Get Answers to all your Questions

header-bg qa

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.

Answers (1)

best_answer


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. 

Posted by

manish

View full answer