Q9. The angle of elevation of the top of the building from the foot of the tower is 30º and the angle of elevation of the top of the tower from the foot of the building is 60º. If the tower is 50 m high, find the height of the building.

Let AB be the building of height h and AC, the horizontal ground through C, the foot of the building. Since the building subtends an angle of 60º at C, hence $$∠ \ ACB \ = \ 30º$$. Let CD be the tower of height 50 m such that $$∠ \ CAB \ = \ 60º$$.

$$∆ \ BAC$$ and $$DCA$$ , we have

$$\frac{AC}{AB} \ = \ cot30º \ => \ \frac{AC}{h} \ = \ \sqrt{3}$$ $$=> \ AC \ = \ \sqrt{3}$$      (1)

and $$\frac{DC}{AC} \ = \ tan60º \ => \ \frac{50}{AC} \ = \ \sqrt{3}$$ $$=> \ AC \ = \ \frac{50}{ \sqrt{3}}$$      (2)

Equating the values of AC from (1) and (2), we get

$$\sqrt{3}h \ = \ 50 \sqrt{3} \ => \ h \ = \ \frac{50}{ \sqrt{3}} × \frac{1}{ \sqrt{3}}$$

$$= \ \frac{50}{3} \ = \ 16.66$$

∴ the height of the building is 16.66 m.