CBSE Class 10 Maths HOTs Heights And Distances Set 02

Multiple Choice Questions

Question. The angle of elevation of the Sun when the shadow of a pole \( h \) m high is \( \sqrt{3} h \) m long is
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer: (b)

Question. If a pole 6 m high casts a 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)

Question. If \( 300\sqrt{3} \) m high tower makes an angle of elevation at a point on ground which is 300 m away from its foot, then the angle of elevation is
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer: (d)

Question. From the top of a 60 m high tower, the angle of depression of a point on the ground is 30°. The distance of the point from the foot of tower is
(a) 180 m
(b) \( 60\sqrt{3} \) m
(c) 150 m
(d) \( 30\sqrt{3} \) m
Answer: (b)

Question. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground, then the height of pole, if the angle made by the rope with the ground level is 30°, is
(a) 5 m
(b) 10 m
(c) 15 m
(d) 20 m
Answer: (b)

Question. A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 9.5 m away from the wall. The length of the ladder is
(a) 10 m
(b) 16 m
(c) 18 m
(d) 19 m
Answer: (d)

Question. A ramp for disabled people in a hospital have slope 30°. If the height of the ramp be 1 m, then the length of ramp is
(a) 2 m
(b) 0.5 m
(c) \( 2\sqrt{3} \) m
(d) 1 m
Answer: (a)

Question. A kite is flying at a height of 80 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with ground is 60°, then the length of the string is
(a) 62.37 m
(b) 92.37 m
(c) 52.57 m
(d) 72.57 m
Answer: (b)

Question. The length of a string between a kite and a point on the ground is 85 m. If the string makes an angle \( \theta \) with the ground level such that \( \tan \theta = \frac{15}{8} \), then the height of kite is
(a) 75 m
(b) 78.05 m
(c) 226 m
(d) None of these
Answer: (a)

Question. A tower stands near an airport. The angle of elevation \( \theta \) of the tower from a point on the ground is such that its tangent is 5/12. The height of the tower, if the distance of the observer from the tower is 120 m is
(a) 40 m
(b) 50 m
(c) 60 m
(d) 70 m
Answer: (b)

Question. The top of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with the horizontal, then the length of the wire is
(a) 12 m
(b) 10 m
(c) 8 m
(d) 6 m
Answer: (a)

Question. An observer, 1.5 m tall is 20.5 m away from a tower 22 m high, then the angle of elevation of the top of the tower from the eye of the observer is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b)

Question. The angle of elevation of the top of the tower from a point, which is 40 m away from the base of the tower in the horizontal level, is 45°. Find the height of the tower.
(a) 70 m
(b) 60 m
(c) 40 m
(d) 30 m
Answer: (c)

Question. The angle of elevation of the top of a building 150 m high, from a point on the ground is 45°. The distance of the point from foot of the building is
(a) 120 m
(b) 130 m
(c) 140 m
(d) 150 m
Answer: (d)

Question. The angle of depression of the car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower is
(a) 150 m
(b) 75 m
(c) \( 150\sqrt{3} \) m
(d) \( \frac{150}{\sqrt{3}} \) m
Answer: (c)

Question. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively, then the height of the tower is
(a) 14.64 m
(b) 28.64 m
(c) 38.64 m
(d) 19.64 m
Answer: (a)

Question. A bridge on a river makes an angle of 45° with its edge. If the length along the bridge from one edge to the other is 150 m, then the width of the river is
(a) 107.75 m
(b) 105 m
(c) 75 m
(d) 106.05 m
Answer: (d)

A group of students of class X visited India Gate on an educational trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 m) in height.

