MCQ 11 Mark
From a light house the angles of depression of two ships on opposite sides of the light house are observed to be $30^\circ$ and $45^\circ$ . If the height of the light house is $h$ metres, the distance between the ships is :
- ✓$(\sqrt{3}+1)\text{h meters}$
- B$(\sqrt{3}-1)\text{h meters}$
- C$\sqrt{3}\text{h meters}$
- D$1+\Big(1+\frac{1}{\sqrt{3}}\Big)\text{h meters}$
Answer
View full question & answer→Correct option: A.
$(\sqrt{3}+1)\text{h meters}$
Let $AB$ be light house and $P$ and $Q$ are two ships on its opposite sides which form angle of elevation of $A$ as $45^\circ $ and $30^\circ $ respectively $AB = h.$
Let $PB = x$ and $QB = y$
Now in right $\triangle\text{APB,}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{AB}}{\text{PB}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}\ .....(\text{i})$
Similarly in right $\triangle\text{AQB,}$
$\tan30^\circ=\frac{\text{AB}}{\text{QB}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \text{y}=\sqrt{3}\text{h}\ ......(\text{ii})$
Adding $(i)$ and $(ii)$
$\text{PQ}=\text{x}+\text{y}=\text{h}+\sqrt{3}\text{h}$
$(\sqrt{3}+1)\text{h meters}$
Let $PB = x$ and $QB = y$
Now in right $\triangle\text{APB,}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{AB}}{\text{PB}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}\ .....(\text{i})$
Similarly in right $\triangle\text{AQB,}$
$\tan30^\circ=\frac{\text{AB}}{\text{QB}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ \text{y}=\sqrt{3}\text{h}\ ......(\text{ii})$
Adding $(i)$ and $(ii)$
$\text{PQ}=\text{x}+\text{y}=\text{h}+\sqrt{3}\text{h}$
$(\sqrt{3}+1)\text{h meters}$
