CBSE Class 10 Mathematics Some Applications of Trigonometry VBQs Set 01

PRACTICE QUESTIONS

Question. The angle of elevation of the top of a tower from a point 20 metres away from its base is 45°. The height of the tower is
(a) 10 m
(b) 20 m
(c) 30 m
(d) \( 20\sqrt{3} \) m
Answer: (b) 20 m

Question. Two poles are 25 m and 15 m high and the line joining their tops makes an angle of 45° with the horizontal. The distance between these poles is
(a) 5 m
(b) 8 m
(c) 9 m
(d) 10 m
Answer: (d) 10 m

Question. A portion of a 60 m long tree is broken by tornado and the top struck up the ground making an angle of 30° with the ground level. The height of the point where the tree is broken is equal to
(a) 30 m
(b) 35 m
(c) 40 m
(d) 20 m
Answer: (d) 20 m

Question. An observer, 1.7 m tall, is \( 20\sqrt{3} \) m away from a tower. The angle of elevation from the eye of observer to the top of tower is 30°. Find the height of tower.  
Answer: Let the height of the tower be H. The height of the observer is 1.7 m. The distance from the tower is \( 20\sqrt{3} \) m. Let the part of the tower above the observer's eye level be \( h \).
In right triangle,
\(\tan 30^\circ = \frac{h}{20\sqrt{3}}\)
\( \implies \) \(\frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}\)
\( \implies \) \(h = 20\) m
Total height of tower \( H = h + 1.7 = 20 + 1.7 = 21.7 \) m

Question. When the length of the shadow of a pole of height 7 m is equal to 7 m then find the elevation of the source of light.
Answer: Let the angle of elevation be \( \theta \).
\(\tan \theta = \frac{\text{Height}}{\text{Shadow}} = \frac{7}{7} = 1\)
\( \implies \) \(\theta = 45^\circ\)

Question. If a pole 6m high throws shadow of \( 2\sqrt{3} \) m, then find the angle of elevation of the sun.
Answer: Let the angle of elevation be \( \theta \).
\(\tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\)
\( \implies \) \(\theta = 60^\circ\)

Question. A person standing on the bank of a river observes that the angle of the elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and the width of the river. (\( \sqrt{3} = 1.732 \))
Answer: Let height of tree be \( h \) and width of river be \( w \).
In triangle 1: \(\tan 60^\circ = \frac{h}{w} \implies h = w\sqrt{3}\)
In triangle 2: \(\tan 30^\circ = \frac{h}{w + 40}\)
\( \implies \) \(\frac{1}{\sqrt{3}} = \frac{w\sqrt{3}}{w + 40}\)
\( \implies \) \(w + 40 = 3w\)
\( \implies \) \(2w = 40 \implies w = 20\) m
Width of river = 20 m
Height of tree \( h = 20\sqrt{3} = 20 \times 1.732 = 34.64 \) m

Question. 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 that instant.  
Answer: Let the height of the second plane be \( H \) and the first plane be \( h = 3125 \) m. Let the horizontal distance to the observer be \( x \).
For the first plane: \(\tan 30^\circ = \frac{3125}{x} \implies x = 3125\sqrt{3}\)
For the second plane: \(\tan 60^\circ = \frac{H}{x} \implies H = x\sqrt{3} = (3125\sqrt{3})\sqrt{3} = 9375\) m
Distance between planes = \( H - h = 9375 - 3125 = 6250 \) m

Question. From the top of a tower 50 m high the angles of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. Find the height of the pole. [Foreign 2016]
Answer: Let tower height \( H = 50 \) m and pole height be \( h \). Let distance between them be \( x \).
\(\tan 60^\circ = \frac{50}{x} \implies \sqrt{3} = \frac{50}{x} \implies x = \frac{50}{\sqrt{3}}\) m
Angle of depression to pole top is 45° \( \implies \) \(\tan 45^\circ = \frac{50 - h}{x}\)
\( \implies \) \(1 = \frac{50 - h}{\frac{50}{\sqrt{3}}}\)
\( \implies \) \(\frac{50}{\sqrt{3}} = 50 - h \implies h = 50 - \frac{50}{\sqrt{3}} = 50(1 - \frac{\sqrt{3}}{3}) \) m

