Q49 of 68 Page 9

The angles of elevation of the top of a rock at the top and foot of a 100 m high tower, at respectively 30° and 45°. Find the height of the rock.


Given: Height of the tower = 100 m


Hence, CD = 100 m = BE


Let the height of the rock = h


Hence, AB = h


In the right ΔABD, we have




⇒ BD = h


⇒ CE = h


In the right ΔAEC, we have




⇒ h = √3(h – 100)


⇒ h = √3h - √3 × 100


⇒ 100×√3 = √3h – h


⇒ 100 × √3 = h(√3 – 1)



Multiplying and divide by the conjugate of √3 – 1





⇒ h = 50(3 + 1.732)


⇒ h = 50(4.732)


⇒ h = 236.6 m


Hence, the height of the rock = 236.6 m (approx.)


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