The angle of elevation of the top of a tower from the two points P and Q at distances of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is √ab where a>b.
Answer:
- Let AB be the tower of height h and ∠APB be θ.
As ∠APB and ∠AQB are complementary angles, ∠AQB=90∘−θ.
The image below represents the given situation. - Now, from right-angled triangle APB, we have
- Now, from right-angled triangle , we have
- On multiplying and , we get
- Therefore, the height of the tower is .