The angle of elevation of the top of a tower from the two points PP and QQ at distances of aa and bb respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is √ab√ab where a>ba>b.
Answer:
- Let ABAB be the tower of height hh and ∠APB∠APB be θ.θ.
As ∠APB∠APB and ∠AQB∠AQB are complementary angles, ∠AQB=90∘−θ.∠AQB=90∘−θ.
The image below represents the given situation. - Now, from right-angled triangle APBAPB, we have [Math Processing Error]
- Now, from right-angled triangle AQB, we have [Math Processing Error]
- On multiplying eq (i) and eq (ii), we get [Math Processing Error]
- Therefore, the height of the tower is √ab.