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The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

Answers (1)

According to question

Case: 1
Here BC is the tower.
Let the height of the tower be H and distance AB = a
In 
\bigtriangleup
ABC

 \tan\theta=\frac{Perpendicular}{Base}

\tan 30^{\circ}= \frac{H}{a}           \left ( \because \theta = 30^{\circ} \right )      

\frac{1}{\sqrt{3}}= \frac{H}{a}\; \cdots \left ( 1 \right )       \left ( \because \tan 30^{\circ} = \frac{1}{\sqrt{3}}\right )                    

Case:2    When height is doubled

Here ED = a
In \bigtriangleup DEF
\tan \theta = \frac{2H}{a}
\tan \theta = \frac{2}{3}   (from (1))
But \tan 60^{\circ}= \sqrt{3}         (If the angle is double)
\sqrt{3}\neq \frac{2}{\sqrt{3}}
Hence the given statement is false.                 

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infoexpert27

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