Question

Two boats approach a light house in mid sea from opposite directions.The angle of elevation of the top of the light house from two boats are 30 degree and 45 degree respectively.If the distance between two boat is 100m.Find the height of light house?

Answer 1

SOLUTION:-

Given:

Two boats approach a light house in mid-sea from opposite directions. The angle of elevation of the top of the light house from two boats are 30° & 45° respectively.

The distance between two boat is 100m.

To find:

The Height of light house.

Explanation:

•Let the boats be at position R & M 100m apart.

•The height of light house is h m &

The angle of elevation of the top of light house the two boats are 30° & 45°.

In ∆ADC,

$tan30 \degree = \frac{CD}{AD} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{h}{x} \\ \\ = > x = \sqrt{3} h.................(1)$

&

In ∆BDC,

$tan45 \degree = \frac{CD}{BD} \\ \\ = > 1 = \frac{h}{100 - x} \\ \\ = > h = 100 - x.................(2)$

Putting the value of x in equation (2), we get;

=) h= 100 - √3h

=) (√3+1)h =100

$= > \frac{100}{ \sqrt{3} + 1 } \\ [Rationalise] \\ \\ = > \frac{100 \times \sqrt{3} - 1 }{ \sqrt{3} + 1 \times \sqrt{3} - 1 } \\ \\ = > \frac{100( \sqrt{3} - 1)}{( { \sqrt{3} )}^{2} - (1) {}^{2} } \\ \\ = > \frac{100( \sqrt{3} - 1) }{3 - 1} \\ \\ = > \frac{100( \sqrt{3} - 1) }{2} \\ \\ = > 50( \sqrt{3} - 1)m$

Thus,

The light of the house is 50(√3-1)m.

Thank you.