Question

An aeroplane at an altitude of $200$ m observes the angle of depression of two opposite points on the two banks of a river to be 45${}^o$ and 60${}^o$. Then, the width of the river is _____.

• A. $\frac{400}{\sqrt{3}}$ m
• B. $400$$\sqrt{3}$ m
• C. $(200-\frac{200}{\sqrt{3}})$ m
• D. $(200+\frac{200}{\sqrt{3}})$ m

Answer 1

The correct option is D $(200+\frac{200}{\sqrt{3}})$ m

From the given figure, A is the aeroplane, and B and C

are the two points of the two banks of the river.
AD = height of aeroplane's altitude $=200$ m

$\because XAB={45}^o\therefore \angle ABD={45}^o$ {as they are alternate interior angles}

and, Similarily,

$\because \angle$YAC = 60${}^o$ $\therefore$ $\angle$ACD = 60${}^o$

Now, In the right-angled triangle ABD

$\tan {45}^o=\frac{AD}{BD}$

$\Rightarrow 1=\frac{200}{BD}$

$\therefore BD=200$ ....(1)

Also,

In the right-angled triangle ADC

$\tan {60}^o=\frac{AD}{DC}$

$\Rightarrow \sqrt{3}=\frac{200}{DC}$

$\therefore DC=\frac{200}{\sqrt{3}}$ ....(2)

Now as we know Width of the river $=BD+DC$

$\therefore From\left(1\right)and\left(2\right)$

$BD+DC$ $=(200+\frac{200}{\sqrt{3}})$ m

So, option D is correct.