Two ships are sailing in the sea on the two sides of a light house. The angle of elevation of the top of the light house as observed from the two ships are 30° and 45° respectively. If the light house is 100m high, the distance between the two ships is :(take $$\sqrt3=1.73$$)

Given : AD is the lighthouse = 100 m

To find : Distance between the ships = BC = ?

Solution : In $$\triangle$$ ADC

=> $$tan(30^\circ)=\frac{AD}{DC}$$

=> $$\frac{1}{\sqrt{3}}=\frac{100}{DC}$$

=> $$DC=100{\sqrt{3}}$$

=> $$DC=100 \times 1.73=173$$ m

Similarly, in $$\triangle$$ ABD

=> $$tan(45^\circ)=\frac{AD}{DB}$$

=> $$1=\frac{100}{DB}$$

=> $$DB=100$$ m

$$\therefore$$ BC = BD + DC

= $$100+173=273$$ m

=> Ans - (C)