Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A man in a boat rowing away from a light house 100m high takes 2 minutes to change the angle of elevation of the top of the light house from $60^{\circ}\: \: to\: \: 30^{\circ}$ . Find the speed of the boat in metres per minute. [Use $\sqrt{3}=1.732$ ].

Answers (1)

In $\triangle ABO,$

$\ tan 60 = \frac{Height}{Base}$

$\sqrt{3}= \frac{100}{x}$

$x\sqrt{3}=100$

$x= \frac{100}{\sqrt{3}}$ ........(1)

In $\triangle ABC,$

$\ tan 30 = \frac{100}{(x+y)}$

$\frac{1}{\sqrt{3}}= \frac{100}{(x+y)}$

$x+y=100\sqrt{3}$ ....(2)

Subtracting eqn(1) from eqn(2)

$x+y-x=100\sqrt{3}-\frac{100}{\sqrt{3}}$

$y=\frac{200}{\sqrt{3}}$

y distance is travelled in 2 minutes.

Speed of Boat = $= \frac{\left ( \frac{200}{\sqrt{3}} \right )}{2}$

$=\frac{100}{\sqrt{3}}$

$=\frac{100}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$=\frac{100\sqrt{3}}{3}$

Thus the speed of boat $=57.66 m/min$

Posted by

Safeer PP

View full answer

Similar Questions

Trending Articles/News

anam-khan

Board Exams Date Sheet 2025 Live: RBSE 10th, 12th exams from March 6; CBSE admit card expected by month-end

Latest Question