SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 94 of 100

Marks: +1, -0

A man on the top of a tower, standing on the sea shore, finds that a boat coming towards him takes 10 minutes for the angle of depression to change from 30° to 60°. How soon the boat reach the seashore ?

5 minutes
7 minutes
10 minutes
15 minutes

Validate

Solution:

(1)

Let speed of boat = v metre/minute
Time taken to reach B from D = t minutes
∠ACB = 30°; ∠ADB = 60°
AB = Tower
In ΔABC,

\tan 30°= { {AB}}/{ {BC}}

⇒{1}/{√{3}}= { {AB}}/{v t+10 {v}}

⇒{AB}= {v t+10 {v}}/{√{3}}

In ΔABD,

\tan 60°= { {AB}}/{ {BD}}

⇒√{3}= { {AB}}/{v t}

⇒√{3} v t= {AB}

⇒√{3} v t= {10 v+v t}/{√{3}}

⇒ 3 t=10+t

⇒ 2 t=10

⇒t=5

minutes

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