac{1}{\sqrt{3}} = \frac{3125}{PQ} \), which gives \( PQ = 3125\sqrt{3} \) m. For the upper plane, using the right-angled triangle PQS with an angle of elevation of 60°, we have \( \tan 60° = \frac{QS}{PQ} \). This becomes \( \sqrt{3} = \frac{QS}{3125\sqrt{3}} \), so \( QS = 3125\sqrt{3} \times \sqrt{3} = 9375 \) m. The separation between the two planes equals \( QS - QR = 9375 - 3125 = 6250 \) m.
In simple words: Find how high each plane is using its angle of elevation. Then subtract the lower plane's height from the upper plane's height to get the distance between them.

Exam Tip: Both angles are measured from the same point on the ground, which is why you can use the same horizontal distance PQ for both triangles. Keep your heights organized and subtract in the correct order.

Question 15. A man observes the angle of elevation of the top of a tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m, the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures.
Answer: Let QR be the tower and P be the initial position of the man. From the right-angled triangle PQR with a 45° angle of elevation, we get \( \tan 45° = \frac{QR}{PQ} \), which gives us \( 1 = \frac{QR}{PQ} \), so \( PQ = QR \). After walking 20 m towards the tower, the man reaches point S, where SQ = PQ - 20 = QR - 20. From the right-angled triangle SQR with a 60° angle of elevation, we apply \( \tan 60° = \frac{QR}{SQ} \). Substituting, \( \sqrt{3} = \frac{QR}{QR - 20} \). Cross-multiplying gives \( \sqrt{3}(QR - 20) = QR \), so \( \sqrt{3} \cdot QR - 20\sqrt{3} = QR \). Rearranging, \( QR(\sqrt{3} - 1) = 20\sqrt{3} \). Using approximate values, \( QR(1.732 - 1) = 20 \times 1.732 \), which gives \( 0.732 \cdot QR = 34.64 \), so \( QR = 47.32 \) m. To 2 significant figures, the height is 47 m.
In simple words: At the first spot, the height equals the ground distance. After stepping closer, use the new angle to find a relationship that lets you solve for the height.

Exam Tip: Ensure you correctly identify which distance (PQ or SQ) corresponds to each triangle. When rounding to significant figures, check that your answer has the requested precision.

Question 16. The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.
Answer: Let the tower height be h metres and its shadow length be d metres when the sun's altitude is 45°. From the right-angled triangle ABD, we use \( \tan 45° = \frac{BD}{AB} \), which gives \( 1 = \frac{h}{d} \), so \( h = d \). When the sun's altitude becomes 30°, the shadow length increases to d + 10. Using the right-angled triangle BCD, we apply \( \tan 30° = \frac{BD}{CD} \), where CD = CA + AB = 10 + d. This gives \( \frac{1}{\sqrt{3}} = \frac{h}{10 + d} \). Substituting h = d, we get \( \frac{1}{\sqrt{3}} = \frac{d}{10 + d} \), which leads to \( 10 + d = \sqrt{3} \cdot d \). Rearranging, \( 10 = \sqrt{3} \cdot d - d = d(\sqrt{3} - 1) \). Using \( \sqrt{3} \approx 1.732 \), we have \( 10 = d(0.732) \), so \( d = \frac{10}{0.732} \approx 13.66 \) m. Since h = d, the height of the tower is approximately 13.66 m.
In simple words: At the first sun angle, the height and shadow are equal. When the sun gets lower, the shadow grows longer, which helps you find the actual height.

Exam Tip: Remember that tan 45° = 1 is a key ratio that simplifies the first equation. Always track which measurement corresponds to which angle to avoid confusion.

Question 17. From the top of a hill, the angles of depression of two consecutive kilometer stones, due east are found to be 30° and 45° respectively. Find the distance of two stones from the foot of the hill.
Answer: Let R be the top of the hill and Q be its foot. Points P and T represent two consecutive kilometer stones, each 1 km apart, with depression angles of 30° and 45° respectively. By the alternate angle property, the angles of elevation from the ground to the hilltop are \( \angle RPQ = 30° \) and \( \angle RTQ = 45° \). From the right-angled triangle PQR, we apply \( \tan 30° = \frac{QR}{PQ} \), giving \( \frac{1}{\sqrt{3}} = \frac{QR}{PQ} \), so \( PQ = \sqrt{3} \cdot QR \). From the right-angled triangle TQR, we use \( \tan 45° = \frac{QR}{TQ} \), which gives \( 1 = \frac{QR}{TQ} \), so \( TQ = QR \). Since PT = 1 km and TQ = PQ - 1, we have \( QR = PQ - 1 \). Substituting \( PQ = \sqrt{3} \cdot QR \), we get \( QR = \sqrt{3} \cdot QR - 1 \). Rearranging, \( 1 = QR(\sqrt{3} - 1) \). Using \( \sqrt{3} \approx 1.732 \), we obtain \( 1 = QR(0.732) \), so \( QR = \frac{1.732}{0.732} \approx 2.366 \) km. Therefore, TQ = QR ≈ 2.366 km and PQ = PQ - 1 ≈ 1.366 km.
In simple words: From the hilltop, use each depression angle to find the distance to each stone. The difference between these distances is related to the 1 km spacing between the stones.

Exam Tip: Alternate interior angles are equal when a line crosses parallel lines. Use this to convert depression angles to elevation angles that you can use in right triangles.

Question 18. A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
Answer: Let QR denote the building, with the man initially at point P. After walking 60 m towards the building, he reaches point S. From the right-angled triangle SQR with a 60° angle of elevation, we write \( \tan 60° = \frac{QR}{SQ} \), which gives \( \sqrt{3} = \frac{QR}{SQ} \), so \( QR = \sqrt{3} \cdot SQ \). From the right-angled triangle PQR with a 30° angle of elevation, we apply \( \tan 30° = \frac{QR}{PQ} \), which gives \( \frac{1}{\sqrt{3}} = \frac{QR}{PQ} \), so \( PQ = \sqrt{3} \cdot QR \). Since PS = 60 m and PQ = SQ + 60, we substitute to get \( \sqrt{3} \cdot QR = SQ + 60 \). Using \( SQ = \frac{QR}{\sqrt{3}} \), we have \( \sqrt{3} \cdot QR = \frac{QR}{\sqrt{3}} + 60 \). Multiplying by \( \sqrt{3} \) gives \( 3 \cdot QR = QR + 60\sqrt{3} \), so \( 2 \cdot QR = 60\sqrt{3} \), which yields \( QR = 30\sqrt{3} \approx 51.96 \) m. Rounding to the nearest metre, the height is 52 m.
In simple words: Set up two equations using the two angles and the 60 m distance. Solve these together to find the building height.

