ABC is triangular park with AB = AC = 100 m. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower atAandBare cot−13.2 and cosec−12.6respectively. The height of the tower is[EAMCET 1992]

DP is a clock tower standing at the middle point D of BC. ∠PAD=α=cot−13.2⇒cotα=3.2 and ∠PBD=β=cosec−12.6⇒cosecβ=2.4 ∴ cotβ=(cosec2β−1)−−−−−−−−−−−√=(5.76)−−−−−√=2.4 In the triangles PAD and PBD, AD=hcotα=3.2hand BD=hcotβ=2.4h In the right angled ΔABD, AB2=AD2+BD2 ⇒ 1002=[(3.2)2+(2.4)2]h2=16h2 ⇒h=25m.