CBSE Class 10 Maths HOTs Heights And Distances Set C

Question. An aeroplane is at an altitude of 1200 m. If two ships are sailing towards it in the same direction. The angles of depression of the ships as observed from the aeroplane are 60° and 30°, respectively. Find the distance between both ships.
Answer: Distance \( = 1200(\cot 30^\circ - \cot 60^\circ) = 1200(\sqrt{3} - 1/\sqrt{3}) = 1200(2/\sqrt{3}) = 800\sqrt{3} \approx 1385.6 \) m.

Question. The angles of depression of two consecutive kilometre stones on the road on right and left of an aeroplane are 60° and 45°, respectively as observed from the aeroplane. Find the height of the aeroplane.
Answer: Let height be \( h \). \( h \cot 60^\circ + h \cot 45^\circ = 1 \). \( h(1/\sqrt{3} + 1) = 1 \). \( h = \sqrt{3}/(\sqrt{3} + 1) \approx 0.634 \) km.

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 flag staff fixed at the top of the tower, at A is 60°, then find the height of the flag staff. [use, \( \sqrt{3} = 1.73 \)] 
Answer: Tower height \( = 120 \tan 45^\circ = 120 \) m. Height including flag staff \( = 120 \tan 60^\circ = 120\sqrt{3} \approx 207.6 \) m. Flag staff height \( = 207.6 - 120 = 87.6 \) m.

Question. A balloon is connected to an electric pole. It is inclined at 60° to the horizontal by a cable of length 215 m. Determine the height of the balloon from the ground. Also, find the height of the balloon, if the angle of inclination is changed from 60° to 30°. 
Answer: Height \( = 215 \sin 60^\circ = 215\sqrt{3}/2 \approx 186.2 \) m. At 30°, height \( = 215 \sin 30^\circ = 107.5 \) m.

Question. A man in a boat rowing away from a light house 100 m high takes 2 min to change the angle of elevation of the light house from 60° to 45°. Find the speed of boat.
Answer: Distance covered \( = 100(\cot 45^\circ - \cot 60^\circ) = 100(1 - 1/\sqrt{3}) \approx 42.26 \) m. Speed \( = 42.26 / 120 \approx 0.35 \) m/s.

Question. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Answer: New angle \( = 45^\circ \). Let height be \( h \). \( h \cot 30^\circ - h \cot 45^\circ = 20 \). \( h(\sqrt{3} - 1) = 20 \). \( h = 20/(\sqrt{3} - 1) = 10(\sqrt{3} + 1) \approx 27.32 \) m.

Question. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Answer: \( h(\cot 30^\circ - \cot 60^\circ) = 50 \). \( h(\sqrt{3} - 1/\sqrt{3}) = 50 \). \( h(2/\sqrt{3}) = 50 \). \( h = 25\sqrt{3} \approx 43.3 \) m.

Question. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are \( \alpha \) and \( \beta \) respectively. Prove that the height of the tower is \( \left( \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \right) \).
Answer: Let tower height be \( H \). Distance \( x = H \cot \alpha = (H+h) \cot \beta \). \( H \tan \beta = (H+h) \tan \alpha \). \( H(\tan \beta - \tan \alpha) = h \tan \alpha \). \( H = \frac{h \tan \alpha}{\tan \beta - \tan \alpha} \). Proved.

Question. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the tower.
Answer: Distance \( x = 30 \cot 60^\circ = 30/\sqrt{3} = 10\sqrt{3} \approx 17.32 \) m. Height of second tower \( h = x \tan 30^\circ = 10\sqrt{3} \times (1/\sqrt{3}) = 10 \) m.

Question. From the top of a tower h m high, angles of depression of two objects, which are in line with the foot of the tower are \( \alpha \) and \( \beta \) \( (\beta > \alpha) \). Find the distance between the two objects.
Answer: Distance \( = h(\cot \alpha - \cot \beta) \).

Question. A ladder against a vertical wall at an inclination \( \alpha \) to the horizontal. Its foot is pulled away from the wall through a distance p, so that its upper end slides a distance q down the wall and then the ladder makes an angle \( \beta \) with the horizontal. Show that \( \frac{p}{q} = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta} \).
Answer: Let ladder length be \( L \). Initially \( x = L \cos \alpha, y = L \sin \alpha \). After move, \( x+p = L \cos \beta, y-q = L \sin \beta \). \( p = L(\cos \beta - \cos \alpha) \) and \( q = L(\sin \alpha - \sin \beta) \). Ratio \( p/q = \frac{\cos \beta - \cos \alpha}{\sin \alpha - \sin \beta} \). Proved.

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 45°. Find the height of the tower.
Answer: Let tower height be \( H \). \( x = H \cot 60^\circ = (H-10) \cot 45^\circ \). \( H/\sqrt{3} = H-10 \). \( H(1 - 1/\sqrt{3}) = 10 \). \( H = \frac{10\sqrt{3}}{\sqrt{3}-1} = 5\sqrt{3}(\sqrt{3}+1) = 5(3+\sqrt{3}) \approx 23.66 \) m.

