Practice MCQ Questions and Answer on Height and Distance

101.

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 $= 30^{\circ}$

102.

A man is watching from the top of tower a boat speeding away from the tower. The boat makes an angle of depression of 45ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° with the manÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. What is the approximate speed of the boat, assuming that it is running in still water?

(A) 32 kmph
(B) 36 kmph
(C) 38 kmph
(D) 40 kmph

Solution: Let AB be the tower and C and D be the two positions of the boats. Then, $∠ ACB = 45^{\circ},$ $∠ A D B = 30^{\circ}$ and AC = 60 m Let, AB = h $\begin{aligned} Then, \frac{A B}{A C} = \tan 45^{\circ} = 1 \\ \Rightarrow A B = A C \\ \Rightarrow h = 60 m \\ And \frac{A B}{A D} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \Rightarrow A D = (A B \times \sqrt{3}) \\ = 60 \sqrt{3} m \\ ∴ C D = (A D - A C) \\ = 60 (\sqrt{3} - 1) m \\ Hence, required speed \\ = [\frac{60 (\sqrt{3} - 1)}{5}] m/s \\ = (12 \times 0.73) m/s \\ = (12 \times 0.73 \times \frac{18}{5}) km/hr \\ = 31 .5km/hr \approx 32km/hr \end{aligned}$

103.

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

104.

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 $= \frac{15 \sqrt{3}}{2} m$

105.

A ladder is placed along a wall such that its upper end is touching the top of the wall. The foot of the ladder is 10 ft away from the wall and the ladder is making an angle of 60ÃÂÃÂÃÂÃÂ° with the ground. When a man starts climbing on it, it slips and now ladder makes an angle of 30ÃÂÃÂÃÂÃÂ° with ground. How much did the ladder slip?

(A) 10 (√3 - 1) ft
(B) 20 (√3 - 1) ft
(C) 30 (√3 - 1) ft
(D) 18(√3 - 1) ft

Solution: $\begin{aligned} A B = 10 \tan 60^{\circ} = 10 \sqrt{3} \\ B C = 10 \sec 60^{\circ} = 20 \end{aligned}$ Since, length of ladder will remain same, hence BC = DE = 20 $\begin{aligned} In Δ A D E, D E = 20 ft \\ \Rightarrow \frac{A D}{D E} = \cos 30^{\circ} \\ \Rightarrow A D = 20 \times \frac{\sqrt{3}}{2} = 10 \sqrt{3} \\ \Rightarrow x = A D - A C \\ \Rightarrow x = 10 \sqrt{3} - 10 \\ \Rightarrow x = 10 (\sqrt{3} - 1) ft \end{aligned}$

106.

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 $\begin{aligned} ∴ E B = 4.5 m, B D = 3 m \\ Now in Δ A E B \\ \tan θ = \frac{A B}{B E} = \frac{1.5}{4.5} = \frac{1}{3} \\ and in Δ C E D \\ \tan θ = \frac{C D}{E D} \Rightarrow \frac{1}{3} = \frac{h}{4.5 + 3} \\ \Rightarrow \frac{1}{3} = \frac{h}{7.5} \\ \Rightarrow h = \frac{7.5}{3} - 2.5 m \\ ∴ Height of lamp - post = 2 .5 m \end{aligned}$

107.

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

108.

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. $\begin{aligned} Required speed \\ = \frac{Distance}{Time} \\ = \frac{100 (\sqrt{3} - 1)}{10} \\ = 10 (1.73 - 1) \\ = 7.3 meter/seconds \\ = 7.3 \times \frac{18}{5} km/hr \\ = 26.28 km/hr \end{aligned}$

109.

The angle of depression of a point situated at a distance of 70m from the base of a tower is 60ÃÂÃÂÃÂÃÂ°. The height of the tower is-

(A) $35 \sqrt{3} m$
(B) $70 \sqrt{3} m$
(C) $\frac{70 \sqrt{3}}{3} m$
(D) $70 m$

Solution: Length of the tower AB = h meter. $\begin{aligned} ∠ D A C = ∠ A C B = 60^{\circ} \\ B C = 70 meter \\ In △ ABC, \\ tan 60^{\circ} = \frac{A B}{B C} \\ \Rightarrow \sqrt{3} = \frac{h}{70} \\ \Rightarrow h = 70 \sqrt{3} meter \end{aligned}$

110.

Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°. If a car crosses these two persons in 10 seconds, what is the speed of the car?

(A) $24 \sqrt{3} km/hr$
(B) $\frac{20 \sqrt{3}}{3} km/hr$
(C) $\frac{24}{\sqrt{3}} km/hr$
(D) None of these

Solution: Let BD be the tower and A and C be the positions of the persons. Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60° $\begin{aligned} From the right Δ A B D, \\ \tan 30^{\circ} = \frac{B D}{B A} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{B A} \\ \Rightarrow B A = 50 \sqrt{3} \\ From the right Δ C B D, \\ \tan 60^{\circ} = \frac{B D}{B C} \\ \Rightarrow \sqrt{3} = \frac{50}{B C} \\ \Rightarrow B C = \frac{50}{\sqrt{3}} \\ = \frac{50 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\ = \frac{50 \sqrt{3}}{3} \end{aligned}$ Distance between the two persons $\begin{aligned} = A C = B A + B C \\ = 50 \sqrt{3} + \frac{50 \sqrt{3}}{3} \\ = \sqrt{3} (50 + \frac{50}{3}) \\ = \frac{200 \sqrt{3}}{3} m \end{aligned}$ i.e., the distance travelled by the car in 10 seconds = $\frac{200 \sqrt{3}}{3} m$ $\begin{aligned} Speed of the car = \frac{Distance}{Time} \\ = \frac{(\frac{200 \sqrt{3}}{3})}{10} = \frac{200 \sqrt{3}}{3} m/s \\ = \frac{200 \sqrt{3}}{3} \times \frac{18}{5} km/hr \\ = 24 \sqrt{3} km/hr \end{aligned}$

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Practice MCQ Questions and Answer on Height and Distance

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