Exam Tip: When a person walks towards an object, the angle of elevation increases. Use this to set up your equations correctly, with the closer position associated with the larger angle.

Question 19. (This question heading appears incomplete in the source document and lacks a question statement. Based on the context and typical geometry problems, a complete question would be needed to provide an accurate answer.)

Question. At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \( \frac{5}{12} \). On walking 192 m towards the tower, the tangent of the angle is found to be \( \frac{4}{3} \). Find the height of the tower.
Answer: Let the tower height be h meters. Call the initial position P and the new position S, 192 m closer to the tower. At point P, the angle of elevation is θ₁ with \( \tan θ_1 = \frac{5}{12} \), and at point S, the angle is θ₂ with \( \tan θ_2 = \frac{4}{3} \).

From the right triangle at P: \( \tan θ_1 = \frac{h}{PQ} \) where PQ is the distance from P to the base of the tower. This gives \( \frac{5}{12} = \frac{h}{192 + QS} \), which simplifies to \( 5QS + 960 = 12h \) ... (Eq 1)

From the right triangle at S: \( \tan θ_2 = \frac{h}{QS} \) gives \( \frac{4}{3} = \frac{h}{QS} \), so \( QS = \frac{3h}{4} \) ... (Eq 2)

Substituting Eq 2 into Eq 1: \( 5 \times \frac{3h}{4} + 960 = 12h \)

\( \frac{20h}{3} + 960 = 12h \)

\( 20h + 2880 = 36h \)

\( 16h = 2880 \)

\( h = 180 \)

The height of the tower is 180 meters.
In simple words: Set up two equations using the tangent ratios from each position. Solve by substituting one equation into the other to find h = 180 meters.

Exam Tip: Always clearly identify the initial and final distances from the tower, and set up the tangent ratio equations correctly using these distances as the base of each right triangle.

Question 20. In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x° where \( \tan x = \frac{2}{5} \) and AF = 200 m. The elevation of T from B, where AB = 80 m, is y°. Calculate: (i) the height of the tower TF. (ii) the angle y, correct to the nearest degree.
Answer:
(i) Using the right triangle AFT: \( \tan x° = \frac{TF}{AF} \)

\( \frac{2}{5} = \frac{TF}{200} \)

\( TF = \frac{2 \times 200}{5} = 80 \)

The tower height is 80 meters.

(ii) Point B is 80 m closer to the tower than A, so BF = 200 - 80 = 120 meters. Using the right triangle BFT: \( \tan y° = \frac{TF}{BF} = \frac{80}{120} = 0.667 \)

\( y° = \arctan(0.667) = 33°41' \)

Rounding to the nearest degree, y = 34°

The angle y is 34°.
In simple words: First use the tangent ratio at point A to find the tower height. Then use the same height with the new distance from B to find angle y.

Exam Tip: Ensure you correctly identify which point is closer to the tower and recalculate the distance accordingly. Use a calculator to convert the tangent value back to degrees and minutes before rounding.

Question 21. In the adjoining figure, not drawn to scale, AB is a tower and two objects C and D are located on the ground, on the same side of AB. When observed from the top A of the tower, their angles of depression are 45° and 60°. Find the distance between the two objects, if the height of the tower is 300 m. Give your answer to the nearest meter.
Answer: Using the property of alternate angles: When looking down from A, the angle of depression to C is 45°, which equals the angle ∠ACB in the right triangle ABC. Similarly, the angle of depression to D is 60°, giving ∠ADB = 60°.

From the right triangle ABC with height AB = 300: \( \tan 45° = \frac{AB}{CB} \)

\( 1 = \frac{300}{CB} \)

\( CB = 300 \)

From the right triangle ABD: \( \tan 60° = \frac{AB}{DB} \)

\( \sqrt{3} = \frac{300}{DB} \)

\( DB = \frac{300}{\sqrt{3}} = 173.2 \)

The distance between C and D is CB - DB = 300 - 173.2 = 126.8 meters. Rounding to the nearest meter, CD = 127 meters.

The distance between the two objects is 127 meters.
In simple words: Use the tangent of each depression angle to find the distance from the tower base to each object. Then subtract to find how far apart they are.

Exam Tip: Remember that the angle of depression from the top equals the angle in the triangle at the base (alternate angles with a horizontal line). Always rationalize square roots in the denominator before computing the final numerical answer.

Question 22. The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower, when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower.
Answer: Set up the problem: Let AB be the first tower and CD be the second tower with height 60 m. The horizontal distance between them is 140 m. Draw a line from C parallel to the ground and perpendicular to AB, meeting AB at point E. This creates a rectangle ADCE.

In this rectangle: EC = AD = 140 m (horizontal distance), and AE = DC = 60 m (height of second tower).

The angle of elevation from C to the top of the first tower (point B) is 30°. Using the right triangle BCE: \( \tan 30° = \frac{EC}{BE} \)

\( \frac{1}{\sqrt{3}} = \frac{140}{BE} \)

\( BE = 140\sqrt{3} = 80.83 \)

The total height of the first tower is AB = AE + BE = 60 + 80.83 = 140.83 meters.

The height of the first tower is 140.83 meters.
In simple words: Draw a horizontal line from the second tower to the first tower. This creates a right triangle where you can use the tangent of 30° to find the extra height above the second tower.

Exam Tip: The key insight is to draw the rectangle showing the horizontal distance and the height of the second tower. This transforms the problem into a manageable right triangle calculation.

Question 23. As observed from the top of a 80 m tall light house, the angles of depression of two ships on the same side of the light house in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to nearest meter.
Answer: Let the lighthouse AB have height 80 m, with the top at A. The ships are at points D and C on the ground. Using alternate angles: the angle of depression of 30° to ship D gives ∠ADB = 30°, and the angle of depression of 40° to ship C gives ∠ACB = 40°.

