Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from two ships are 30° and 45°, respectively. If the lighthouse is 100 m high, the distance between two ships is approximately:

Let AB be the lighthouse

Also, let C and D be the points where the ships were present such that angle of elevation of the top of the lighthouse are 30° and 45°respectively.

So, $$\angle\ ACD=30^{\circ\ }$$ and $$\angle\ ADC=45^{\circ\ }$$

Now, $$AB=100$$ m (given)

Triangle ABC is a right angled triangle, so $$BC=\dfrac{AB}{\tan30^{\circ\ }}=\dfrac{100}{\dfrac{1}{\sqrt{\ 3}}}=100\sqrt{\ 3}$$ m

Triangle ABD is a right angled triangle, so $$BD=\dfrac{AB}{\tan45^{\circ\ }}=\dfrac{100}{1}=100$$ m

So, distance between two ships = CD = $$100\sqrt{\ 3}+100$$ m = $$100\left(\sqrt{\ 3}+1\right)=100\left(1.732+1\right)=100\left(2.732\right)=273.2$$ m

So, option C is the correct approximate answer.