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# The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

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Let the height of the tower be $h$ m.
we have PB = 4m and QB = 9 m
Suppose $\angle BQA = \theta$ , so$\angle APB =90- \theta$

According to question,

In triangle $\Delta ABQ$,

$\\\tan \theta = \frac{h}{9}\\ \therefore h = 9 \tan \theta$..............(i)

In triangle $\Delta ABP$,

$\\\tan (90-\theta)=\cot \theta = \frac{h}{4}\\ \therefore h = 4\cot \theta$.....................(ii)

multiply the equation (i) and  (ii), we get

$\\h^2 = 36\\ \Rightarrow h = 6 m$

Hence the height of the tower is 6 m.

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