Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

31.

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60ÃÂÃÂÃÂÃÂ° with the wall, then the height of the wall is

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

Solution: Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground ∴ In ∆ABC, ∠B = 90° Let height of wall AB = h Then $\sin θ = \frac{A B}{A C} \Rightarrow \sin 60^{\circ} = \frac{h}{15}$ $\begin{aligned} \Rightarrow \frac{\sqrt{3}}{2} = \frac{h}{15} \\ \Rightarrow h = \frac{15 \sqrt{3}}{2} m \end{aligned}$ ∴ Height of the wall $= \frac{15 \sqrt{3}}{2} m$

32.

From the top of a 12 m high building, The angle of elevation of the top of a tower is 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and the angle of depression of the foot of the tower is ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ¸. Such that tan ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ¸ = $\frac{3}{4}$ , what is the height of the tower (ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 = 1.73)?

(A) 39.68 m
(B) 41.41 m
(C) 37.95 m
(D) 36.22 m

Solution: Let AB be the building and EC be the tower. $\begin{aligned} In Δ A D E \\ \tan 60^{\circ} = \frac{E D}{A D} \\ \sqrt{3} = \frac{E D}{16} \\ E D = 16 \sqrt{3} = 16 \times 1.73 = 27.68 m \end{aligned}$ Hence the height of the tower EC = CD + ED = 12 + 27.68 = 39.68 m

33.

There is a tree between houses of A and B. If the tree leans on AÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs House, the tree top rests on his window which is 12 m from ground. If the tree leans on BÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs House, the tree top rests on his window which is 9 m from ground. If the height of the tree is 15 m, what is distance between AÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs and BÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs house?

(A) 21 m
(B) 25 m
(C) 16 m
(D) 12 m

Solution: In $Δ$ STR, by Pythagoras theorem $\begin{aligned} R T^{2} = S T^{2} + R S^{2} \\ ∴ S T^{2} = 15^{2} - 9^{2} = 144 \\ ∴ S T = 12 m \end{aligned}$ In $Δ$ TQP, by Pythagoras theorem $\begin{aligned} P T^{2} = T Q^{2} + P Q^{2} \\ ∴ T Q^{2} = 15^{2} - 12^{2} = 81 \\ ∴ T Q = 9 m \end{aligned}$ Distance between houses ⇒ SQ = ST + TQ ⇒ SQ = 12 + 9 ⇒ SQ = 21 m

34.

A tree is cut partially and made to fall on ground. The tree however does not fall completely and is still attached to its cut part. The tree top touches the ground at a point 10m from foot of the tree making an angle of 30ÃÂÃÂÃÂÃÂ°. What is the length of the tree?

(A) $10 \sqrt{3} m$
(B) $\frac{10}{\sqrt{3}} m$
(C) $\frac{(\sqrt{2} - 1)}{10} m$
(D) $\frac{10}{\sqrt{2}} m$

Solution: $\begin{aligned} in Δ M N Q, \tan 30^{\circ} = \frac{M Q}{N Q} \\ ∴ \frac{1}{\sqrt{3}} = \frac{M Q}{10} \\ ∴ M Q = \frac{10}{\sqrt{3}} \\ Also by Pythagoras theorem \\ M N^{2} = M Q^{2} + N Q^{2} \\ ∴ L^{2} = \frac{100}{3} + 100 \\ ∴ L = \frac{20}{\sqrt{3}} \\ ∴ Height of tree \\ = L + M Q \\ = \frac{20}{\sqrt{3}} + \frac{10}{\sqrt{3}} \\ = \frac{30}{\sqrt{3}} \\ = \frac{3 \times 10}{\sqrt{3}} \\ = 10 \sqrt{3} m \end{aligned}$

35.

Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.

(A) 141 m and 282 m
(B) 70.5 m and 141 m
(C) 65 m and 130 m
(D) 130 m and 260 m

Solution: Let AB and CD be the poles with heights h and 2h respectively. Given that distance between the poles, BD = 200 m Let E be the middle point of BD, ∠ AEB = $θ$ ∠ CED = (90 - $θ$ ) (∵ given that angular elevations are complementary) Since E is the middle point of BD, we have BE = ED = 100 m $\begin{aligned} From the right Δ A B E, \\ \tan θ = \frac{A B}{B E} \\ \tan θ = \frac{h}{100} \\ h = 100 \tan θ . . . . . . (1) \\ From the right Δ E D C, \\ \tan (90 - θ) = \frac{C D}{E D} \end{aligned}$ $\cot θ = \frac{2 h}{100}$ $[∵ \tan (90 - θ) = \cot θ]$ $\begin{aligned} 2 h = 100 \cot θ . . . . . . (2) \\ (1) \times (2) \end{aligned}$ $\Rightarrow 2 h^{2} = 100^{2}$ $[∵ \tan θ \times \cot θ = \tan θ \times \frac{1}{\tan θ} = 1]$ $\begin{aligned} \Rightarrow \sqrt{2} h = 100 \\ \Rightarrow h = \frac{100}{\sqrt{2}} \\ = \frac{100 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\ = 50 \sqrt{2} \\ = 50 \times 1.41 \\ = 70.5 \\ \Rightarrow 2 h = 2 \times 70.5 = 141 \end{aligned}$ i.e. the height of the poles are 70.5 m and 141 m.