Question. As observed from the top of a light-house, 100 m high above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the ship during the period of observation. (Use \( \sqrt{3} = 1.732 \))  
Answer: Height of light-house \( h = 100 \) m.
Distance at 30°: \( x_1 = 100\sqrt{3} \)
Distance at 60°: \( x_2 = \frac{100}{\sqrt{3}} \)
Distance travelled = \( x_1 - x_2 = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200 \times 1.732}{3} \approx 115.47 \) m

Question. Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. [Use \( \sqrt{3} = 1.73 \)]  
Answer: Height \( h = 200 \) m. Let distances be \( x_1 \) and \( x_2 \).
\( \tan 60^\circ = \frac{200}{x_1} \implies x_1 = \frac{200}{\sqrt{3}} \approx 115.47 \) m
\( \tan 45^\circ = \frac{200}{x_2} \implies x_2 = 200 \) m
Distance between ships = \( x_1 + x_2 = 115.47 + 200 = 315.47 \) m

Question. The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of \( 3000\sqrt{3} \) m, find the speed of the aeroplane.  
Answer: Height \( h = 3000\sqrt{3} \) m. Time \( t = 30 \) s.
Horizontal distance 1: \( x_1 = \frac{3000\sqrt{3}}{\tan 60^\circ} = \frac{3000\sqrt{3}}{\sqrt{3}} = 3000 \) m
Horizontal distance 2: \( x_2 = \frac{3000\sqrt{3}}{\tan 30^\circ} = (3000\sqrt{3})\sqrt{3} = 9000 \) m
Distance covered \( D = 9000 - 3000 = 6000 \) m
Speed = \( \frac{D}{t} = \frac{6000}{30} = 200 \) m/s

Question. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 30 m high, find the height of the building. 
Answer: Tower height \( H = 30 \) m. Building height \( h \). Let distance be \( d \).
From tower: \(\tan 45^\circ = \frac{30}{d} \implies d = 30\) m
From building: \(\tan 30^\circ = \frac{h}{d} \implies h = d \tan 30^\circ = 30 \times \frac{1}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \) m

Question. The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of \( 1500\sqrt{3} \) m, find the speed of the plane in km/hr.  
Answer: Height \( h = 1500\sqrt{3} \) m. Time \( t = 15 \) s.
\( x_1 = 1500 \) m, \( x_2 = 4500 \) m.
Distance \( D = 4500 - 1500 = 3000 \) m
Speed = \( \frac{3000 \text{ m}}{15 \text{ s}} = 200 \) m/s = \( 200 \times 3.6 = 720 \) km/hr

Question. The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of \( 3600\sqrt{3} \) m, find the speed in km/hr of the plane.
Answer: Distance = \( 3600\sqrt{3}(\sqrt{3} - \frac{1}{\sqrt{3}}) = 3600(3 - 1) = 7200 \) m.
Speed = \( \frac{7200}{30} = 240 \) m/s = \( 240 \times 3.6 = 864 \) km/hr

Question. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.
Answer: Let height of tower be \( H \) and horizontal distance \( x \).
\(\tan 60^\circ = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}\)
\(\tan 30^\circ = \frac{H - 10}{x} \implies \frac{1}{\sqrt{3}} = \frac{H - 10}{\frac{H}{\sqrt{3}}} \implies \frac{H}{3} = H - 10\)
\( \implies \) \( \frac{2H}{3} = 10 \implies H = 15 \) m

Question. The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building. 
Answer: Building height = 60 m. Let distance be \( d \) and lighthouse height be \( H \).
\(\tan 60^\circ = \frac{60}{d} \implies d = \frac{60}{\sqrt{3}} = 20\sqrt{3} \approx 34.64 \) m
(ii) Distance = 34.64 m.
Difference \( h' = d \tan 30^\circ = 20\sqrt{3} \times \frac{1}{\sqrt{3}} = 20 \) m
(i) Difference = 20 m. (Lighthouse height = 60 + 20 = 80 m)

Question. From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 45°. Determine the height of the tower and the distance between the tower and the building.
Answer: Let building height = 15 m, tower height = \( H \), distance = \( d \).
At bottom: \(\tan 45^\circ = \frac{H}{d} \implies H = d\)
At top: \(\tan 30^\circ = \frac{H - 15}{d} \implies \frac{1}{\sqrt{3}} = \frac{H - 15}{H}\)
\( \implies \) \(H = \sqrt{3}H - 15\sqrt{3} \implies H(\sqrt{3} - 1) = 15\sqrt{3}\)
\( H = \frac{15\sqrt{3}}{\sqrt{3} - 1} \approx 35.49 \) m. Distance \( d = 35.49 \) m.

