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 tanθ=ABPB⟹tanθ=ha⟹h=atanθ…(i)
- Now, from right-angled triangle AQB, we have tan(90∘−θ)=ABBQ⟹cotθ=hb[tan(90∘−θ)=cotθ]⟹h=b cotθ=btanθ…(ii)
- On multiplying eq (i) and eq (ii), we get h2=abh=√ab
- Therefore, the height of the tower is √ab.