36.

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

37.

The length of shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. The angle of elevation of sun is

(A) 45°
(B) 30°
(C) 60°
(D) 90°

Solution: Let AB be tower and BC be its shadow ∴ Let AB = x $\begin{aligned} Then B C = \sqrt{3} \times x = \sqrt{3} x \\ ∴ \tan θ = \frac{A B}{B C} \\ = \frac{x}{\sqrt{3} x} \\ = \frac{1}{\sqrt{3}} \\ = \tan 30^{\circ} \\ ∴ θ = 30^{\circ} \end{aligned}$ ∴ Angle of elevation of the sun $= 30^{\circ}$

38.

A tree is cut partially and made to fall on ground. The tree however does not fall completely and is still attached to its cut part. The tree top touches the ground at a point 10m from foot of the tree making an angle of 30ÃÂÃÂÃÂÃÂ°. What is the length of the tree?

(A) $10 \sqrt{3} m$
(B) $\frac{10}{\sqrt{3}} m$
(C) $\frac{(\sqrt{2} - 1)}{10} m$
(D) $\frac{10}{\sqrt{2}} m$

Solution: $\begin{aligned} in Δ M N Q, \tan 30^{\circ} = \frac{M Q}{N Q} \\ ∴ \frac{1}{\sqrt{3}} = \frac{M Q}{10} \\ ∴ M Q = \frac{10}{\sqrt{3}} \\ Also by Pythagoras theorem \\ M N^{2} = M Q^{2} + N Q^{2} \\ ∴ L^{2} = \frac{100}{3} + 100 \\ ∴ L = \frac{20}{\sqrt{3}} \\ ∴ Height of tree \\ = L + M Q \\ = \frac{20}{\sqrt{3}} + \frac{10}{\sqrt{3}} \\ = \frac{30}{\sqrt{3}} \\ = \frac{3 \times 10}{\sqrt{3}} \\ = 10 \sqrt{3} m \end{aligned}$

39.

A lower subtends an angle of 30ÃÂÃÂÃÂÃÂ° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60ÃÂÃÂÃÂÃÂ°. The height of the tower is

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

Solution: Let CD is the tower and A is a point such that the angle of elevation of C is 30° B is and their point h m high of A and angle of depression of D is 60° $\begin{aligned} The A B = h m \\ Let C D = H m and A D = x \\ Now in right Δ A B D, \\ \tan θ = \frac{A B}{A D} \\ \Rightarrow \tan 60^{\circ} = \frac{h}{x} \\ \Rightarrow \sqrt{3} = \frac{h}{x} \\ \Rightarrow x = \frac{h}{\sqrt{3}} . . . . . . . (i) \\ Similarly in right Δ A C B, \\ \tan 30^{\circ} = \frac{C D}{A D} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{H}{x} \\ \Rightarrow x = \sqrt{3} H . . . . . . . (ii) \\ From (i) and (ii) \\ \sqrt{3} H = \frac{h}{\sqrt{3}} \\ H = \frac{h}{\sqrt{3} \times \sqrt{3}} = \frac{h}{3} \\ ∴ Height of tower = \frac{h}{3} \end{aligned}$

40.

An observer 2 m tall is $10 \sqrt{3}$ ÃÂÃÂ m away from a tower. The angle of elevation from his eye to the top of the tower is 30ÃÂÃÂÃÂÃÂº. The height of the tower is:

(A) 14 m
(B) 12 m
(C) 10 m
(D) None of these

Solution: $\begin{aligned} S R = P Q = 2 m \\ P S = Q R = 10 \sqrt{3} m \\ \tan 30^{\circ} = \frac{T S}{P S} \\ \frac{1}{\sqrt{3}} = \frac{T S}{10 \sqrt{3}} \\ T S = \frac{10 \sqrt{3}}{\sqrt{3}} = 10 m \\ T R = T S + S R = 10 + 2 = 12 m \end{aligned}$

Practice MCQ Questions and Answer on Height and Distance

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