Question

An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. The width of the river is

(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(b) $\left(200-\frac{200}{\sqrt{3}}\right)\mathrm{m}$
(c) $400\sqrt{3}\mathrm{m}$
(d) $\frac{400}{\sqrt{3}}\mathrm{m}$

Answer 1

(a) $\left(200+\frac{200}{\sqrt{3}}\right)\mathrm{m}$
Let $A$ be the position of the aeroplane and $XAY$ be the horizontal line.
Let $BDC$ be the line on the ground such that $AD$ ⊥ $BC$ and $AD=200$ m.
It observes an angle of depression such that ∠$XAB={45}^o$ and ∠$YAC={60}^o$.
Also,
∠$ABD={45}^o$ and ∠$ACD={60}^o$ (Alternate angles)

In ∆$ABD$, we have:
$\frac{BD}{AD}=cot{45}^o=1$

⇒ $\frac{BD}{200}=1$
⇒ $BD=200$ m
​Now, in ∆$ACD$, we have:
$\frac{CD}{AD}=cot{60}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{CD}{200}=\frac{1}{\sqrt{3}}$
⇒ $CD=\frac{200}{\sqrt{3}}$ m
∴ Width of the river = $BC=BD+CD=200+\frac{200}{\sqrt{3}}=200(1+\frac{1}{\sqrt{3}})$ m