Question. A parachutist is descending vertically and makes angles of depression of 45° and 60° at two observation points 100 m apart from each other on the left side of himself. Find, in metres, the approximate height from which he falls and also find, in metres the approximate distance of the point where he falls on the ground from the first observation point.
Answer: Let height be \( h \) and distance from first point be \( x \). Points are 100m apart.
\( \tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3} \)
\( \tan 45^\circ = \frac{h}{x + 100} \implies h = x + 100 \)
\( x\sqrt{3} = x + 100 \implies x(\sqrt{3} - 1) = 100 \implies x \approx 136.6 \) m
Height \( h \approx 136.6 + 100 = 236.6 \) m

Question. The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, from A is 60°, then find the height of the flagstaff. [Use \( \sqrt{3} = 1.73 \)]  
Answer: Distance \( d = 120 \) m.
Tower height \( h = 120 \tan 45^\circ = 120 \) m
Total height \( H = 120 \tan 60^\circ = 120\sqrt{3} \approx 207.6 \) m
Flagstaff height = \( H - h = 207.6 - 120 = 87.6 \) m

Question. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A. 
Answer: Let height of cloud above water be \( h \). Observation point is 20m above water.
Horizontal distance \( x \).
\(\tan 30^\circ = \frac{h - 20}{x}\)
\(\tan 60^\circ = \frac{h + 20}{x}\)
\( \implies \) \(\frac{\sqrt{3}}{1/\sqrt{3}} = \frac{h + 20}{h - 20} \implies 3 = \frac{h + 20}{h - 20}\)
\( \implies \) \(3h - 60 = h + 20 \implies 2h = 80 \implies h = 40\) m
Distance from A = \( \frac{h - 20}{\sin 30^\circ} = \frac{20}{0.5} = 40 \) m

Question. A contract was awarded to construct a vertical pillar at a horizontal distance of 100 m from a fixed point. It was decided that angle of elevation of the top of the complete pillar from this point to be 60°. Contractor finished the job by making a pillar such that angle of elevation of its top was 45°. Find the height of the pillar to be increased as per the terms of contract.
Answer: Distance = 100 m. Final height \( H = 100 \tan 60^\circ = 100\sqrt{3} \approx 173.2 \) m.
Initial height \( h = 100 \tan 45^\circ = 100 \) m.
Increase needed = \( 173.2 - 100 = 73.2 \) m.

Question. A highway leads to the foot of 300 m high tower. An observatory is set at the top of the tower. It sees a car moving towards it at an angle of depression of 30°. After 15 seconds angle of depression becomes 60°. Find the distance travelled by the car during this time.
Answer: Height \( h = 300 \) m.
Distance 1: \( x_1 = 300 \sqrt{3} \approx 519.6 \) m
Distance 2: \( x_2 = \frac{300}{\sqrt{3}} \approx 173.2 \) m
Distance travelled = \( 519.6 - 173.2 = 346.4 \) m

Question. A fire in a building B is reported on telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60° to the road and Q observes that it is at an angle of 45° to the road. Which station should send its team and how much will this team have to travel?
Answer: Let distance from P to the foot of the perpendicular from B to the road be \( x \) and height of building be \( h \).
\(\tan 60^\circ = h/x \implies h = x\sqrt{3}\)
\(\tan 45^\circ = h/(20 - x) \implies h = 20 - x\)
\(x\sqrt{3} = 20 - x \implies x = \frac{20}{\sqrt{3} + 1} \approx 7.32 \) km
Distance PB = \( x / \cos 60^\circ = 2x \approx 14.64 \) km.
Distance QB = \( (20 - x) / \cos 45^\circ \approx 17.93 \) km.
Station P should send the team. Travel distance = 14.64 km.

