Question

A 1.5 mtall boy is looking at the top of a temple which is 30 meter in height from a point
at certain distance. The angle of elevation from his eye to the top of the crown of the
temple increases from 30° to 60° as he walks towards the temple. Find the distance he
walked towards the temple
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Answer 1

Question

A1.5m tall boy is looking at the top of a temple which is 30 meter in height from a point at certam distance. The angle of elevation from his eye to the top of the crown of the temple increases from 30 to 60 as he walks towards the temple. Find the distance he walked towards the temple?

Find

Find the distance he walked towards the temple=?

Given

Height of the boy =1.5m

Height of the temple =30m

$so \: (30 - 1.5)m \\ \implies 28.5 \\ \implies \frac{57}{2} m \: \ \\ : \ \bf in \: { \triangle}PAR \\ \frac{PR}{AD} = \tan( {30}^{o} ) \\\frac{57}{2AR} = \frac{1}{ \sqrt{3} } \\ AR = \frac{57}{2} \sqrt{3}$

IN ∆PRB

Solution

According to Question

$\bf \large \: PR=PQ=RQ$

Now in the ∆PRB

$\frac{PR}{BR} = \tan( {60}^{o} ) \\ \frac{57}{2BR} = \sqrt{3} \\ BR = \frac{57}{2 \sqrt{3} } \\ = \frac{19 \sqrt{3} }{2} m$

Now we know that ST=AB

$\large \ \: AR-BR = \frac{57 \sqrt{3} }{2} - \frac{19 \sqrt{3} }{2} \\ \\ \implies\frac{38 \sqrt{3} }{2} m \\ \\ = 19 \sqrt{3} m$

therefore,the distance he walked towards the temple=19√3m