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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Solution

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively.

The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ΔAQR,

In ΔAQS,

On multiplying equations (i) and (ii), we obtain

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

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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.

Q. The angles of elevations of the top of the tower from two points in the same straight line and at a distance of

9 m

, and

16 m

from the base of the tower are complementary. The height of the tower is

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