Question. On a horizontal plane there is a vertical tower with a flag pole on the top of the tower. At a point 9 metres away from the foot of the tower the angles of elevation of the top and bottom of the flag pole are 60° and 30° respectively. Find the heights of the tower and flag pole mounted on it. (\( \sqrt{3} = 1.732 \))
Answer: Distance = 9 m.
Tower height \( h = 9 \tan 30^\circ = 9 \times \frac{1}{\sqrt{3}} = 3\sqrt{3} \approx 5.196 \) m.
Total height \( H = 9 \tan 60^\circ = 9\sqrt{3} \approx 15.588 \) m.
Flagpole height = \( 15.588 - 5.196 = 10.392 \) m.

Question. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of the tower PQ and the distance XQ.
Answer: Let horizontal distance be \( d \). Tower height \( H \).
\(\tan 60^\circ = H/d \implies H = d\sqrt{3}\)
\(\tan 45^\circ = (H - 40)/d \implies d = H - 40\)
\(d = d\sqrt{3} - 40 \implies d(\sqrt{3} - 1) = 40 \implies d = 20(\sqrt{3} + 1) \approx 54.64 \) m
Height \( H = 54.64 + 40 = 94.64 \) m.
Distance \( XQ = \sqrt{d^2 + H^2} \approx 109.28 \) m.

Question. At the foot of a mountain, the elevation of its summit is 45°. After ascending 1000  m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain. (use \( \sqrt{3} = 1.732 \)) 
Answer: Vertical height reached = \( 1000 \sin 30^\circ = 500 \) m. Horizontal distance covered = \( 1000 \cos 30^\circ = 500\sqrt{3} \) m.
Let mountain height be \( H \). Original distance \( D = H \tan 45^\circ = H \).
New distance \( d = D - 500\sqrt{3} = H - 500\sqrt{3} \).
New height from current level \( h' = H - 500 \).
\(\tan 60^\circ = (H - 500) / (H - 500\sqrt{3})\)
\(\sqrt{3}H - 1500 = H - 500 \implies H(\sqrt{3} - 1) = 1000\)
\( H = 1000 / 0.732 \approx 1366.12 \) m.

MORE PRACTICE QUESTIONS

Question. If at some time, the length of the shadow of a tower is \( \sqrt{3} \) times its height, then the angle of elevation of the Sun, at that time, is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer: (b) 30°

Question. The angle of elevation of the Sun, if the length of the shadow of a tower of height 20 m is \( 20\sqrt{3} \) m is
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer: (a) 30°

Question. A person standing on the bank of a river finds that the angle of elevation of the top of a tower on the opposite bank is 45°. Which of the following statements is correct?
(a) Breadth of the river is twice the height of the tower.
(b) Breadth of the river is half of the height of the tower.
(c) Breadth of the river is equal to the height of the tower.
(d) None of the options.
Answer: (c) Breadth of the river is equal to the height of the tower.

Question. If the elevation of the Sun is 30°, then the length of the shadow cast by a tower of 150 feet height is
(a) 150 feet
(b) \( 50\sqrt{3} \) feet
(c) \( 150\sqrt{3} \) feet
(d) 200 feet
Answer: (c) \( 150\sqrt{3} \) feet

Question. When a point is observed, the angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level is known as
(a) angle of elevation
(b) angle of depression
(c) right angle
(d) None of the options
Answer: (b) angle of depression

Question. A straight highway leads to the foot of a 100 m tall tower. From the top of the tower, angle of depression of a car on the highway is 30°. The distance of the car from foot of the tower is
(a) \( 100\sqrt{3} \) m
(b) \( 50\sqrt{3} \) m
(c) \( 200\sqrt{3} \) m
(d) 100 m
Answer: (a) \( 100\sqrt{3} \) m

Question. The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then the value of x : y is  
(a) 3 : 1
(b) 1 : 3
(c) 2 : 3
(d) 3 : 2
Answer: (b) 1 : 3

Question. From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. Then the height of the tower is
(a) \( 10\sqrt{3} \) m
(b) \( 20\sqrt{3} \) m
(c) \( 30\sqrt{3} \) m
(d) \( 40\sqrt{3} \) m
Answer: (b) \( 20\sqrt{3} \) m