JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Heights and Distances

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A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is

3 0^{\circ}

(Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is the level of water), the boat has reached a point B, where the angle of depression is

4 5^{\circ}

. Then the time taken (in second) by the boat from

B

to reach the base of the tower is :

(A) $10 (3 - 1)$
(B) $10 (3 + 1)$ ✓ Correct
(C)10
(D) $103$

Answer: (B)

Solution

Let the height of the tower be

h

and the boat's speed be

v

. From point A,

tan 3 0^{\circ} = \frac{h}{d _{A}}

, so

d_{A} = h 3

. From point B,

tan 4 5^{\circ} = \frac{h}{d _{B}}

, so

d_{B} = h

. The distance AB is

d_{A} - d_{B} = h 3 - h = h (3 - 1)

. Since this distance is covered in 20 seconds,

v = \frac{h ( 3 - 1 )}{20}

. The remaining distance from B to the base is

d_{B} = h

. Time taken is

t = \frac{h}{v} = \frac{h}{\frac{h ( 3 - 1 )}{20}} = \frac{20}{3 - 1} = \frac{20 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )} = \frac{20 ( 3 + 1 )}{2} = 10 (3 + 1)

seconds.

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Heights and Distances

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