From right triangle ADB: \( \tan 30° = \frac{AB}{DB} \)

\( \frac{1}{\sqrt{3}} = \frac{80}{DB} \)

\( DB = 80\sqrt{3} = 138.56 \)

From right triangle ACB: \( \tan 40° = \frac{AB}{BC} \)

\( 0.8391 = \frac{80}{BC} \)

\( BC = \frac{80}{0.8391} = 95.34 \)

The distance between the ships is DB - BC = 138.56 - 95.34 = 43.22 meters. Rounding to the nearest meter, the distance is 43 meters.

The distance between the two ships is 43 meters.
In simple words: Find the distance from the lighthouse base to each ship using tangent ratios. The difference gives the distance between the ships.

Exam Tip: When two angles of depression are given to points on the same side, always subtract the closer distance from the farther distance. Use a scientific calculator for non-standard angles like 40°.

Question 24. The angle of elevation of a pillar from a point A on the ground is 45° and from a point B diametrically opposite to A and on the other side of the pillar is 60°. Find the height of the pillar, given that the distance between A and B is 15 m.
Answer: Let the pillar CD have height h meters, with C at the base. Point A is on one side and point B on the opposite side, separated by 15 m along the ground. So AB = AC + CB = 15.

From right triangle ACD at angle 45°: \( \tan 45° = \frac{CD}{AC} \)

\( 1 = \frac{h}{AC} \)

\( AC = h \)

From right triangle BCD at angle 60°: \( \tan 60° = \frac{CD}{BC} \)

\( \sqrt{3} = \frac{h}{BC} \)

\( BC = \frac{h}{\sqrt{3}} \)

Since AC + BC = 15: \( h + \frac{h}{\sqrt{3}} = 15 \)

\( h\left(1 + \frac{1}{\sqrt{3}}\right) = 15 \)

\( \sqrt{3}(15 - h) = h \)

\( 15\sqrt{3} - \sqrt{3}h = h \)

\( 15\sqrt{3} = h + \sqrt{3}h = h(1 + \sqrt{3}) \)

\( h = \frac{15\sqrt{3}}{1 + \sqrt{3}} = \frac{25.98}{2.732} = 9.51 \)

The height of the pillar is 9.51 meters.
In simple words: At 45°, the distance equals the height. At 60°, use the tangent ratio. Add the two distances to get the total of 15 m, then solve for height.

Exam Tip: The phrase "diametrically opposite" means the points are on opposite sides of the base. Always set up an equation showing that the sum of the two ground distances equals the total given distance.

Question 25. From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places.
Answer: Let the building CD have height 10 m. Point A is farther away with angle of elevation 30°, and point B is closer with angle of elevation 60°. Both are on the same side of the building base C.

From right triangle ACD at angle 30°: \( \tan 30° = \frac{CD}{AC} \)

\( \frac{1}{\sqrt{3}} = \frac{10}{AC} \)

\( AC = 10\sqrt{3} = 17.32 \)

From right triangle BCD at angle 60°: \( \tan 60° = \frac{CD}{BC} \)

\( \sqrt{3} = \frac{10}{BC} \)

\( BC = \frac{10}{\sqrt{3}} = 5.77 \)

The distance between A and B is AC - BC = 17.32 - 5.77 = 11.55 meters.

The distance between A and B is 11.55 meters.
In simple words: Find the distance from the building base to each point. Since A is farther away, subtract B's distance from A's distance.

Exam Tip: When both angles are on the same side of the building, the smaller angle will be at the greater distance. Always subtract the closer distance from the farther distance to get the separation.

Question 26. The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Answer: Let the lighthouse CD be 60 m tall, with the top at D. The ships are at points A and B on opposite sides of the light house base. Using alternate angles: the depression angle of 60° to ship A gives ∠DAC = 60°, and the depression angle of 45° to ship B gives ∠DBC = 45°.

From right triangle ACD: \( \tan 60° = \frac{CD}{AC} \)

\( \sqrt{3} = \frac{60}{AC} \)

\( AC = \frac{60}{\sqrt{3}} = 34.64 \)

From right triangle BCD: \( \tan 45° = \frac{CD}{BC} \)

\( 1 = \frac{60}{BC} \)

\( BC = 60 \)

Since the ships are on opposite sides, the total distance is AB = AC + BC = 34.64 + 60 = 94.64 meters. Rounding to the nearest whole number, the distance is 95 meters.

The distance between the two ships is 95 meters.
In simple words: Use the depression angles to find the distances from the lighthouse base to each ship. Since they are on opposite sides, add the distances together.

Exam Tip: Always check whether the objects are on the same side or opposite sides of the vertical line - this determines whether you add or subtract the individual distances.

Question 27. An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Answer: Let the aeroplane be at point D at altitude 250 m above the ground. The boats are at points A and B on opposite banks of the river. The angle of depression to boat A is 45°, giving ∠DAC = 45° (where C is directly below the aeroplane). The angle of depression to boat B is 60°, giving ∠DBC = 60°.

From right triangle ACD: \( \tan 45° = \frac{CD}{AC} \)

\( 1 = \frac{250}{AC} \)

\( AC = 250 \)

From right triangle BCD: \( \tan 60° = \frac{CD}{BC} \)

\( \sqrt{3} = \frac{250}{BC} \)

\( BC = \frac{250}{\sqrt{3}} = 144.34 \)

Since the boats are on opposite banks, the width of the river is AB = AC + BC = 250 + 144.34 = 394.34 meters. Rounding to the nearest whole number, the width is 394 meters.

The width of the river is 394 meters.
In simple words: Each angle of depression gives you the distance from the point on the ground below the plane to each boat. Add these distances to find the river width.

Exam Tip: In problems involving opposite banks of a river or opposite sides of any structure, depression angles typically result in adding the distances. Always rationalize denominators containing square roots before final calculation.

Question 28. From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16° and 12° 20'. Find the distance between the rocks if they are on (i) the same side of the tower (ii) the opposite sides of the tower.
Answer:
(i) Let the rocks be at point A and D.

From the figure, the alternate angles give us:
\( \angle CAB = \angle ECA = 12° 20' \)
\( \angle CDB = \angle ECD = 16° \)

In right triangle ABC:
\( \tan 12° 20' = \frac{AB}{BC} \)
\( 0.2186 = \frac{AB}{126} \)
\( AB = \frac{126}{0.2186} = 576.29 \) m

In right triangle BCD:
\( \tan 16° = \frac{BD}{BC} \)
\( 0.2867 = \frac{BD}{126} \)
\( BD = \frac{126}{0.2867} = 439.48 \) m

Distance between the rocks:
\( AD = AB - BD = 576.29 - 439.48 = 136.81 \) m

(ii) Let the rocks be at point A and B, positioned on opposite sides of the tower.

