Question

From the top of a 300 M high Lighthouse the angles of depression of two ships which are due south of the observer and in a straight line with it base are 60 and 30. find the distance apart

Answer 1

The distance of ships is 200√3 meters.

Given:

• From the top of a 300 M high Lighthouse.
• The angles of depression of two ships which are due south of the observer and in a straight line with it base are 60° and 30°.

To find:

• Find the distance apart.

Solution:

Concept to be used:

• Draw the suitation and apply trigonometric ratio.
• $tan \: 60^{\circ} = \sqrt{3}$
• $tan \: 30^{\circ} = \frac{1}{\sqrt{3}}$

Step 1:

Draw the situation as shown in attachment.

here,

AB is light house.

One ship is at point S and other at point T.

Distance of ship S is say x meters, from base of light house and that of ship T is y meters.

Step 2:

Apply trigonometric ratio.

In right ∆ABS

$tan \: 60^{\circ} = \frac{AB}{AS}$

or

$\sqrt{3} = \frac{300}{x}$

or

$x = \frac{300}{ \sqrt{3} } \: meters$

or

$\bf x = 100 \sqrt{3} \: meters$

Step 3:

Apply trigonometric ratio in ∆ABT.

$tan \: 30^{\circ} = \frac{AB}{AT}$

or

$\frac{1}{ \sqrt{3} } = \frac{300}{y}$

or

$\bf y = 300 \sqrt{3} \: meters$

Step 4:

Distance between ships is given by y-x.

$= 300 \sqrt{3} - 100 \sqrt{3}$

or

Distance between ships$\bf= 200 \sqrt{3} \: meters$

Thus,

Both ships are 200√3 meters apart.

_________________________

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