Let AB = h m be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30° Now from right triangle ABC, tan‌45°=
h
60
⇒1=
h
60
∴h=60m Again from right triangle ABD; tan‌30°=
h
x+60
⇒
1
√3
=
60
x+60
⇒x+60=60√3 ∴ x = 60 (1.73 – 1) = 43.8m Hence, speed of boat =