Practice MCQ Questions and Answer on Height and Distance

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 $\begin{aligned} \Rightarrow H + 300 \sqrt{3} - 300 = \sqrt{3} H \\ \Rightarrow 300 \sqrt{3} - 300 = \sqrt{3} H - H \\ \Rightarrow (\sqrt{3} - 1) H = 300 (\sqrt{3} - 1) \\ \Rightarrow H = 300 m \end{aligned}$

Due to sun, a 6ft man casts a shadow of 4ft, whereas a pole next to the man casts a shadow of 36ft. What is the height of the pole?

(A) 63 ft
(B) 72 ft
(C) 54 ft
(D) 48 ft

Solution: Both the man and pole are near each other and are illuminated by same sun from same direction. So angle of elevation for sun is same for both. So ration of object to shadow will be same for all objects. (Proportionality Rule) $\frac{Object height}{Shadow length} = \frac{6}{4} = \frac{H}{36}$ ∴ H = 54 ft = Height of pole

If the angles of elevation of a tower from two points distance a and b (a > b) from its foot and in the same straight line from it are 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°, then the height of the tower is?

(A) $\sqrt{a + b}$
(B) $\sqrt{a b}$
(C) $\sqrt{a - b}$
(D) $\sqrt{\frac{a}{b}}$

Solution: Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively $\begin{aligned} Let A B = h \\ Now in right Δ A P B, \\ \tan θ = \frac{Perpendicular}{Base} = \frac{A B}{P B} \\ \Rightarrow \tan 30^{\circ} = \frac{h}{a} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{a} . . . . . . . . . . . (i) \\ Similarly in right Δ A Q B, \\ \tan 60^{\circ} = \frac{A B}{Q B} \\ \Rightarrow \sqrt{3} = \frac{h}{b} . . . . . . . . . . . (i i) \\ Multiplying (i) and (ii) \\ \frac{1}{\sqrt{3}} \times \sqrt{3} = \frac{h}{a} \times \frac{h}{b} \\ \Rightarrow 1 = \frac{h^{2}}{a b} \\ \Rightarrow h^{2} = a b \\ \Rightarrow h = \sqrt{a b} \\ ∴ Height of the tower = \sqrt{a b} \end{aligned}$

A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole?

(A) $60 \sqrt{5} m$
(B) $15 \sqrt{5} m$
(C) $15 \sqrt{3} m$
(D) $60 \sqrt{3} m$

Solution: Let CB be the pole and point D divides it such that BD : DC = 1 : 9 Given that AB = 15 m Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $θ$ From "Angle Bisector Theorem", we have $\frac{B D}{D C} = \frac{A B}{A C}$ $\Rightarrow \frac{1}{9} = \frac{15}{A C}$ [∵ BD : DC = 1 : 9 and AB = 15(given)] $\Rightarrow A C = 15 \times 9 m . . . . . . (e q : 1)$ $From the right Δ A B C,$ $C B = \sqrt{A C^{2} - A B^{2}}$ (∵ Pythagorean theorem) $= \sqrt{{(15 \times 9)}^{2} - 15^{2}}$ [∵AC=15 × 9(eq : 1) and AB=15 m(given)] $\begin{aligned} = \sqrt{15^{2} \times 9^{2} - 15^{2}} \\ = \sqrt{15^{2} (9^{2} - 1)} \\ = \sqrt{15^{2} \times 80} \\ = \sqrt{15^{2} \times 16 \times 5} \\ = 15 \times 4 \times \sqrt{5} \\ = 60 \sqrt{5} m \end{aligned}$

If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is

(A) 1.5 m
(B) 2 m
(C) 2.5 m
(D) 2.8 m

Solution: Let AB is girls and CD is lamp-post AB = 1.5 which casts her shadow EB $\begin{aligned} ∴ E B = 4.5 m, B D = 3 m \\ Now in Δ A E B \\ \tan θ = \frac{A B}{B E} = \frac{1.5}{4.5} = \frac{1}{3} \\ and in Δ C E D \\ \tan θ = \frac{C D}{E D} \Rightarrow \frac{1}{3} = \frac{h}{4.5 + 3} \\ \Rightarrow \frac{1}{3} = \frac{h}{7.5} \\ \Rightarrow h = \frac{7.5}{3} - 2.5 m \\ ∴ Height of lamp - post = 2 .5 m \end{aligned}$

