Question

From the point on a bridge across a river, the angles of depressions of the banks on opposite sides of the river are ${30}^{\circ }$ and ${45}^{\circ }$, respectively. If the bridge is at a height of $3m$ from the banks, find the width of the river.

• A. $3(\sqrt{3}-1)$ m
• B. $3(\sqrt{3}+1)$ m
• C. $(3+\sqrt{3})$ m
• D. $(3-\sqrt{3})$ m

Answer 1

The correct option is A $3(\sqrt{3}-1)$ m

Let width of river =AB

And bridge is at height of $3m$ from banks

So, $DP=3m$

Angel of depression of banks on the opposite sides of river are $={30}^0,{45}^0$

So,$\angle QPA={30}^0$

$\angle RPB={45}^0$

We need to find AB=?

Since , PD height so it will be perpendicular at AB

$\angle PDA=\angle PDB={90}^0$

And line QR is parallel to line AB

$\angle PAD=\angle QPA={30}^0$ (Alternate angle)

Similarly,

$\angle QPB=\angle PBD={45}^0$ (Alternate angle)

Now, in triangle PAD ,

$\tan 30=\frac{PD}{AD}$

$\frac{1}{\sqrt{3}}=\frac{3}{AD}$

$AD=3\sqrt{3}$

Now in triangle PBD ,

$\tan {45}^0=\frac{PD}{DB}$

$1=\frac{3}{DB}$

$DB=3$

AB=AD+DB

$=3\sqrt{3}+3$

$=3(\sqrt{3}+1)$

Hence width of river is $3(\sqrt{3}+1)$.