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. The height of the tower is

Given : CD is the tower, BD = 4 m and AD = 4 + 5 = 9 m

To find : CD = $$h$$ = ?

Solution : $$\angle$$ DBC and $$\angle$$ DAC are complementary

=> $$\angle$$ DAC = $$\theta$$ and $$\angle$$ DBC = $$(90^\circ-\theta)$$

In $$\triangle$$ BCD,

=> $$tan(90^\circ-\theta)=\frac{CD}{DB}$$

=> $$cot(\theta)=\frac{h}{4}$$ -----------(i)

Similarly, in $$\triangle$$ ACD,

=> $$tan(\theta)=\frac{CD}{DA}$$

=> $$tan(\theta)=\frac{h}{9}$$ -----------(ii)

Multiplying equations (i) and (ii), we get :

=> $$tan(\theta)cot(\theta)=\frac{h}{4} \times \frac{h}{9}$$

=> $$1=\frac{h^2}{36}$$

=> $$h^2=36$$

=> $$h=\sqrt{36}=6$$ m

=> Ans - (D)