Question

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

Answer 1

The height of the light house is 50(√3-1).

Answer 2

Answer:

36.6 m

Step-by-step explanation:

Refer the attached figure

AC is the height of the tower

The angles of elevation of top of light house from two boats are 30°and 45° respectively.i.e.∠ABC = 30° and ∠ADC = 45°

The distance between the two boats is 100 m. i.e. BD = 100 m

Let BC be x

So, CD =100-x

InΔABC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 30^{\circ}= \frac{AC}{BC}$

$\frac{1}{\sqrt{3}}= \frac{AC}{x}$

$\frac{1}{\sqrt{3}}x= AC$   ---1

InΔADC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ}= \frac{AC}{CD}$

$1= \frac{AC}{100-x}$

$100-x= AC$  ---2

Equating 1 and 2

$\frac{1}{\sqrt{3}}x= 100-x$

$\frac{1}{\sqrt{3}}x+x= 100$

$(\frac{1}{\sqrt{3}}+1)x= 100$

$x= \frac{100}{(\frac{1}{\sqrt{3}}+1)}$

$x= 63.397$

AC = 100-x = 100-63.397 = 36.6 m

Hence the height of the tower is 36.6 m