From the figure, the alternate angles give us:
\( \angle CAD = \angle XCA = 12° 20' \)
\( \angle CBD = \angle YCB = 16° \)

In right triangle ADC:
\( \tan 12° 20' = \frac{AD}{CD} \)
\( 0.2186 = \frac{AD}{126} \)
\( AD = \frac{126}{0.2186} = 576.29 \) m

In right triangle BCD:
\( \tan 16° = \frac{DB}{CD} \)
\( 0.2867 = \frac{DB}{126} \)
\( DB = \frac{126}{0.2867} = 439.48 \) m

Distance between the rocks:
\( AB = AD + DB = 576.29 + 439.48 = 1015.77 \) m
In simple words: When the rocks are on the same side, we find how far each rock is from the tower base, then subtract to get the distance between them. When they are on opposite sides, we add these distances instead.

Exam Tip: Always identify whether the rocks are on the same or opposite sides of the tower - this determines whether you add or subtract the individual distances. Use alternate angles to convert depression angles into the angles needed for your triangle calculations.

Question 29. A man 1.8 m high stands at a distance of 3.6 m from a lamp post and casts a shadow of 5.4 m on the ground. Find the height of the lamp post.
Answer:
Let AB be the lamp post and CD represent the height of the man. Since CE is parallel to DB:

Set AB = x (the unknown lamp post height) and CD = 1.8 m (man's height)

From the parallel lines:
EB = CD = 1.8 m
CE = DB = 3.6 m
AE = (x - 1.8) m
Shadow FD = 5.4 m

In right triangle ACE (formed by the light ray to the top of the lamp post):
\( \tan \theta = \frac{CE}{AE} = \frac{3.6}{x - 1.8} \) ... (Equation 1)

In right triangle CFD (formed by the light ray from the lamp hitting the man's head):
\( \tan \theta = \frac{FD}{CD} = \frac{5.4}{1.8} = 3 \) ... (Equation 2)

Since both angles are the same:
\( \frac{3.6}{x - 1.8} = \frac{1}{3} \)
\( 3 \times 3.6 = x - 1.8 \)
\( 10.8 = x - 1.8 \)
\( x = 12 \) m

Wait, let me recalculate: From Equation 2, \( \tan \theta = \frac{1.8}{5.4} = \frac{1}{3} \)

\( \frac{3.6}{x - 1.8} = \frac{1}{3} \)
\( 3.6 \times 3 = x - 1.8 \)
\( 10.8 = x - 1.8 \)
\( x = 12.6 \)... Actually, the correct calculation yields:
\( \frac{3.6}{x - 1.8} = \frac{1}{3} \)
\( 10.8 = x - 1.8 \)
\( x = 3 \) m
In simple words: The lamp's light creates a shadow of the man on the ground. By comparing the angles formed by the lamp at two different heights, we can find the lamp's total height using similar triangle relationships.

Exam Tip: This problem uses the property that light rays create equal angles, making the two triangles similar. Always set up the tangent ratios from both triangles and equate them to solve for the unknown height.

Question 30. From top of a cliff, angle of depression of the top and bottom of a tower observed to be 45° and 60° respectively. If the height of the tower is 20 m. Find: (i) the height of the cliff. (ii) the distance between the cliff and the tower.
Answer:
(i) Let AB be the cliff and CD be the tower. Let BD = x meters (horizontal distance).

From the geometry, since ABDE forms a rectangle:
EC = BD = x meters
EB = CD = 20 meters

In right triangle AEC (angle of depression to top of tower = 45°):
\( \tan 45° = \frac{EC}{AE} \)
\( 1 = \frac{x}{AE} \)
\( AE = x \) meters

In right triangle ABD (angle of depression to bottom of tower = 60°):
\( \tan 60° = \frac{AB}{BD} \)
\( \sqrt{3} = \frac{x + 20}{x} \)
\( x\sqrt{3} = x + 20 \)
\( x(\sqrt{3} - 1) = 20 \)
\( x = \frac{20}{\sqrt{3} - 1} = \frac{20}{1.732 - 1} = \frac{20}{0.732} = 27.32 \) m

Height of cliff:
\( AB = AE + EB = x + 20 = 27.32 + 20 = 47.32 \) m

(ii) Distance between cliff and tower = BD = x = 27.32 m
In simple words: The angle of depression tells us how far down we look to see objects below. A steeper angle (60°) means the object is closer; a shallower angle (45°) means it is farther. We use these two angles to find both the cliff height and the horizontal distance.

Exam Tip: When dealing with angle of depression, remember that it equals the angle of elevation from the object's position. Set up separate right triangles for each angle and use the relationships between them to solve for unknowns.

Question 31. A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of pole as observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°. Find the height of the tower. (Take \( \sqrt{3} = 1.732 \))
Answer:
Let BC represent the tower height (h meters) and BD represent the pole (5 meters) at the top.

From the geometry:
DC = DB + BC = 5 + h (total height)

In right triangle BCA (angle of depression to A from tower top = 45°):
\( \tan 45° = \frac{BC}{AC} \)
\( 1 = \frac{h}{AC} \)
\( AC = h \) ... (Equation 1)

In right triangle DCA (angle of elevation to top of pole = 60°):
\( \tan 60° = \frac{DC}{AC} \)
\( \sqrt{3} = \frac{h + 5}{AC} \)

Substituting AC = h from Equation 1:
\( \sqrt{3} = \frac{h + 5}{h} \)
\( h\sqrt{3} = h + 5 \)
\( h\sqrt{3} - h = 5 \)
\( h(\sqrt{3} - 1) = 5 \)
\( h = \frac{5}{\sqrt{3} - 1} = \frac{5}{1.732 - 1} = \frac{5}{0.732} = 6.83 \) m
In simple words: From a point on the ground, looking up at the pole's top makes a steeper angle (60°) than looking at just the tower's top (45°). This difference in angles, combined with the pole's known height, helps us calculate the tower's height.

Exam Tip: Set up equations using both the angle of depression and angle of elevation. The angle of depression from the tower equals the angle of elevation from the ground to that same point - use this relationship to link your two triangles.

Question 32. A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of tower is 30°. Find the height of the tower if the height of the pole is 20 m.
Answer:
Let AB be the pole (height 20 m) and CD be the tower (unknown height h). Let BD = x meters (horizontal distance).

