Practice MCQ Questions and Answer on Height and Distance

51.

From a tower of 80 m high, the angle of depression of a bus is 30ÃÂÃÂÃÂÃÂ°. How far is the bus from the tower?

(A) 40 m
(B) 138.4 m
(C) 46.24 m
(D) 160 m

Solution: Let AC be the tower and B be the position of the bus. Then BC = the distance of the bus from the foot of the tower. Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30° ∠ABC = ∠DAB = 30° (because DA || BC) $\begin{aligned} \tan 30^{\circ} = \frac{A C}{B C} \\ \Rightarrow \tan 30^{\circ} = \frac{80}{B C} \\ \Rightarrow B C = \frac{80}{\tan 30^{\circ}} \\ = \frac{80}{(\frac{1}{\sqrt{3}})} \\ = 80 \times 1.73 = 138.4 m \end{aligned}$ i.e., Distance of the bus from the foot of the tower = 138.4 m

52.

The angle of elevation of a ladder leaning against a wall is 60ÃÂÃÂÃÂÃÂº and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

(A) 2.3 m
(B) 4.6 m
(C) 7.8 m
(D) 9.2 m

Solution: Let AB be the wall and BC be the ladder. Then, ∠ACB = 60° = AC = 4.6m $\frac{A C}{B C} = \cos 60^{\circ} = \frac{1}{2}$ ⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

53.

The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°. What is the distance between these two points?

(A) 45 m
(B) 30 m
(C) 103.8 m
(D) 94.6 m

Solution: Let BD be the lighthouse and A and C be the two points on ground. Then, BD, the height of the lighthouse = 60 m ∠BAD = 45°, ∠BCD = 60° $\begin{aligned} \tan 45^{\circ} = \frac{B D}{B A} \\ \Rightarrow 1 = \frac{60}{B A} \\ \Rightarrow B A = 60 m . . . . . . . . (i) \\ \tan 60^{\circ} = \frac{B D}{B C} \\ \Rightarrow \sqrt{3} = \frac{60}{B C} \\ \Rightarrow B C = \frac{60}{\sqrt{3}} \\ = \frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\ = \frac{60 \sqrt{3}}{3} \\ = 20 \sqrt{3} \\ = 20 \times 1.73 \\ = 34.6 m . . . . . . . . . . (ii) \end{aligned}$ Distance between the two points A and C = AC = BA + BC = 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)] = 94.6 m

54.

A poster is on top of a building. A person is standing on the ground at a distance of 50 m from the building. The angles of elevation to the top of the poster and bottom of the poster are 45ÃÂÃÂÃÂÃÂ° and 30ÃÂÃÂÃÂÃÂ°, respectively. What is 200% of the height (in cm) of the poster?

(A) $\frac{25}{3} (3 - \sqrt{3})$
(B) $\frac{75}{3} (3 - \sqrt{3})$
(C) $\frac{50}{3} (3 - \sqrt{3})$
(D) $\frac{100}{3} (3 - \sqrt{3})$

Solution: $\begin{aligned} \tan 30^{\circ} = \frac{B D}{B C} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{B D}{50} \\ \Rightarrow B D = \frac{50}{\sqrt{3}} \\ \tan 45^{\circ} = \frac{A B}{B C} \\ \Rightarrow 1 = \frac{A B}{50} \\ \Rightarrow A B = 50 \\ ∴ A D = A B - B D \\ = 50 - \frac{50}{\sqrt{3}} \\ = \frac{50 (\sqrt{3} - 1) \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\ = \frac{50}{3} (3 - \sqrt{3}) m \\ 200 % of A D = \frac{50}{3} (3 - \sqrt{3}) \times \frac{200}{100} \\ = \frac{100}{3} (3 - \sqrt{3}) m \end{aligned}$

55.

Two houses are in front of each other. Both have chimneys on their top. The line joining the chimneys makes an angle of 45ÃÂÃÂÃÂÃÂ° with the ground. How far are the houses from each other if one house is 25m and other is 10m in height?

(A) 18 m
(B) 12 m
(C) 7.5 m
(D) 15 m

56.

A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30ÃÂÃÂÃÂÃÂ° to 45ÃÂÃÂÃÂÃÂ°, how soon after this will the car reach the observation tower?

