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

Answer:

Distance Between two ships is 200$\sqrt{3}$ m.  

Step-by-step explanation:

Given: Height of light house = 300 m And Angle of depression of 2

           ships are 60° & 30°.

To find: Distance between the ships.

Figure is attached.

Angle of Depression is equal to angle of elevation because of alternate internior angles between parallel lines.

Height of Light house, AB = 300 m

let, Distance between both ships be CD

Using Angle elevation

In Δ ABC,

by trignometric ratio,

$tan\, 60^{\circ}=\frac{AB}{BC}$

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

⇒ $BC=\frac{300}{\sqrt{3}}$

In Δ ABD,

by trignometric ratio,

$tan \,30^{\circ}=\frac{AB}{BD}$

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

⇒ $BD=300\times\sqrt{3}$

from figure,

CD = BD - BC

      = $300\times\sqrt{3}-\frac{300}{\sqrt{3}}$                  

      = $300\times(\sqrt{3}-\frac{1}{\sqrt{3}})$                  

      = $300\times(\frac{\sqrt{3}\times\sqrt{3}-1}{\sqrt{3}})$

      = $300\times(\frac{3-1}{\sqrt{3}})$

      = $300\times(\frac{2}{\sqrt{3}})$

      = $\frac{600}{\sqrt{3}}$

      = $200\sqrt{3}$  (∵ 600 = 200 × 3 & $3=\sqrt{3}\times\sqrt{3}$)

Therefore, Distance Between two ships is 200$\sqrt{3}$ m.

Answer 2

"From the triangle, ADC

tan30 = DA/AC = 300 / AC

$AC = \frac {300} {tan30}$

    $= \frac {300} \sqrt 3$

$= 100 \sqrt3$

From the triangle, ADB

$tan60 = \frac {DA} {AB} = \frac {300} {AB}$

$AB = \frac {300} {tan60}$

   $= 300 \sqrt 3$

AC = AB + BC

$100 \sqrt3 =300 \sqrt3 + BC$

$BC = 300 \sqrt 3 - 100 \sqrt 3 = 200 \sqrt 3$"