Question

If the angle of depression from a lighthouse of two ships situated in the same line with the lighthouse are ${60}^{\circ }\text{and}{30}^{\circ }$ and if the nearer ship is $150m$ away from the lighthouse, then find the distance of the other ship from the lighthouse. $\left[4\text{Marks}\right]$

Answer 1

Let $AB$ be the lighthouse and two ships are situated at $C\text{and}D$. We have to find the distance $BD$. $\left[2\text{Marks}\right]$

From the figure, $\tan \left(\angle ACB\right)=\tan {60}^{\circ }=\frac{AB}{BC}\text{and}\tan \left(\angle ADB\right)=\tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow AB=BC\tan {60}^{\circ }=150\sqrt{3}\text{and}AB=BD\tan {30}^{\circ }=\frac{BD}{\sqrt{3}}$

$\Rightarrow \frac{BD}{\sqrt{3}}=150\sqrt{3}$

$\Rightarrow BD=150\times 3=450m$

The distance of the 2nd ship from the lighthouse is $450m$. $\left[2\text{Marks}\right]$