Practice MCQ Questions and Answer on Height and Distance

111.

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

112.

Find the angle of elevation of the sun when the shadow of a pole of 18 m height is $6 \sqrt{3}$ m long?

(A) 30°
(B) 60°
(C) 45°
(D) None of these

Solution: Let RQ be the pole and PQ be the shadow Given that RQ = 18 m and PQ = $6 \sqrt{3}$ m Let the angle of elevation, ∠RPQ = $θ$ $\begin{aligned} From the right Δ P Q R, \\ tan θ = \frac{R Q}{P Q} \\ = \frac{18}{6 \sqrt{3}} \\ = \frac{3}{\sqrt{3}} \\ = \sqrt{3} \\ \Rightarrow θ = \tan^{- 1} (\sqrt{3}) = 60^{\circ} \end{aligned}$

113.

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

114.

The ratio of the length of a rod and its shadow is 1 : $\sqrt{3}$ The angle of elevation of the sum is

(A) 30°
(B) 45°
(C) 60°
(D) 90°

Solution: Let AB be rod and BC be its shadow So that AB : BC = 1 : $\sqrt{3}$ Let $θ$ be the angle of elevation $\begin{aligned} ∴ \tan θ = \frac{A B}{B C} = \frac{1}{\sqrt{3}} = \tan 30^{\circ} \\ (∵ \tan 30^{\circ} = \frac{1}{\sqrt{3}}) \\ ∴ θ = 30^{\circ} \end{aligned}$ ∴ Hence angle of elevation $= 30^{\circ}$

115.

A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of 22.5ÃÂÃÂÃÂÃÂ° and 67.5ÃÂÃÂÃÂÃÂ°. What is the ratio of the height of the taller tower to the height of the shorter tower? (Given that tan 22.5ÃÂÃÂÃÂÃÂ° = $\sqrt{2} - 1$ )

(A) $1 - 2 \sqrt{2} : 1$
(B) $1 + 2 \sqrt{2} : 1$
(C) $3 + 2 \sqrt{2} : 1$
(D) $3 - 2 \sqrt{2} : 1$

Solution: Let ED be the taller tower and AB be the shorter tower. Let C be the point of observation Given that ∠ ACB = 22.5° and ∠ DCE = 67.5° Given that C is the midpoint of BD Hence, BC = CD $\begin{aligned} From the right Δ A B C, \\ \tan {22.5}^{\circ} = \frac{A B}{B C} . . . . . (e q : 1) \\ From the right Δ C D E, \\ \tan {67.5}^{\circ} = \frac{E D}{C D} . . . . . (e q : 2) \\ \frac{(e q : 2)}{(e q : 1)} \Rightarrow \frac{\tan {67.5}^{\circ}}{\tan {22.5}^{\circ}} = \frac{(\frac{E D}{C D})}{(\frac{A B}{B C})} \\ = \frac{E D}{A B} (∵ C D = B C) \\ \Rightarrow \frac{\tan (90^{\circ} - {22.5}^{\circ})}{\tan {22.5}^{\circ}} = \frac{E D}{A B} \\ \Rightarrow \frac{\cot {22.5}^{\circ}}{\tan {22.5}^{\circ}} = \frac{E D}{A B} \\ [∵ \tan (90 - θ) = \cot θ] \\ \Rightarrow \frac{(\frac{1}{\tan {22.5}^{\circ}})}{\tan {22.5}^{\circ}} = \frac{E D}{A B} \\ [∵ \cot θ = \frac{1}{\tan θ}] \\ \Rightarrow \frac{E D}{A B} = \frac{1}{{(\tan {22.5}^{\circ})}^{2}} \\ = \frac{1}{{(\sqrt{2} - 1)}^{2}} \\ = {(\frac{1}{\sqrt{2} - 1})}^{2} \\ = {[\frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)}]}^{2} \\ = {[\frac{(\sqrt{2} + 1)}{(2 - 1)}]}^{2} \\ = {[\frac{(\sqrt{2} + 1)}{1}]}^{2} \\ = {(\sqrt{2} + 1)}^{2} \\ = (2 + 2 \sqrt{2} + 1) \\ = (3 + 2 \sqrt{2}) \\ Required Ratio \\ = E D : A B \\ = (3 + 2 \sqrt{2}) : 1 \end{aligned}$

116.

A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45ÃÂÃÂÃÂÃÂ° with the man's eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30ÃÂÃÂÃÂÃÂ°. What is the approximate speed of the boat, assuming that it is running in still water?