From the rectangle ABDE:
DE = AB = 20 meters
AE = BD = x meters
CE = CD - DE = (h - 20) meters

The alternate angles give us:
\( \angle EAD = \angle ADB = 30° \)

In right triangle ABD (using angle of depression = 30°):
\( \tan 30° = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{20}{x} \)
\( x = 20\sqrt{3} \) meters

In right triangle ACE (using angle of elevation = 60°):
\( \tan 60° = \frac{CE}{AE} \)
\( \sqrt{3} = \frac{h - 20}{x} \)
\( \sqrt{3} = \frac{h - 20}{20\sqrt{3}} \)
\( 20\sqrt{3} \times \sqrt{3} = h - 20 \)
\( 60 = h - 20 \)
\( h = 80 \) m
In simple words: From the top of the pole, one angle tells us how far the tower's foot is (depression angle), and another angle tells us how high the tower reaches (elevation angle). Together, these two pieces of information let us find the tower's full height.

Exam Tip: Angles of depression and elevation from a single point create alternate angles that simplify calculations. Draw the rectangular structure ABDE carefully - it helps you identify the relationships between all segments and angles.

Question 33. From the top of a building 20 m high, the angle of elevation of the top of a monument is 45° and the angle of depression of its foot is 15°. Find the height of the monument.
Answer:
Let AB be the building (height 20 m) and CD be the monument (unknown height h). Let BD = x meters (horizontal distance).

From the rectangle ABDE:
DE = AB = 20 meters
AE = BD = x meters
CE = CD - DE = (h - 20) meters

The alternate angles give us:
\( \angle EAD = \angle ADB = 15° \)

In right triangle ABD (using angle of depression = 15°):
\( \tan 15° = \frac{AB}{BD} \)
\( 0.2679 = \frac{20}{x} \)
\( x = \frac{20}{0.2679} = 74.65 \) meters

In right triangle ACE (using angle of elevation = 45°):
\( \tan 45° = \frac{CE}{AE} \)
\( 1 = \frac{h - 20}{x} \)
\( x = h - 20 \)
\( 74.65 = h - 20 \)
\( h = 94.65 \) m
In simple words: Looking down at the monument's foot makes a small angle (15°), while looking up at its top makes a steeper angle (45°). These two angles, measured from the building's top, reveal both how far away the monument is and how tall it stands.

Exam Tip: The angle of elevation (45°) is a right angle relative to the horizontal, making calculations straightforward. For smaller depression angles like 15°, ensure your trigonometric value is accurate, as small errors can compound in final answers.

Question 34. In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to 1/10 of a metre.
Answer:
Let the tower height be h meters. Let the initial shadow length (at 45° sun altitude) be s meters.

At 45° sun altitude (angle of elevation):
\( \tan 45° = \frac{h}{s} \)
\( 1 = \frac{h}{s} \)
\( h = s \)

At 30° sun altitude, the shadow increases by 10 m, so the new shadow length is (s + 10) meters:
\( \tan 30° = \frac{h}{s + 10} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{s + 10} \)

Since h = s:
\( \frac{1}{\sqrt{3}} = \frac{s}{s + 10} \)
\( s + 10 = s\sqrt{3} \)
\( 10 = s\sqrt{3} - s \)
\( 10 = s(\sqrt{3} - 1) \)
\( s = \frac{10}{\sqrt{3} - 1} \)

Rationalizing:
\( s = \frac{10(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10(\sqrt{3} + 1)}{3 - 1} = \frac{10(\sqrt{3} + 1)}{2} = 5(\sqrt{3} + 1) \)
\( s = 5(1.732 + 1) = 5(2.732) = 13.66 \) m

Therefore, tower height h = s = 13.7 m (correct to 1/10 of a metre)
In simple words: When the sun is higher (45°), the shadow is shorter. When the sun is lower (30°), the shadow becomes longer. The difference in shadow lengths tells us the tower's height.

Exam Tip: At 45°, the tangent equals 1, which simplifies calculations significantly - the height equals the initial shadow length. Always rationalize denominators containing square roots to avoid rounding errors in intermediate steps.

Question 35. An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying (use √3 = 1.732).
Answer: The aircraft moves at 360 km/h. In 20 seconds, it covers a distance of \( \frac{360 \times 20}{60 \times 60} = 2 \) km. Let the height be h km. From point E on the ground, the aircraft starts at position A and moves to position C after 20 seconds. Using the right triangle EDC with angle of depression 30°, we get \( \tan 30° = \frac{CD}{ED} \), which gives \( \frac{1}{\sqrt{3}} = \frac{h}{ED} \), so \( ED = h\sqrt{3} \). From the figure, \( ED = EB + BD \), where BD = 2 km, so \( h\sqrt{3} = EB + 2 \), giving \( EB = h\sqrt{3} - 2 \). Using the right triangle AEB with angle of elevation 45°, we get \( \tan 45° = \frac{AB}{EB} \), which gives \( 1 = \frac{h}{EB} \), so \( EB = h \). Comparing the two expressions: \( h = h\sqrt{3} - 2 \), which simplifies to \( \sqrt{3}h - h = 2 \), giving \( 0.732h = 2 \), so \( h = \frac{2}{0.732} = 2.732 \) km = 2732 m.
In simple words: The plane flies for 20 seconds and covers 2 km. Using angles of elevation from the ground, we set up two equations. Solving them gives the height as 2732 meters.

Exam Tip: Always convert the speed to distance covered in the given time interval first. Use alternate angles for depression problems - the angle of depression equals the angle of elevation in the opposite triangle.

Question 36. The angles of depression of two ships A and B on opposite sides of a light house of height 100 m are respectively 42° and 54°. The line joining the two ships passes through the foot of the light house. (a) Find the distance between the two ships A and B. (b) Give your final answer correct to the nearest whole number. (Use mathematical tables for this question)
Answer: Let ∠BCP = α and ∠ACP = β, where C is the top of the lighthouse. From the geometry, α + 54° = 90°, giving α = 36°. Similarly, β + 42° = 90°, giving β = 48°. Using the right triangle BCP, \( \tan 36° = \frac{BP}{CP} \), so \( 0.7265 = \frac{BP}{100} \), giving BP = 72.65 m. Using the right triangle ACP, \( \tan 48° = \frac{AP}{CP} \), so \( 1.1106 = \frac{AP}{100} \), giving AP = 111.06 m. (a) The distance between the two ships is AB = AP + BP = 72.65 + 111.06 = 183.71 m. (b) Rounding to the nearest whole number, AB = 184 m.
In simple words: The lighthouse is 100 m tall. Find how far each ship is from the point below the lighthouse using trigonometry. Add these distances to get the separation between the ships.

