Question

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are ${30}^{\circ }$ and ${45}^{\circ }$, respectively. If the width of the river is 12 m , find the height at which the bridge is

• A. 6(√3 -1) m
• B. 6 m
• C. 6√3 m
• D. 3 m

Answer 1

The correct option is A

6(√3 -1) m

The situation can be represented by the figure below

$\angle PAD=\angle ADCand\angle QAB=\angle ABC\left(alternateangles\right)$
$tan\angle ADC=\frac{AC}{DC}=tan{45}^{\circ }$
$\Rightarrow BD=12cmLetthelengthofBCbex,thenCD=12-x$

$\Rightarrow tan{45}^{\circ }=\frac{AC}{12-x}$
$\Rightarrow AC=tan{45}^{\circ }(12-x)$
$Alsotan\angle ABC=\frac{AC}{BC}$
$\Rightarrow tan\angle ABC=\frac{AC}{BC}$
$\Rightarrow AC=xtan{30}^{\circ }$
$\Rightarrow AC=\frac{x}{\sqrt{3}}$

$\Rightarrow AC=BCSincetan45=1$
$\therefore 12-x=\frac{x}{\sqrt{3}}$
$\Rightarrow x=6\sqrt{3}(\sqrt{3}-1)$
$\therefore AC=xtan{30}^{\circ }=\frac{6\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}}$
$\therefore AC=6(\sqrt{3}-1)$
$\therefore heightofbridgeis6(\sqrt{3}-1)$