If the angle of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is

(A) $a b$
(B) $\sqrt{a b}$
(C) $\frac{a}{b}$
(D) $\sqrt{\frac{a}{b}}$

Solution: Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $θ$ and (90° – $θ$ ) Let height of tower = h $\begin{aligned} Then in right Δ A P B, \\ \tan θ = \frac{Perpendicular}{Base} = \frac{A B}{P B} \\ = \frac{h}{a} . . . . . . . . . . . . (i) \\ Similarly in right Δ A Q B, \\ \tan (90^{\circ} - θ) = \frac{A B}{Q B} = \frac{h}{b} \\ \Rightarrow \cot θ = \frac{h}{b} . . . . . . . . . . . (ii) \\ Multiplying (i) and (ii) \\ \tan θ \cot θ = \frac{h}{a} \times \frac{h}{b} \\ \Rightarrow 1 = \frac{h^{2}}{a b} \\ \Rightarrow h^{2} = a b \\ \Rightarrow h = \sqrt{a b} \\ ∴ Height of tower = \sqrt{a b} \end{aligned}$

The angle of elevation of a cloud from a point 200 m above a lake is 30ÃÂÃÂÃÂÃÂ° and the angle of depression of its reflection in the lake is 60ÃÂÃÂÃÂÃÂ°. The height of the cloud is-

(A) 200 m
(B) 300 m
(C) 400 m
(D) 600 m

Solution: $\begin{aligned} Let, A C = H m \\ O A = a m \\ \tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{H}{a} \end{aligned}$ $\tan 60^{\circ} = \sqrt{3}$ $= \frac{H + 200 + 200}{a}$ $\begin{aligned} \Rightarrow \frac{H + 400}{\sqrt{3} H} = \sqrt{3} \\ \Rightarrow H + 400 = 3 H \\ \Rightarrow 2 H = 400 m \\ \Rightarrow H = 200 m \end{aligned}$ Height of cloud above lake = 200 m

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

A and B are standing on ground 50 meters apart. The angles of elevation for these two to the top of a tree are 60ÃÂÃÂÃÂÃÂ° and 30ÃÂÃÂÃÂÃÂ°. What is height of the tree?

(A) $50 \sqrt{3} m$
(B) $\frac{25}{\sqrt{3}} m$
(C) $25 \sqrt{3} m$
(D) $\frac{25}{\sqrt{3} - 1} m$

Solution: $\begin{aligned} In Δ P B Q, \tan 60^{\circ} = \frac{P Q}{B Q} \\ ∴ B Q = \frac{P Q}{\sqrt{3}} \\ In Δ P A Q, \tan 30^{\circ} = \frac{P Q}{A Q} \\ ∴ \frac{1}{\sqrt{3}} = \frac{P Q}{50 + B Q} \\ ∴ P Q = \frac{50 + B Q}{\sqrt{3}} \\ = \frac{50 + \frac{P Q}{\sqrt{3}}}{\sqrt{3}} \\ = \frac{50 \sqrt{3} + P Q}{\sqrt{3} \times \sqrt{3}} \\ ∴ 3 P Q = 50 \sqrt{3} + P Q \end{aligned}$ $∴ P Q = 25 \sqrt{3} m =$ Height of tree

10.

A balloon leaves from a point P rises at a uniform speed. After 6 minutes, an observer situated at a distance of 450ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 meter from point P observes that angle of elevation of the balloon is 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. Assume that point of observation and point P are on the same level. What is the speed (in m/s) of the balloon?

(A) 4.25
(B) 3.75
(C) 4.5
(D) 3.45

Solution: $\begin{aligned} \tan 60^{\circ} = \frac{A P}{P O} \\ \sqrt{3} = \frac{A P}{450 \sqrt{3}} \\ A P = 450 \sqrt{3} \times \sqrt{3} \\ A P = 1350 m \\ Speed = \frac{Distance}{Time} \\ = \frac{1350 m}{6 \times 60 \sec} \\ = \frac{15}{4} \\ = 3.75 m/s \end{aligned}$

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Practice MCQ Questions and Answer on Height and Distance

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