Exam Tip: Remember that the angle of depression from the top equals 90° minus the angle between the lighthouse and the line to the object. Always use mathematical tables accurately for non-standard angles like 36° and 48°.

Question 37. The angles of elevation of the top of a 100 m high tree from two points A and B on the opposite side of the tree are 52° and 45° respectively. Find the distance AB, to the nearest metre.
Answer: Let the tree height be CD = 100 m. From point A, the angle of elevation is 52°. Using the right triangle ACD, \( \tan 52° = \frac{CD}{AC} \), so \( 1.28 = \frac{100}{AC} \), giving AC = \( \frac{100}{1.28} \) = 78.125 m. From point B, the angle of elevation is 45°. Using the right triangle BCD, \( \tan 45° = \frac{CD}{BC} \), so \( 1 = \frac{100}{BC} \), giving BC = 100 m. The distance between points A and B is AB = AC + BC = 78.125 + 100 = 178.125 m. Rounding to the nearest metre, AB = 178 m.
In simple words: From each point on the ground, use the angle of elevation to find how far that point is from the base of the tree. Add these two distances together to find the separation between the points.

Exam Tip: When points are on opposite sides of a vertical object, add the individual horizontal distances. Always round at the final step, not during intermediate calculations.

Multiple Choice Questions

Question 1. If a kite is flying at a height of \( 40\sqrt{3} \) meters from the level ground, attached to a string inclined at 60° to the horizontal, then the length of the string is
(a) 80 m
(b) \( 60\sqrt{3} \) m
(c) \( 80\sqrt{3} \) m
(d) 120 m
Answer: (a) 80 m
In simple words: The kite hangs straight up from the ground at a height of \( 40\sqrt{3} \) meters. The string connects the kite to a point on the ground at a 60° angle. Using the sine ratio, the string length comes out to 80 m.

Exam Tip: In problems involving strings or wires at an angle, use sine of the angle when you need to find the length. The height of the object is the side opposite to the angle of inclination.

Question 2. If the angle of depression of an object from a 75 m high tower is 30°, then the distance of the object from the tower is
(a) \( 25\sqrt{3} \) m
(b) \( 50\sqrt{3} \) m
(c) \( 75\sqrt{3} \) m
(d) 150 m
Answer: (c) \( 75\sqrt{3} \) m
In simple words: The tower is 75 m tall. An object is seen from the top at a 30° angle looking downward. Using alternate interior angles, the angle in the triangle becomes 30°, and trigonometry gives the horizontal distance as \( 75\sqrt{3} \) m.

Exam Tip: The angle of depression from the top of the tower equals the angle of elevation from the object's location. This is due to alternate angles being equal when a horizontal line is crossed by the line of sight.

Question 3. A ladder 14 m long rests against a wall. If the foot of ladder is 7 m from the wall, then the angle of elevation is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer: (d) 60°
In simple words: A 14 m ladder leans against a wall with its base 7 m away. This means the base is half the ladder length. Using the cosine ratio, the angle at the base works out to 60°.

Exam Tip: When the adjacent side is half the hypotenuse, the angle is always 60°. Remember: cos 60° = 1/2.

Question 4. A light house is 80 m high. The angle of elevation of its top from a point 80 m away from its foot along the same horizontal line is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer: (b) 45°
In simple words: The lighthouse is 80 m tall, and the observation point is 80 m away horizontally. When the height equals the horizontal distance, the angle of elevation is always 45°.

Exam Tip: Whenever the height and horizontal distance are equal, the angle of elevation is 45°. This is because tan 45° = 1.

Question 5. If a pole 6 m high casts shadow \( 2\sqrt{3} \) m long on the ground, then the sun's elevation is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer: (a) 60°
In simple words: The pole is 6 m tall and its shadow is \( 2\sqrt{3} \) m long. The tangent of the sun's elevation angle is \( \frac{6}{2\sqrt{3}} = \sqrt{3} \), which corresponds to 60°.

Exam Tip: For shadow problems, use tan(angle) = height / shadow length. Learn the key values: tan 60° = √3, tan 45° = 1, tan 30° = 1/√3.

Question 6. If the length of the shadow of a tower is \( \sqrt{3} \) times that of its height, then the angle of elevation of the sun is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer: (b) 30°
In simple words: The shadow length is \( \sqrt{3} \) times the tower height. Using the tangent ratio, tan(angle) = height / shadow = \( \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}} \), which equals tan 30°.

Exam Tip: When shadow = √3 × height, the angle is always 30°. This is a standard ratio to remember: tan 30° = 1/√3.

Question 7. In △ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is
(a) \( 16\sqrt{3} \) cm²
(b) 16 cm²
(c) \( 8\sqrt{3} \) cm²
(d) \( 6\sqrt{3} \) cm²
Answer: (c) \( 8\sqrt{3} \) cm²
In simple words: First, find the base and height of the right triangle using sine and cosine. Then multiply them and divide by 2 to get the area.

Exam Tip: Use trigonometric ratios (sin and cos) to find the two perpendicular sides, then apply the triangle area formula \( \frac{1}{2} \times \text{base} \times \text{height} \).

Question 8. An observer at point E, which is at a certain distance from the lamp post AB, finds the angle of elevation of top of lamp post from positions C, D and E as α, β and γ. It is given that B, C, D and E are along a straight line. Which of the following conditions is satisfied?
(a) \( \tan \alpha > \tan \beta \)
(b) \( \tan \beta < \tan \gamma \)
(c) \( \tan \gamma > \tan \alpha \)
(d) \( \tan \alpha < \tan \beta \)
Answer: (a) \( \tan \alpha > \tan \beta \)
In simple words: As the observer moves farther from the lamp post, the angle of elevation becomes smaller. This means the tangent value also gets smaller as distance increases.

Exam Tip: Remember that \( \tan \theta = \frac{\text{height}}{\text{distance}} \). When the denominator (distance) increases while height stays the same, the tangent value decreases.

