Q23 of 55 Page 1

From the top of a vertical tower, the angles of depression of two cars, in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. [Use √3 = 1.73].


Given: Distance between the cars = CD = 100 m


Let height of the tower = AB = h m


Let CB = x m


In right angled triangle ABC,


tan 60° = h/x


⇒ x = h/tan 60°


⇒ x = h/√3 ………………(1)


In right angled triangle ABD,


tan 45° = h/(100 + x)


⇒ 1 = h/(100 + x)


⇒ 100 + x = h


Put the values from equation (1) and solve further:


⇒ 100 + h/√3 = h


⇒ 100 = h – (h/√3)


⇒ 100 = h


⇒ h =


⇒ h =


⇒ h =


⇒ h = 50√3(√3 + 1)


⇒ h = 50 × (3 + √3)


⇒ h = 50 × (3 + 1.73)


⇒ h = 50 × (4.73)


⇒ h = 236.5


∴ Height of the tower = h = 236.5 m


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