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  • The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Answers (1)

Answer.       [27.322 m]      
Solution. 
                 
The angle of elevation of the top of a tower AB from certain point C is 30°
Let observer moves from C to D that is CD = 20m
Now angle of elevation increased by 15° that is 45° on point D
In \bigtriangleupABD
\tan 45^{\circ}= \frac{AB}{BD}        \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
1= \frac{AB}{BD}
BD = AB        …(1)
In  \bigtriangleupABC
\tan 30^{\circ}= \frac{AB}{BC}      \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
\frac{1}{\sqrt{3}}= \frac{AB}{BD+DC}      \begin{bmatrix} \because \tan 30= \frac{1}{\sqrt{3}} & \\ BC= BD+DC & \end{bmatrix}

\frac{1}{\sqrt{3}}= \frac{AB}{BD+20}        \left ( \because DC= 20 \right )
By cross multiplication we get
BD+20= \sqrt{3}AB
Now put the value of BD from equation (1) we have

AB+20= \sqrt{3}AB
20= \sqrt{3}AB-AB
20= AB\left ( \sqrt{3}-1 \right )
AB= \frac{20}{\sqrt{3}-1}
AB= \frac{20}{1\cdot 732-1}= \frac{20}{0\cdot 732}
AB = 27.322
Hence the height of the tower is 27.322 m

 

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