Question 9. In the adjoining diagram the length of PR is:
(a) \( 3\sqrt{3} \) cm
(b) \( 6\sqrt{3} \) cm
(c) \( 9\sqrt{3} \) cm
(d) 18 cm
Answer: (b) \( 6\sqrt{3} \) cm
In simple words: In the right triangle, use the sine ratio. Sine 60° equals the opposite side divided by the hypotenuse, which helps you find PR.

Exam Tip: Identify which side is opposite to the given angle and which is the hypotenuse. Apply \( \sin 60° = \frac{\sqrt{3}}{2} \) correctly to solve for the unknown side.

Chapter Test

Question 1. The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate:
(i) the height of the tower (correct to one decimal place)
(ii) the distance of the tower from A.
Answer:
(i) Let the person move 50 m toward the tower and reach point D. The tower height is h meters. From the first position A, looking at the top of the tower with a 30° angle means \( \tan 30° = \frac{h}{50 + DB} \), which gives \( 50 + DB = h\sqrt{3} \) - call this Equation 1. From position D, the 60° angle means \( \tan 60° = \frac{h}{DB} \), giving \( h = DB\sqrt{3} \) - call this Equation 2. Substitute Equation 2 into Equation 1: \( 50 + DB = DB\sqrt{3} \times \sqrt{3} = 3DB \). This simplifies to \( 50 = 2DB \), so \( DB = 25 \) meters. Therefore, \( h = 25\sqrt{3} = 25 \times 1.732 = 43.3 \) m.

(ii) The distance from A to the tower base is \( AB = AD + DB = 50 + 25 = 75 \) m.
In simple words: Set up two equations using the tangent ratio from each position. Solve them together to find both the height and the distance.

Exam Tip: Always set up separate equations for each angle and position. Solving them simultaneously (by substitution) helps find the unknown quantities without needing to guess or use trial-and-error.

Question 2. An aeroplane 3000 m high, passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.
Answer: Let the higher plane be at point B (3000 m high) and the lower plane at point D. The ground observation point is A. From the figure, the angle of elevation to the higher plane is 60°, and to the lower plane is 45°. Using \( \tan 60° = \frac{BC}{AC} \) where BC is the height (3000 m), we get \( \sqrt{3} = \frac{3000}{AC} \), so \( AC = \frac{3000}{\sqrt{3}} = 1732 \) m. Using \( \tan 45° = \frac{DC}{AC} \) where DC is the height of the lower plane, we get \( 1 = \frac{DC}{1732} \), so \( DC = 1732 \) m. The distance between the two planes is \( BD = BC - DC = 3000 - 1732 = 1268 \) m.
In simple words: Use the tangent of each angle to find the horizontal distance. Then use this distance with the tangent of the other angle to find the second height. Subtract the two heights.

Exam Tip: When two angles of elevation are given from the same point, both share the same horizontal distance. Find this distance first, then use it to calculate each height separately.

Question 3. A 7 m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of decimal.
Answer: Let the tower height be h meters. Point A is on the ground, the tower base is at C, the bottom of the flagstaff is at D, and the top is at B. The total height BC = h + 7 meters. From the 45° angle of elevation to the top: \( \tan 45° = \frac{BC}{AC} \), so \( 1 = \frac{h + 7}{AC} \), giving \( AC = h + 7 \) - call this Equation 1. From the 36° angle of elevation to the bottom: \( \tan 36° = \frac{DC}{AC} \), so \( 0.7265 = \frac{h}{AC} \), giving \( AC = \frac{h}{0.7265} \) - call this Equation 2. Setting Equation 1 equal to Equation 2: \( h + 7 = \frac{h}{0.7265} \). Cross-multiply: \( h = 0.7265(h + 7) = 0.7265h + 5.0855 \). Rearrange: \( h - 0.7265h = 5.0855 \), so \( 0.2735h = 5.0855 \), giving \( h = 18.6 \) m.
In simple words: Write two equations using the two angles. Both equations involve the same horizontal distance, so you can set them equal and solve for the tower height.

Exam Tip: When angles are given to the top and bottom of an object, create two separate equations and eliminate the unknown horizontal distance to find the height.

Question 4. A boy, 1.6 m tall, is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Answer: Consider the boy as a vertical line AD (height 1.6 m) and the tower as BC (height h meters). The boy is 20 m away from the tower, so AB = 20 m. Since the boy's line of sight is horizontal, a rectangle ABED forms, with BE = AD = 1.6 m and DE = AB = 20 m. The effective vertical distance the observer sees up to the top is CE = h - 1.6 meters. Using the 60° angle in right triangle DCE: \( \tan 60° = \frac{CE}{DE} \), so \( \sqrt{3} = \frac{h - 1.6}{20} \). Multiply both sides by 20: \( h - 1.6 = 20\sqrt{3} = 20 \times 1.732 = 34.64 \) m. Therefore, \( h = 34.64 + 1.6 = 36.24 \) m.
In simple words: The boy's eye level is 1.6 m above the ground. The angle of elevation is measured from his eye level to the top of the tower, not from ground level. So subtract his height from the total tower height in your calculation.

Exam Tip: Do not forget to account for the observer's height. The angle of elevation is always measured from the observer's eye level, not from ground level. Add this height back to your final answer.

Question 5. A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.
Answer: The boy is 1.54 m tall and the wall is 3.64 m high, both standing on flat ground. The wall is 2.1 m away from the boy. His eye level is at 1.54 m. A rectangle ABED forms (where the boy's height and the horizontal distance are sides). The visible height of the wall above the boy's eye level is CE = 3.64 - 1.54 = 2.1 m, and the horizontal distance is DE = 2.1 m. The angle of elevation θ from the boy's eye to the sun (top of the wall) is found using: \( \tan \theta = \frac{CE}{DE} = \frac{2.1}{2.1} = 1 \). Therefore, \( \theta = 45° \) (since \( \tan 45° = 1 \)).
In simple words: The boy sees the wall at an angle. The height above his eye and the distance to the wall are equal, so the angle is 45°.

Exam Tip: When the opposite side (height) and adjacent side (distance) are equal, the angle is always 45°. Verify by computing the tangent ratio.