Question. What is the angle of elevation if they are standing at a distance of 42 m away from the monument?
(a) 30°
(b) 45°
(c) 60°
(d) 0°
Answer: (b)

Question. They want to see the tower at an angle of 60°. So, they want to know the distance where they should stand and hence find the distance.
(a) 25.24 m
(b) 20.12 m
(c) 42 m
(d) 24.24 m
Answer: (d)

Question. If the altitude of the Sun is at 60°, then the height of the vertical tower that will cast a shadow of length 20 m is
(a) \( 20\sqrt{3} \) m
(b) \( \frac{20}{\sqrt{3}} \) m
(c) \( \frac{15}{\sqrt{3}} \) m
(d) \( 15\sqrt{3} \) m
Answer: (a)

Question. The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (b)

Question. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is
(a) corresponding angle
(b) angle of elevation
(c) angle of depression
(d) complete angle
Answer: (c)

Short Answer Type Questions

Question. If the height of a tower and the distance of the point of observation from its foot, both are increased by 10%, then the angle of elevation of its top remains unchanged. Explain.
Answer: Let height be \( h \) and distance be \( x \). \( \tan \theta = h/x \). New height \( h' = 1.1h \) and new distance \( x' = 1.1x \). New angle \( \tan \theta' = (1.1h)/(1.1x) = h/x = \tan \theta \). Hence, angle remains unchanged.

Question. A straight tree is broken due to thunderstorm. The broken part is bent in such a way that the peak of the tree touches the ground at an angle of 60° at a distance of \( 2\sqrt{3} \) m. Find the whole height of the tree.
Answer: Let vertical part be \( h \) and broken part be \( l \). \( \tan 60^\circ = h/(2\sqrt{3}) \Rightarrow \sqrt{3} = h/(2\sqrt{3}) \Rightarrow h = 6 \) m. \( \cos 60^\circ = (2\sqrt{3})/l \Rightarrow 1/2 = (2\sqrt{3})/l \Rightarrow l = 4\sqrt{3} \) m. Total height \( = 6 + 4\sqrt{3} \) m.

Question. Determine the height of a mountain, if the elevation of its top at an unknown distance from the base is 30° and at a distance 10 km farther off from the mountain, along the same line, the angle of elevation is 15°. (take \( \tan 15^\circ = 0.27 \))
Answer: Let height be \( h \) and initial distance be \( x \). \( x = h \cot 30^\circ = h\sqrt{3} \). For 15°, \( x + 10 = h \cot 15^\circ \). \( h\sqrt{3} + 10 = h/0.27 \). Solving for \( h \), \( h \approx 5.09 \) km.

Question. There is a flag staff on a tower of height 20 m. At a point on the ground, the angles of elevation of the foot and top of the flag are 45° and 60°, respectively. Find the height of the flag staff.
Answer: Distance from foot \( x = 20 \tan 45^\circ = 20 \) m. Height of top \( H = 20 \tan 60^\circ = 20\sqrt{3} \). Height of flag staff \( = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \) m.

Question. If the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is also increasing. Why or why not?
Answer: No, it is decreasing. \( \tan \theta = \text{height}/\text{shadow} \). As shadow increases, \( \tan \theta \) decreases, so \( \theta \) decreases.

Question. A window in a building is at a height of 10 m from the ground. The angle of depression of a point P on the ground from the window is 30°. The angle of elevation of the top of the building from the point P is 60°. Find the height of the building. 
Answer: Distance to P, \( x = 10 \cot 30^\circ = 10\sqrt{3} \). Height of building \( H = x \tan 60^\circ = 10\sqrt{3} \times \sqrt{3} = 30 \) m.

Question. A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball.
Answer: Distance \( = 20 \cot 60^\circ = 20/\sqrt{3} \) m.

Question. If two towers of height \( x \) m and \( y \) m subtend angles of 30° and 60°, respectively at the centre of the line joining their feet, then find \( x : y \). 
Answer: Let midpoint distance be \( d \). \( x = d \tan 30^\circ = d/\sqrt{3} \) and \( y = d \tan 60^\circ = d\sqrt{3} \). \( x/y = (d/\sqrt{3}) / (d\sqrt{3}) = 1/3 \). Ratio is 1 : 3.

Question. If a man standing on a platform 3 m above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection. State true or false. Justify.
Answer: False. The reflection is as deep as the cloud is high above the lake surface. Since the observer is 3m above the lake, the distances are different, making the angles unequal.

