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

In the given figure, let AB be the tower. Let P and C be the points on the ground 4 m and 9 m from the foot of the tower. Join B, P and C. So, BC = 9m, BP = 4m. Again join P and C with A. Then we get two right-angled triangles ABP and ABC with right angle at B. Also, APB and ACB are given to be complementary. So, APB + ACB = 90°. We have to show that the height of the tower is 6 m.
We find the value of AB from ∆ABC using tan and also using the fact that tan(90-θ) = cotθ.

In ∆ABC,

or,

We will now use the above-found value of AB in ∆APB.

From ∆APB,

or,

or,

or,

Now,

Hence, proved.

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