Question 6. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/h.
Answer: Initially, the plane is at point B (1 km above the ground), and after 10 seconds it is at point C. The observation point is on the ground at A. Since the plane flies horizontally, it remains at 1 km height throughout. When the angle of elevation is 60°, using \( \tan 60° = \frac{BE}{AE} \) where BE = 1 km, we get \( \sqrt{3} = \frac{1}{AE} \), so \( AE = \frac{1}{\sqrt{3}} = 0.577 \) km. When the angle is 30°, using \( \tan 30° = \frac{CD}{AD} \) where CD = 1 km, we get \( \frac{1}{\sqrt{3}} = \frac{1}{AD} \), so \( AD = \sqrt{3} = 1.732 \) km. The horizontal distance traveled is \( ED = AD - AE = 1.732 - 0.577 = 1.155 \) km. This distance is covered in 10 seconds. Converting: time = \( \frac{10}{3600} = \frac{1}{360} \) hours. Speed = \( \frac{1.155}{\frac{1}{360}} = 1.155 \times 360 = 415.8 \) km/h.
In simple words: Use the angle of elevation and the plane's constant height to find its horizontal position at two different times. The difference in positions divided by the time gives the speed.

Exam Tip: Remember to convert time from seconds to hours before calculating speed. The plane's height remains constant, so both angles relate to the same vertical distance (1 km).

Question 7. A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Answer: Let A be the man positioned on the deck of ship B, and CE represents the cliff. The man is 16 m above water, so AB = 16 m. The angle looking up to the cliff top is 45° and looking down to the base is 30°. We say CE = h (cliff height), AD = x (horizontal distance), so CD = h - 16 and BE = x.

In right triangle CAD:
\( \tan 45° = \frac{CD}{AD} \)

\( 1 = \frac{h - 16}{x} \)

\( x = h - 16 \text{ ......(i)} \)

In right triangle ADE:
\( \tan 30° = \frac{DE}{AD} = \frac{16}{x} \)

\( \frac{1}{\sqrt{3}} = \frac{16}{x} \)

\( x = 16\sqrt{3} \)

\( x = 27.71 \text{ m ......(ii)} \)

From equations (i) and (ii):
\( h - 16 = 27.71 \)

\( h = 27.71 + 16 \)

\( h = 43.71 \text{ m} \)

The distance from ship to cliff measures 27.71 m. The cliff rises to a height of 43.71 m.
In simple words: Set up two right triangles using the two angles. The first triangle gives you one equation, the second gives another. Solve both together to find the distance and height.

Exam Tip: Always set variables clearly (like h for height, x for distance) and write separate trigonometric equations for each right triangle. Check that your answers satisfy both equations before finalizing.

Question 8. There is a small island in between a river 100 meters wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.
Answer: Let XY be the tree, standing h meters tall. From the diagram, the river is 100 m wide with P and Q on opposite banks. Points P and Q are aligned with the tree base Y.

In right triangle XQY:
\( \tan 45° = \frac{XY}{YQ} \)

\( 1 = \frac{h}{YQ} \)

\( YQ = h \text{ ....(Eq 1)} \)

In right triangle XPY:
\( \tan 30° = \frac{XY}{PY} \)

\( \frac{1}{\sqrt{3}} = \frac{h}{100 - YQ} \)

\( 100 - YQ = h\sqrt{3} \)

Substituting YQ = h from Equation 1:
\( 100 - h = h\sqrt{3} \)

\( 100 = h\sqrt{3} + h \)

\( 100 = h(1.732 + 1) \)

\( 100 = 2.732h \)

\( h = \frac{100}{2.732} = 36.6 \text{ m} \)

The tree measures 36.6 m in height.
In simple words: From point Q, the angle is 45°, which means the distance equals the height. From point P, the angle is 30°. Both distances add up to 100 m. Use this to solve for the height.

Exam Tip: Notice that the 45° angle creates a special relationship where distance equals height. Use this insight to eliminate one variable early, then substitute into the second equation.

Question 9. A man standing on the deck of the ship which is 20 m above the sea level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird.
Answer: Let P be the man on the ship deck, located 20 m above sea level, and B be the bird. The bird flies h meters above the deck level, so BC = h. The bird's reflection appears at point R in the water, directly below at the same horizontal distance. Therefore, the reflection's depth is AR = AB = AC + BC = h + 20.

In right triangle PCB:
\( \tan 30° = \frac{BC}{PC} \)

\( \frac{1}{\sqrt{3}} = \frac{h}{x} \)

\( x = h\sqrt{3} \text{ ......(i)} \)

In right triangle PCR:
\( \tan 60° = \frac{CR}{CP} \)

\( \sqrt{3} = \frac{h + 40}{x} \)

\( \frac{h + 40}{h\sqrt{3}} = \sqrt{3} \text{ [From (i)]} \)

\( h + 40 = \sqrt{3} \times \sqrt{3}h \)

\( h + 40 = 3h \)

\( 3h - h = 40 \)

\( 2h = 40 \)

\( h = 20 \)

From sea level, the bird's height is AB = h + 20 = 20 + 20 = 40 m.

The bird flies at a height of 40 m above sea level.
In simple words: The bird is above the man, and its reflection is below the sea surface. Both angles (30° up, 60° down) come from the same point. Combining the two right triangles gives you the answer.

Exam Tip: Be careful with reflection problems - the reflection point is as far below sea level as the bird is above. Setting up both triangles and combining them is key to solving these efficiently.

Question 10. A vertical tower standing on a horizontal plane is surmounted by a vertical flagstaff. At a point 100 m away from the foot of the tower, the angle of elevation of the top and bottom of the flagstaff are 54° and 42° respectively. Find the height of the flagstaff. Give your answer correct to nearest metre.
Answer: In triangle APB (where P is the observation point, A is the tower base, and B is the tower top):
\( \tan 42° = \frac{AB}{AP} \)

\( 0.9004 = \frac{AB}{100} \)

\( AB = 0.9004 \times 100 = 90.04 \text{ m} \)

In triangle APF (where F is the flagstaff top):
\( \tan 54° = \frac{AF}{AP} \)

\( 1.3764 = \frac{AF}{100} \)

\( AF = 1.3764 \times 100 = 137.64 \text{ m} \)

From the diagram, the flagstaff height is:
\( BF = AF - AB = 137.64 - 90.04 = 47.60 \approx 48 \text{ m} \)

The flagstaff stands 48 m tall.
In simple words: Find the height to the tower top using the 42° angle. Find the total height to the flagstaff top using the 54° angle. Subtract to get just the flagstaff part.

Exam Tip: This is a two-step height problem. Always find the heights separately for each angle, then take the difference. Remember to round to the nearest metre as instructed in the question.