Question

A ship of height 24m is sighted from a lighthouse . From the top of the lighthouse , the angles of depression to the top of the mast and base of the ship are 30°&45° respectively. How far is the ship from lighthouse?

Answer 1

I think so this is the right answer

Answer 2

Answer:

56.784 m

Step-by-step explanation:

Refer the attached figure

Height of ship i.e. AB = DC =24 m

The angle of depression to the top of the mast i.e. ∠EAD = 30°

The angle of depression to the base of the ship i.e. ∠EBC = 45°

We are supposed to find the distance between the base of ship and lighthouse i.e. BC =AD

Let ED be x

In ΔAED

Using trigonometric ratio

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

$Tan30^{\circ} = \frac{ED}{AD}$

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

In ΔEBC

Using trigonometric ratio

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

$Tan45^{\circ} = \frac{EC}{BC}$

$BC=24+x$  ---2

Since BC = AD

So, equate 1 and 2

$24+x= \frac{x}{\frac{1}{\sqrt{3}}}$

$24+x= \sqrt{3}x$

$24= \sqrt{3}x-x$

$24=x (\sqrt{3}-1)$

$24=0.732050x$

$\frac{24}{0.732050}=x$

$32.784=x$

Substitute the value of x in 2

$BC=24+32.784$

$BC=56.784$

Hence the ship is 56.784 m from lighthouse.