3 Tutor System
Starting just at 265/hour

# From a point on the ground the angles of elevation of the bottom and top of a tower fixed at the top of a 20 m high building are 45º and 60º respectively. Find the height of the tower.

Let BC be the building of height 20 m and CD be the tower of height x metres. Let A be a point on the ground at a distance of y metres away from the foot B of the building.

In $$∆ \ ABC$$, we have

$$\frac{BC}{AB} \ = \ tan45º \$$
$$=> \ \frac{20}{y} \ = \ 1$$
$$=> \ y \ = \ 20$$

In $$∆ \ ABD$$, we have

$$\frac{BD}{AB} \ = \ tan60º \$$
$$=> \ \frac{20+x}{20} \ = \ \sqrt{3}$$

$$=> 20+x \ = \ 20 \sqrt{3} \$$
$$=> \ x \ = \ 20( \sqrt{3} -1)$$ $$=> x \ = \ 20(1.732 - 1) \$$
$$=> \ x \ = \ 14.64$$

$$\therefore$$ the height of the tower is 14.64 metres.