(A) 26.28 km/hr
(B) 32.42 km/hr
(C) 24.22 km/hr
(D) 31.25 km/hr

Solution: Consider the diagram shown above. Let AB be the tower. Let C and D be the positions of the boat Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m $\begin{aligned} \tan 45^{\circ} = \frac{A B}{B C} \\ \Rightarrow 1 = \frac{A B}{100} \\ \Rightarrow A B = 100 . . . . . . (e q : 1) \end{aligned}$ $\tan 30^{\circ} = \frac{A B}{B D}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{B D}$ (∵ Substituted the value of AB from equation 1) $\Rightarrow B D = 100 \sqrt{3}$ $\begin{aligned} C D = (B D - B C) \\ = (100 \sqrt{3} - 100) \\ = 100 (\sqrt{3} - 1) \end{aligned}$ It is given that the distance CD is covered in 10 seconds. i.e., the distance $100 (\sqrt{3} - 1)$ is covered in 10 seconds. $\begin{aligned} Required speed \\ = \frac{Distance}{Time} \\ = \frac{100 (\sqrt{3} - 1)}{10} \\ = 10 (1.73 - 1) \\ = 7.3 meter/seconds \\ = 7.3 \times \frac{18}{5} km/hr \\ = 26.28 km/hr \end{aligned}$

117.

From the top of a tower, the angles of depression of two objects P and Q (situated on the ground on the same side of the tower) separated at a distance of 100 $(3 - \sqrt{3})$ ÃÂÃÂ m are 45ÃÂÃÂÃÂÃÂ° and 60 ÃÂÃÂÃÂÃÂ° respectively. The height of the tower is-

(A) 200 m
(B) 250 m
(C) 300 m
(D) None of these

Solution: $\begin{aligned} Let, O P = a \\ tan 60^{\circ} = \frac{H}{a} \\ \Rightarrow H = \sqrt{3} a \\ \Rightarrow \frac{H}{\sqrt{3}} = a . . . . . (i) \end{aligned}$ $t a n 45^{\circ} = 1$ $= \frac{H}{a + 100 (3 - \sqrt{3})}$ $\Rightarrow a + 100 (3 - \sqrt{3}) = H$ From (i) $\frac{H}{\sqrt{3}} +$ $100 (3 - \sqrt{3})$ = H $\begin{aligned} \Rightarrow H + 300 \sqrt{3} - 300 = \sqrt{3} H \\ \Rightarrow 300 \sqrt{3} - 300 = \sqrt{3} H - H \\ \Rightarrow (\sqrt{3} - 1) H = 300 (\sqrt{3} - 1) \\ \Rightarrow H = 300 m \end{aligned}$

118.

The top of a 15 metre high tower makes an angle of elevation of 60ÃÂÃÂÃÂÃÂ° with the bottom of an electric pole and angle of elevation of 30ÃÂÃÂÃÂÃÂ° with the top of the pole. What is the height of electric pole ?

(A) 5 meters
(B) 8 meters
(C) 10 meters
(D) 12 meters

Solution: Let AB be the tower and CD be the electric pole. Then, $∠ A C B = 60^{\circ},$ $∠ E D B = 30^{\circ}$ and AB = 15 m Let CD = h Then, BE = (AB - AE) = (AB - CD) = (15 - h) $\begin{aligned} \frac{A B}{A C} = \tan 60^{\circ} = \sqrt{3} \\ \Rightarrow A C = \frac{A B}{\sqrt{3}} = \frac{15}{\sqrt{3}} \\ And, \frac{B E}{D E} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \Rightarrow D E = (B E \times \sqrt{3}) \\ = \sqrt{3} (15 - h) \\ So, A C = D E \\ \Rightarrow \frac{15}{\sqrt{3}} = \sqrt{3} (15 - h) \\ \Rightarrow 3 h = (45 - 15) \\ \Rightarrow h = 10 m \end{aligned}$

119.

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

120.

A ladder leaning against a wall makes an angle $θ$ with the horizontal ground such that $\tan θ = \frac{12}{5} .$ ÃÂÃÂ If the height of the top of the ladder from the wall is 24 m, then what is the distance (in m) of the food the ladder from the wall?

(A) 18
(B) 19.5
(C) 10
(D) 7.5

Solution: $\begin{aligned} \tan θ = \frac{12}{5} \\ 12 unit \to 24 \\ 1 unit \to 2 \\ B C = 5 unit = 5 \times 2 = 10 m \end{aligned}$

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Practice MCQ Questions and Answer on Height and Distance

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