Practice MCQ Questions and Answer on Height and Distance

91.

If the altitude of the sun is at 60ÃÂÃÂÃÂÃÂ°, then the height of the vertical tower that will cast a shadow of length 30 m is

(A) $30 \sqrt{3} m$
(B) $15 m$
(C) $\frac{30}{\sqrt{3}} m$
(D) $15 \sqrt{2} m$

Solution: Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60° $\begin{aligned} Let height of tower A B = h \\ Then in right Δ A P B, \\ \tan θ = \frac{Perpendicular}{Base} = \frac{A B}{P B} \\ \Rightarrow \tan 60^{\circ} = \frac{h}{30} \\ \Rightarrow \sqrt{3} = \frac{h}{30} \\ \Rightarrow h = 30 \sqrt{3} \end{aligned}$ ∴ Height of the tower $= 30 \sqrt{3} m$

92.

From the top and bottom of a straight hill; the angle of depression and elevation of the top of a pillar of 10 m height are observed to be 60ÃÂÃÂÃÂÃÂ° and 30ÃÂÃÂÃÂÃÂ° respectively. The height (in metres) of the hill is

(A) 30
(B) 80
(C) 60
(D) 40

93.

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30ÃÂÃÂÃÂÃÂº with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60ÃÂÃÂÃÂÃÂº. What is the distance between the base of the tower and the point P?

(A) 4 √3 units
(B) 8 units
(C) 12 units
(D) Data inadequate

94.

There is a tower of 10m between two parallel roads. The angles of depression of the roads from the top of the tower are 30ÃÂÃÂÃÂÃÂ° and 45ÃÂÃÂÃÂÃÂ°. How far are the roads from each other?

(A) 27.32 m
(B) 29.56 m
(C) $20 \sqrt{3} m$
(D) $\frac{10}{\sqrt{3}} m$

Solution: Angle of Depression = Angle of Elevation Tower PS = 10 m in height $\begin{aligned} tan 45^{\circ} = 1 = \frac{P S}{R S} \\ ∴ P S = R S = 10 \\ \tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{P S}{S Q} = \frac{10}{S Q} \\ ∴ S Q = 10 \sqrt{3} \\ R Q = R S + S Q \\ = 10 + 10 \sqrt{3} \\ = 27.32 m \end{aligned}$

95.

From the top of hill 240 m high, the angles the angles of depression of the top and bottom of a pole are 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:

(A) 120(2 - √3)
(B) 120(√3 - 1)
(C) 80(√3 - 1)
(D) 80(2 - √3)

Solution: $\begin{aligned} 3 \to 240 \\ \sqrt{3} \to \frac{240}{3} \times \sqrt{3} = 80 \sqrt{3} \\ Height of the pole \\ = 160 - 80 \sqrt{3} \\ = 80 (2 - \sqrt{3}) \end{aligned}$

96.

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)] $\begin{aligned} = \sqrt{15^{2} \times 9^{2} - 15^{2}} \\ = \sqrt{15^{2} (9^{2} - 1)} \\ = \sqrt{15^{2} \times 80} \\ = \sqrt{15^{2} \times 16 \times 5} \\ = 15 \times 4 \times \sqrt{5} \\ = 60 \sqrt{5} m \end{aligned}$

97.

As observed from the top a light house, 120ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 m above the sea level, the angle of depression of a ship sailing towards it changes from 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° to 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. The distance travelled by the during the period of observation is:

(A) 240 m
(B) 240√3 m
(C) 180√3 m
(D) 180 m

Solution: $\begin{aligned} \sqrt{3} \to 120 \sqrt{3} \\ 1 u \to 120 \\ 2 u \to 120 \times 2 = 240 m \end{aligned}$

98.

The angle of elevation of a ladder leaning against a wall is 60ÃÂÃÂÃÂÃÂº and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

(A) 2.3 m
(B) 4.6 m
(C) 7.8 m
(D) 9.2 m

Solution: Let AB be the wall and BC be the ladder. Then, ∠ACB = 60° = AC = 4.6m $\frac{A C}{B C} = \cos 60^{\circ} = \frac{1}{2}$ ⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

99.

Mohan looks at a tree top and the angle made is 45ÃÂÃÂÃÂÃÂ°. He moves 10 cm back and again looks at the tree top but this time angle made is 30ÃÂÃÂÃÂÃÂ°. How high is the tree top from ground?

(A) $\frac{10}{\sqrt{3} + 1} cm$
(B) $20 \sqrt{3} cm$
(C) $20 cm$
(D) $\frac{10}{\sqrt{3} - 1} cm$

Solution: Let PQ be three and M and N be positions where Mphan stands. $\begin{aligned} Now, tan 45^{\circ} = 1 = \frac{P Q}{M Q} \\ ∴ P Q = M Q \\ \tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{P Q}{N Q} \\ = \frac{P Q}{10 + M Q} \\ ∴ 10 + M Q = \sqrt{3} P Q \end{aligned}$ $∴ 10 + P Q = \sqrt{3} P Q$ (As, PQ = MQ) $∴ P Q = \frac{10}{\sqrt{3} - 1} cm$

100.

P and Q are two points on the ground on either side of a pole. The angles of elevation of the top of the pole as observed from P and Q are 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, respectively and the distance between them is 84ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ3 m. What is the height (in m) of the pole?

(A) 60
(B) 63
(C) 52.5
(D) 73.5

Solution: $60^{\circ} = \frac{\sqrt{3}}{1} \begin{matrix} \to & h \\ \to & \frac{h}{\sqrt{3}} \end{matrix}$ $P N = \frac{h}{\sqrt{3}}$ $30^{\circ} = \frac{1}{\sqrt{3}} \begin{matrix} \to & h \\ \to & \sqrt{3} h \end{matrix}$ $\begin{aligned} N Q = \sqrt{3} h \\ P N + N Q = 84 \sqrt{3} \\ \frac{h}{\sqrt{3}} + \sqrt{3} h = 84 \sqrt{3} \\ h (1 + 3) = 84 \times \sqrt{3} \times \sqrt{3} \\ h = \frac{84 \times 3}{4} \\ h = 63 m \end{aligned}$

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Practice MCQ Questions and Answer on Height and Distance

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