(A) 8 min 17 second
(B) 10 min 57 second
(C) 14 min 34 second
(D) 12 min 23 second

Solution: Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car. Then, ∠ ADC = 30° , ∠ ACB = 45° Let AB = h, BC = x, CD = y $\begin{aligned} \tan 45^{\circ} = \frac{A B}{B C} = \frac{h}{x} \\ \Rightarrow 1 = \frac{h}{x} \\ \Rightarrow h = x . . . . . . (1) \\ \tan 30^{\circ} = \frac{A B}{B D} \\ = \frac{A B}{(B C + C D)} \\ = \frac{h}{x + y} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + y} \\ \Rightarrow x + y = \sqrt{3} h \\ \Rightarrow y = \sqrt{3} h - x \end{aligned}$ $\Rightarrow y = \sqrt{3} h - h$ (∵ Substituted the value of x from equation 1 ) $\Rightarrow y = h (\sqrt{3} - 1)$ Given that distance y is covered in 8 minutes. i.e, distance $h (\sqrt{3} - 1)$ is covered in 8 minutes. Time to travel distance x = Time to travel distance h (∵ Since x = h as per equation 1). Let distance h is covered in t minutes. since distance is proportional to the time when the speed is constant, we have $\begin{aligned} h (\sqrt{3} - 1) \propto 8 . . . . . . (A) \\ h \propto t . . . . . . . . . . . . . . (B) \\ \frac{(A)}{(B)} \Rightarrow \frac{h (\sqrt{3} - 1)}{h} = \frac{8}{t} \\ \Rightarrow (\sqrt{3} - 1) = \frac{8}{t} \\ \Rightarrow t = \frac{8}{(\sqrt{3} - 1)} \\ = \frac{8}{(1.73 - 1)} \\ = \frac{8}{.73} \\ = \frac{800}{73} minutes \\ = 10 \frac{70}{73} minutes \\ \approx 10 minutes 57 seconds \end{aligned}$

57.

If the altitude of the sun is at 60ÃÂÃÂÃÂÃÂ°, then the height of the vertical tower that will cast a shadow of length 30 m is

(A) $30 \sqrt{3} m$
(B) $15 m$
(C) $\frac{30}{\sqrt{3}} m$
(D) $15 \sqrt{2} m$

Solution: Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60° $\begin{aligned} Let height of tower A B = h \\ Then in right Δ A P B, \\ \tan θ = \frac{Perpendicular}{Base} = \frac{A B}{P B} \\ \Rightarrow \tan 60^{\circ} = \frac{h}{30} \\ \Rightarrow \sqrt{3} = \frac{h}{30} \\ \Rightarrow h = 30 \sqrt{3} \end{aligned}$ ∴ Height of the tower $= 30 \sqrt{3} m$

58.

An aeroplane when 900 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 60ÃÂÃÂÃÂÃÂ° and 45ÃÂÃÂÃÂÃÂ° respectively. Approximately, how many meters higher is the one than the other?

(A) 381 m
(B) 169 m
(C) 254 m
(D) 211 m

Solution: Let C and D be the position of the aeroplanes. Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45° $\begin{aligned} From the right Δ A B C, \\ \tan 60^{\circ} = \frac{C B}{A B} \\ \sqrt{3} = \frac{900}{A B} \\ A B = \frac{900}{\sqrt{3}} \\ = \frac{900 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\ = \frac{900 \sqrt{3}}{3} \\ = 300 \sqrt{3} \\ From the right Δ A B D, \\ \tan 45^{\circ} = \frac{D B}{A B} \\ 1 = \frac{D B}{A B} \\ D B = A B = 300 \sqrt{3} \\ Required height \\ = C D \\ = (C B - D B) \\ = (900 - 300 \sqrt{3}) \\ = (900 - 300 \times 1.73) \\ = (900 - 519) \\ = 381 m \end{aligned}$

59.

Exactly midway between the foot of two towers P and Q, the angles of elevation of their tops are 45ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°, respectively. The ratio of the heights of P and Q is:

(A) 1 : √3
(B) 3 : 1
(C) 1 : 3
(D) √3 : 1

60.

From the top of a hill 200 m high the angle of depression of the top and the bottom of a tower are observed to be 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°. The height of the tower is (in m);

(A) $\frac{400 \sqrt{3}}{3}$
(B) $166 \frac{2}{3}$
(C) $133 \frac{1}{3}$
(D) $200 \sqrt{3}$

Solution: $\begin{aligned} A B = hill = 200 metre \\ C D = tower \\ In Δ A P C \\ \tan 30^{\circ} = \frac{A P}{P C} \\ \frac{1}{\sqrt{3}} = \frac{A P}{P C} \\ \Rightarrow A P : P C = 1 : \sqrt{3} . . . . (i) \\ In Δ A B D \\ \tan 60^{\circ} = \frac{A B}{B D} \\ \sqrt{3} = \frac{A B}{B D} \\ \Rightarrow A B : B D = \sqrt{3} : 1 . . . . (ii) \\ P B = C D and P C = B D \\ Now \end{aligned}$ $\begin{matrix} A B & : & B D & : & A P \\ \sqrt{3} & : & 1 \\ \sqrt{3} & : & 1 \\ 3 & : & \sqrt{3} & : & 1 \end{matrix}$ $\begin{aligned} C D = P B \Rightarrow A B - A P \\ C D = 3 - 1 \\ C D = 2 units \\ A B = 3 units = 200 metre \\ C D = 2 units = \frac{200}{3} \times 2 = 133 \frac{1}{3} metre \end{aligned}$

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Practice MCQ Questions and Answer on Height and Distance

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