Two ships are approaching the lighthouse in the opposite direction. The angle of depression of the two ships from the top of the light house is ${{35}^{\circ }}$ and ${{45}^{\circ }}$ . if the distance between the two ships is 100m. find the height of the light house. $\left( \sqrt{3}=1.73 \right)$

Hint: Now first we will draw the figure of the given conditions where we get two right angle triangles. Now we are given an angle of depression from the lighthouse. We will consider the height of the Lighthouse to be h. Now take tan ratios in both the triangles of the known angles and hence we will get two equations. Now we know that the ships are 100m apart. Hence we will substitute the values obtained from the 2 equations in this condition and find the value of h.

Complete step by step answer:
Now Let the two ships be A and B and L be the light house. We know that A and B are 100m apart. Let h be the height of the light house. Now let us draw the figure representing the conditions in the problem

Now we know that $\Delta LOB$ and $\Delta LOA$ are right angle triangles such that $\angle O={{90}^{\circ }}$.
Now first let us consider $\angle OLA$. We are given that $\angle OLA={{30}^{\circ }}$
Now we know that in a right angle triangle tan is the ratio of opposite side and adjacent side.
Hence we get, $\tan \angle OLA=\dfrac{OA}{OL}$
Now substituting the values we get,
$\begin{align}
  & \Rightarrow \tan {{30}^{\circ }}=\dfrac{OA}{h} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{OA}{h} \\
 & \Rightarrow OA=\dfrac{h}{\sqrt{3}}...................\left( 1 \right) \\
\end{align}$
Now consider the triangle $\Delta LOB$
Now again we know that $\angle OLB={{45}^{\circ }}$
Now again taking the tan ratio we get
$\begin{align}
  & \Rightarrow \tan \angle OLB=\dfrac{OB}{OL} \\
 & \Rightarrow \tan {{45}^{\circ }}=\dfrac{OB}{h} \\
 & \Rightarrow 1=\dfrac{OB}{h} \\
 & \Rightarrow OB=h............\left( 2 \right) \\
\end{align}$
Now we are given that the ships are 100m apart.
Hence we get $OB+OA=100$ . Now substituting the values from equation (1) and equation (2) we get,
\[\begin{align}
  & \Rightarrow h+\dfrac{h}{\sqrt{3}}=100 \\
 & \Rightarrow h\left( 1+\dfrac{1}{\sqrt{3}} \right)=100 \\
 & \Rightarrow h\left( 1+0.58 \right)=100 \\
 & \Rightarrow 1.58h=100 \\
 & \Rightarrow h=\dfrac{100}{1.58} \\
 & \Rightarrow h=63.29 \\
 & \Rightarrow h=63.29 \\
\end{align}\]

Hence the height of the light house is 63.29m

Note: Now note that whenever we have a right angle triangle with the angle known we can always use the trigonometric ratios to find the relation between sides of the triangle. Since here in the problem we were known with conditions of adjacent and opposite sides of the triangle we used the ratio of tan. Similarly depending on the problem we can select an appropriate trigonometric ratio to find missing values.