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From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression $30^{\circ}$ and $45^{\circ}$ . Find the distance between the cars. [Take $\sqrt{3}=1.732$ ]

Answers (1)

Let tower $AB=100\; m$

Car 1 is at point C.

Car 2 is at point D.

$\angle ACB=30^{\circ}$

& $\angle ADB=45^{\circ}$

To find : length CD

In $\bigtriangleup ABC, \angle ABC=90^{\circ}$

$\Rightarrow \tan 30=\frac{AB}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{BC}\; \; \; \; \; \left ( \tan 30=\frac{1}{\sqrt{3}} \right )$

$\Rightarrow BC=100\sqrt{3}\; m$

In $\bigtriangleup ABD, \angle ABD=90^{\circ}$

$\Rightarrow \tan 45 =\frac{AB}{BD}$

$\Rightarrow \tan 45 =\frac{1}{BD} \: \: \: \: \: \: \: \left ( \tan 45=1 \right )$

$\Rightarrow BD=100\: m$

$CD=BC+BD=100\sqrt{2}+100=100(\sqrt{3}+1)$

$=100(1.732+1)=100(2.732)=273.2\; m$

Distance between cars is 273.2 metres.

Posted by

Ravindra Pindel

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