Question. A window of a house is \( h \) m above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be \( \alpha \) and \( \beta \), respectively. Prove that the height of the other house is \( h(1 + \tan \alpha \cot \beta) \) m.
Answer: Distance across lane \( x = h \cot \beta \). Height of top above window \( y = x \tan \alpha = h \cot \beta \tan \alpha \). Total height \( = y + h = h \cot \beta \tan \alpha + h = h(1 + \tan \alpha \cot \beta) \) m. Proved.

Question. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At any instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
Answer: Let balloon be \( H \) m high. \( x = (H-2) \cot 60^\circ = (H-6) \cot 30^\circ \). \( (H-2)/\sqrt{3} = (H-6)\sqrt{3} \). \( H-2 = 3H - 18 \). \( 2H = 16 \). \( H = 8 \) m.

Case Based Questions

A girl 8 m tall spots a parrot sitting on the top of a building of height 58 m from the ground. The angle of elevation of the parrot from the eyes of girl at any instant is 60°. The parrot flies away horizontally in such a way that it remained at a constant height from the ground. After 8 s, the angle of elevation of the parrot from the same point is 30°.

Question. Find the distance of first position of the parrot from the eyes of the girl.
Answer: Net vertical height \( = 50 \) m. Distance from eyes \( = 50 / \sin 60^\circ = 100/\sqrt{3} \approx 57.73 \) m.

Question. If the distance between the position of parrot increases, then the angle of elevation decreases. Justify with girl.
Answer: Yes, as the horizontal distance \( x \) increases for a constant height \( h \), \( \tan \theta = h/x \) decreases, meaning the angle \( \theta \) decreases.

Question. Find the distance between the girl and the building.
Answer: Distance \( = 50 \cot 60^\circ = 50/\sqrt{3} \approx 28.87 \) m.

Question. How much distance covers parrot covers?
Answer: Distance \( = 50(\cot 30^\circ - \cot 60^\circ) = 50(\sqrt{3} - 1/\sqrt{3}) = 100/\sqrt{3} \approx 57.73 \) m.

Question. Find the speed of the parrot in 8s.
Answer: Speed \( = 57.73 / 8 \approx 7.22 \) m/s.

Multiple Choice Questions

Question. A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied at the ground. The height of the pole is 12 m and the angle made by the rope with ground level is 30°. The distance covered by the artist in climbing to the top of the pole is
(a) 12 m
(b) 6 m
(c) 24 m
(d) 32 m
Answer: (c)

Question. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is
(a) 30 m
(b) \( \frac{15}{2} \) m
(c) 15 m
(d) 25 m
Answer: (b)

Question. A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, then the angle of elevation of the kite at the ground is
(a) 30°
(b) 45°
(c) 60°
(d) None of these
Answer: (a)

Question. The tops of two poles of height 30 m and 24 m are connected by a wire. If the wire makes an angle of 45° with the horizontal, then find the length of the wire.
(a) 14 m
(b) 3 m
(c) 4 m
(d) \( 6\sqrt{2} \) m
Answer: (d)

Question. An observer 3.5 m tall is 38.5 m away from a tower 42 m high. The angle of elevation of the top of the tower from the eye of the observer is
(a) 30°
(b) 90°
(c) 45°
(d) 60°
Answer: (c)

Case Based MCQs

A boy is standing on the top of mountain. He observed that boat P and boat Q are approaching towards mountain from opposite directions. He finds that angle of depression of boat P is 60° and angle of depression of boat Q is 45°. He also knows that height of the mountain is 50 m.

Question. Measure of \( \angle ACD \) is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer: (c)

Question. If \( \angle YAB = 45^\circ \), then \( \angle ABD \) is also 45°, Why?
(a) vertically opposite angles
(b) alternate interior angles
(c) alternate exterior angles
(d) corresponding angles
Answer: (b)

Question. Length of \( CD \) is equal to
(a) 90 m
(b) \( 50\sqrt{3} \) m
(c) \( 50/\sqrt{3} \) m
(d) 100 m
Answer: (c)

Question. Length of \( BD \) is equal to
(a) 50 m
(b) 100 m
(c) \( 100\sqrt{2} \) m
(d) \( 100\sqrt{3} \) m
Answer: (a)

Question. Length of \( AC \) is equal to
(a) \( 100/\sqrt{3} \) m
(b) \( 100\sqrt{3} \) m
(c) 50 m
(d) 100 m
Answer: (a)

Short Answer Type Questions

Question. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled. State true or false. Explain.
Answer: False

Question. A peacock is sitting on the top of a tree. It observes a serpent on the ground making an angle of depression of 30°. The peacock catches the serpent in 12 s with the speed of 300 m/min. What is the height of the tree? 
Answer: 30 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 the difference between the heights of the light house and building.
Answer: 20 m

Question. As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships.
Answer: \( 100 (\sqrt{3} - 1) \) m

Long Answer Type Questions

Question. Two ships are sailing in the sea on either side of the light house. The angles of depression of two ships as observed from the top of the light house are 60° and 45°, respectively. If the distance between the ships is \( 100 \left( \frac{\sqrt{3} + 1}{\sqrt{3}} \right) \) m, then find the height of the light house.
Answer: 100 m