Q7. 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

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

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

∴ the height of the tower is 14.64 metres.