Question

Two boats approach a lighthouse in mid-sea from opposite directions. The angles of elevation of the top of the lighthouse from the two boats are ${30}^0$ and ${60}^0$ respectively. If the distance between two boats is $100$ m, then find the height of the lighthouse.

Answer 1

$\Rightarrow$ Here, $CD$ is the height of the lighthouse of height $h$ and $A$ and $B$ are the two boats $100m$ far from each other.

$\Rightarrow$ let say $x$ be the distance between $A$ and lighthouse, then $100-x$ would be the distance between $B$ and the lighthouse.

$\Rightarrow$ In $△ACD,\tan {30}^o=\frac{CD}{AD}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{h}{x}$

$\Rightarrow$ $x=\sqrt{3}hm$ ----- ( 1 )

$\Rightarrow$ In $△BCD,\tan {60}^o=\frac{CD}{BD}$

$\Rightarrow$ $\sqrt{3}=\frac{h}{100-x}$

$\Rightarrow$ $h=\sqrt{3}(100-x)$

$\Rightarrow$ $h=\sqrt{3}(100-\sqrt{3}h)$ { from (1) }

$\Rightarrow$ $h+3h=100\sqrt{3}$

$\Rightarrow$ $4h=100\sqrt{3}$

$\therefore$ $h=25\sqrt{3}m$

$\therefore$ $h=43.30m$ $(Take\sqrt{3}=1.73)$