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  • The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is sqroot st

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is\sqrt{st}

Answers (1)

Solution.         According to question
                        
Let the height of tower = h
the distance of the first point from its foot = s
the distance of the second point from its foot = t
\tan \theta = \frac{h}{s}\cdots \left ( 1 \right )   \left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )
\tan \left ( 90-\theta \right )= \frac{h}{t}
\cot \theta = \frac{h}{t}\cdots \left ( 2 \right )
Multiply equation (1) and (2) we get
\tan \theta \times \cot \theta= \frac{h}{t}\times \frac{h}{s}
\tan \theta \frac{1}{ \tan \theta }= \frac{h^{2}}{st}          \left ( \because \cot \theta = \frac{1}{\tan \theta } \right )
1= \frac{h^{2}}{st}
st= h^{2}
h= \sqrt{st}
Hence proved.

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