Question

From the top of a Lighthouse a man observes two ships coming towards the tower at an angle of depression of two be 45 degree and 60 degree. The ships are on either side of the lighthouse.The distance between the ship is 400 metre Find the height of the tower.​

Answer 1

Step-by-step explanation:

From the top of a Lighthouse a man observes two ships coming towards the tower at an angle of depression of two be 45 degree and 60 degree. The ships are on either side of the lighthouse.The distance between the ship is 400 metre Find the height of the tower

Answer 2

$\sf \Large Solution!!$

AC is the lighthouse.

h is the height.

B and D are the 2 ships.

Distance between B and D is 400 m.

Angle of depression of B and D is 45° and 60°, respectively.

Let CD = 400 - x

Let BC = x

In ∆ABD

tan 45° = h/x

x = h

In ∆ADC

tan 60° = $\sf \dfrac{h}{400-x}$

$\sf \sqrt{3}=\dfrac{h}{400-x}$

$\sf \sqrt{3}(400-x)=h$

$\sf 400\sqrt{3}-\sqrt{3}x=h$

$\sf 400\sqrt{3}-\sqrt{3}h=h\;\;(\because x=h)$

$\sf 400\sqrt{3}=h+\sqrt{3}h$

$\sf 400\sqrt{3}=h(1+\sqrt{3})$

$\sf h=\dfrac{400\sqrt{3}}{1+\sqrt{3}}\times \dfrac{1-\sqrt{3}}{1-\sqrt{3}}$

$\sf h=\dfrac{400\sqrt{3}-1200}{1-3}$

$\sf h=\dfrac{-400(3-\sqrt{3})}{-2}$

$\sf h=200(3-\sqrt{3})$