Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

111.

The angle of elevation of the sun, when the length of the shadow of a tree ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 times the height of the tree, is:

(A) 30º
(B) 45º
(C) 60º
(D) 90º

Solution: Let AB be the tree and AC be its shadow. Let ∠ACB = θ $\frac{A C}{A B} = \cot θ$ $\frac{A C}{A B} = \sqrt{3}$ $\cot 30^{\circ} = \sqrt{3}$ ∴ θ = 30°

112.

A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ° respectively. What is the height of the tower?

(A) 9 m
(B) 10.40 m
(C) 15.57 m
(D) 12 m

Solution: Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation. Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30° Let DC be h $\begin{aligned} \tan 30^{\circ} = \frac{D C}{B C} \\ \frac{1}{\sqrt{3}} = \frac{h}{B C} \\ h = \frac{B C}{\sqrt{3}} . . . . . . (e q : 1) \\ \tan 60^{\circ} = \frac{A C}{B C} \\ \sqrt{3} = \frac{18 + h}{B C} \\ 18 + h = B C \times \sqrt{3} . . . . . . (e q : 2) \\ \frac{e q : 1}{e q : 2} \Rightarrow \frac{h}{18 + h} = \frac{(\frac{B C}{\sqrt{3}})}{(B C \times \sqrt{3})} \\ = \frac{1}{3} \\ \Rightarrow 3 h = 18 + h \\ \Rightarrow 2 h = 18 \\ \Rightarrow h = 9 m \end{aligned}$ i.e., the height of the tower = 9 m

113.

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30ÃÂÃÂÃÂÃÂº and 45ÃÂÃÂÃÂÃÂº respectively. If the lighthouse is 200 m high, the distance between the two ships is:

(A) 600 m
(B) 273 m
(C) 546 m
(D) 446 m

Solution: Let BD be the lighthouse and A and C be the positions of the ships. Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45° $\begin{aligned} \tan 30^{\circ} = \frac{B D}{B A} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{200}{B A} \\ \Rightarrow B A = 200 \sqrt{3} \\ \tan 45^{\circ} = \frac{B D}{B C} \\ \Rightarrow 1 = \frac{200}{B C} \\ \Rightarrow B C = 200 \end{aligned}$ Distance between the two ships $\begin{aligned} = A C = B A + B C \\ = 200 \sqrt{3} + 200 \\ = 200 (\sqrt{3} + 1) \\ = 200 (1.73 + 1) \\ = 200 \times 2.73 \\ = 546 m \end{aligned}$

114.

The respective ratio between the height of tower and the point at some distance from its foot is 57 : 19ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3. What is the angle (in degrees) of elevation of the top of the tower?

(A) 30°
(B) 45°
(C) 60°
(D) 75°

Solution: $\begin{aligned} In Δ A B C \\ \tan θ = \frac{A B}{B C} \\ \tan θ = \frac{57}{19 \sqrt{3}} \\ \tan θ = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \tan θ = \sqrt{3} \\ \tan θ = \tan 60^{\circ} \\ ∴ θ = 60^{\circ} \end{aligned}$

115.

The distance between two parallel poles is 40ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 m. The angle of depression of the top of the second pole when seen from the top of first pole is 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. What will be the height of second tower if the first pole is 100 m long

(A) 50√3 m
(B) 80 m
(C) 35√3 m
(D) 60 m

Solution: $\begin{aligned} \tan 30^{\circ} = \frac{x}{40 \sqrt{3}} \\ \frac{1}{\sqrt{3}} = \frac{x}{40 \sqrt{3}} \\ x = 40 \\ So, y = 100 - 40 = 60 \end{aligned}$

116.

A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ° respectively. What is the height of the tower?

(A) 9 m
(B) 10.40 m
(C) 15.57 m
(D) 12 m

Solution: Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation. Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30° Let DC be h $\begin{aligned} \tan 30^{\circ} = \frac{D C}{B C} \\ \frac{1}{\sqrt{3}} = \frac{h}{B C} \\ h = \frac{B C}{\sqrt{3}} . . . . . . (e q : 1) \\ \tan 60^{\circ} = \frac{A C}{B C} \\ \sqrt{3} = \frac{18 + h}{B C} \\ 18 + h = B C \times \sqrt{3} . . . . . . (e q : 2) \\ \frac{e q : 1}{e q : 2} \Rightarrow \frac{h}{18 + h} = \frac{(\frac{B C}{\sqrt{3}})}{(B C \times \sqrt{3})} \\ = \frac{1}{3} \\ \Rightarrow 3 h = 18 + h \\ \Rightarrow 2 h = 18 \\ \Rightarrow h = 9 m \end{aligned}$ i.e., the height of the tower = 9 m

117.

The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60ÃÂÃÂÃÂÃÂ° and 30ÃÂÃÂÃÂÃÂ° respectively. What is the height of the tree?

(A) 22 m
(B) 44 m
(C) 33 m
(D) None of these

Solution: Let DC be the wall, AB be the tree. Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m $\begin{aligned} \tan 30^{\circ} = \frac{D C}{B C} \\ \frac{1}{\sqrt{3}} = \frac{11}{B C} \\ B C = 11 \sqrt{3} m \\ A E = B C = 11 \sqrt{3} m . . . . . (1) \\ \tan 60^{\circ} = \frac{E D}{A E} \end{aligned}$ $\sqrt{3} = \frac{E D}{11 \sqrt{3}}$ [∵ Substituted value of AE from (1)] $\begin{aligned} E D = 11 \sqrt{3} \times \sqrt{3} \\ = 11 \times 3 \\ = 33 \\ Height of the tree \\ = A B = E C = (E D + D C) \\ = 33 + 11 \\ = 44 m \end{aligned}$

118.

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

119.

If the angle of elevation of a tower from a distance of 100 metres from its foot is 60?, the height of the tower is

(A) $100 \sqrt{3} m$
(B) $\frac{100}{\sqrt{3}} m$
(C) $50 \sqrt{3} m$
(D) $\frac{200}{\sqrt{3}} m$

Solution: Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60° Let height of the tower = h $\begin{aligned} Then in right Δ A B P \\ \tan θ = \frac{Perpendicular}{Base} = \frac{A B}{P B} \\ \Rightarrow \tan 60^{\circ} = \frac{h}{100} \Rightarrow \sqrt{3} = \frac{h}{100} \\ \Rightarrow h = 100 \sqrt{3} \\ ∴ Height of tower = 100 \sqrt{3} \end{aligned}$

120.

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 = $θ$ $\begin{aligned} From the right Δ P Q R, \\ tan θ = \frac{R Q}{P Q} \\ = \frac{18}{6 \sqrt{3}} \\ = \frac{3}{\sqrt{3}} \\ = \sqrt{3} \\ \Rightarrow θ = \tan^{- 1} (\sqrt{3}) = 60^{\circ} \end{aligned}$

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