Q30 of 55 Page 1

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower.


Let AC be the tower and B be the point which is 10 m vertically above the bottom of the tower.


Let y be the height of the tower.


∴ AC = y m


BC = 10 m


Let CD = BE = x m


Thus, AB = (y – 10) m


Now, in right angled triangle ABE,


tan 30° = AB/BE


1/√3 = (y - 10)/x


⇒ x = √3(y - 10) ………………………..(1)


Also, in right angled triangle ADC,


tan 60° = AC/CD


√3 = y/x


Put the value of x in this equation:


∴√3 =


⇒ √3 × √3(y - 10) = y


⇒ 3y – 30 = y


⇒ 2y = 30


⇒ y = 15


Therefore, height of the tower = 15 m


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