Question. From the top of a hill, the angles of depression of two consecutive kilometre stones due East are found to be 30° and 45°. Find the height of the hill. 
Answer: Let height be \( h \). Distances are \( h \cot 45^\circ = h \) and \( h \cot 30^\circ = h\sqrt{3} \). Difference \( h\sqrt{3} - h = 1 \) km. \( h = 1/(\sqrt{3} - 1) = (\sqrt{3} + 1)/2 \approx 1.366 \) km.

Question. The shadow of a tower is 30 m long, when the Sun’s angle of elevation is 30°. What is the length of the shadow, when Sun’s elevation is 60°?
Answer: Height \( h = 30 \tan 30^\circ = 30/\sqrt{3} = 10\sqrt{3} \). New shadow \( = h \cot 60^\circ = 10\sqrt{3} \times (1/\sqrt{3}) = 10 \) 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 base of the light house are in the same straight line. The angle 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, then find the distance between the two ships. 
Answer: Distance \( = 200 \cot 60^\circ + 200 \cot 45^\circ = 200/\sqrt{3} + 200 = 200(1/\sqrt{3} + 1) \) m.

Question. From the top of a tower of height 50 m, the angles of depression of the top and bottom of a pole are 30° and 45°, respectively. Find 
(i) how far the pole is from the bottom of the tower.
(ii) the height of the pole. [take, \( \sqrt{3} = 1.732 \)]
Answer: (i) Distance \( = 50 \cot 45^\circ = 50 \) m. (ii) Height difference \( = 50 \tan 30^\circ = 50/\sqrt{3} \approx 28.87 \). Height of pole \( = 50 - 28.87 = 21.13 \) m.

Question. A man standing on the deck of a ship, which is 10 m above the water level. He observes that the angle of elevation of the top of a hill is 60° and the angle of depression of the base of the hill is 30°. Calculate the distance of the hill from the ship and height of the hill. 
Answer: Distance \( = 10 \cot 30^\circ = 10\sqrt{3} \approx 17.32 \) m. Height above deck \( = 10\sqrt{3} \tan 60^\circ = 30 \). Total height \( = 30 + 10 = 40 \) m.

Question. The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 s, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 2500 m, then find the average speed of the aeroplane. 
Answer: Distance covered \( = 2500(\cot 30^\circ - \cot 45^\circ) = 2500(\sqrt{3} - 1) \approx 1830 \) m. Speed \( = 1830/15 = 122 \) m/s.

Question. The shadow of a flag staff is three times as long as the shadow of the flag staff, when the Sun rays meet the ground at an angle of 60°. Find the angle between the Sun rays and the ground at the time of longer shadow.
Answer: \( h = s_1 \tan 60^\circ \Rightarrow s_1 = h/\sqrt{3} \). Longer shadow \( s_2 = 3s_1 = 3h/\sqrt{3} = h\sqrt{3} \). \( \tan \theta = h/s_2 = h/(h\sqrt{3}) = 1/\sqrt{3} \). Angle is 30°.

Question. An aeroplane, when flying at a height of 4000 m from the ground, passes vertically above another aeroplane at an instant when the angles of elevation of two planes from the same point on the ground are 60° and 45°, respectively. Find the vertical distance between the aeroplanes at that instant.
Answer: Distance from point \( x = 4000 \cot 60^\circ = 4000/\sqrt{3} \). Height of lower plane \( h = x \tan 45^\circ = 4000/\sqrt{3} \). Vertical distance \( = 4000 - 4000/\sqrt{3} = 4000(1 - 1/\sqrt{3}) \) m.

Question. There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 30° and 45°, then find the height of the tree. [take, \( \sqrt{3} = 1.732 \)]
Answer: Let height be \( h \). \( h \cot 30^\circ + h \cot 45^\circ = 100 \). \( h(\sqrt{3} + 1) = 100 \). \( h = 100/(\sqrt{3} + 1) \approx 36.6 \) m.