Question

The angle of elevation and depression of the top and bottom of a tower AB from the top of a building CD of height 51 m are 30 and 45 degrees respectively. Find the height of the tower

Answer 1

SOLUTION:-

Given:

The angle of elevation & depression of the top & bottom of a tower AB from the top of a building CD of height 51m are 30° & 45° respectively.

To find:

The height of the tower.

Explanation:

•Let the height of the building be CD=51m

•Let the height of the tower be AB= h m

In ∆CDB,

$tan45 \degree = \frac{CD}{BD} \\ \\ = > 1 = \frac{51}{BD} \\ \\ = > BD = 51m$

&

In ∆AEC,

$tan30 \degree = \frac{AE}{EC} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{AE}{51} \\ \\ = > \sqrt{3} AE = 51 \\ \\ = > AE = \frac{51}{ \sqrt{3} } \\ [rationalise] \\ = > \frac{51 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ = > \frac{51 \sqrt{3} }{3} \\ \\ = > 17 \sqrt{3} \\ \\ = > 17 \times 1.732 \: \: \: \: \: \: ( \sqrt{3} = 1.732) \\ \\ = >AE = 29.44m$

Thus,

The height of a tower is 29.44m.

Thank you.