Question

An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of river to be 30° and 45°, then the width of the river is ​

Answer 1

(Diagram is given in attachment)

EF || BC

∴ ∠EAB = 45° (Alternate Interior Angles)

Similarly,

∠ACD = 60° (Alternate Interior Angles)

We need to find the width of river that is BD + DC

Let

• BD = x
• DC = y

$\sf{In \: \triangle ABD,}\\\\\sf{tan \: 45^{\circ} = \dfrac{AD}{BD} }\\\\\\\implies \sf{tan \: 45^{\circ} = \dfrac{200}{x}}\\\\\\\implies \sf{1 = \dfrac{200}{x}}\\\\\\\implies \sf{x = 200 \: m}\\\\\\\therefore \boxed{\bf{BD = 200 \: m}}$

$\sf{In \: \triangle ACD,}\\\\\sf{tan \: 60^{\circ} = \dfrac{AD}{DC}}\\\\\\\implies \sf{\sqrt{3} = \dfrac{200}{y} }\\\\\\\implies \sf{\sqrt{3}y = 200}\\\\\\\implies \sf{y = \dfrac{200}{\sqrt{3}}}\\\\\\\\\therefore \boxed{\bf{DC = \dfrac{200}{\sqrt{3} }}}$

$\rm{Total \: Width \: of \: the \: river = BD + CD}\\\\\\\longrightarrow \sf{Total\:Width\:of\:river = 200 + \dfrac{200}{\sqrt{3}}}\\\\\\\longrightarrow \sf{BC = \dfrac{200 \sqrt{3}+200}{\sqrt{3}}}\\\\\\\\\longrightarrow \sf{BC = \dfrac{200(\sqrt{3} + 1)}{\sqrt{3}}}\\\\\\\\\longrightarrow \sf{BC = \dfrac{200(1.732 + 1)}{1.732}}\\\\\\\\\longrightarrow \sf{BC = \dfrac{200(2.732)}{1.732}}\\\\\\\\\therefore \large{\boxed{\boxed{\bf{Width \: of \: river = 315.47 m}}}}$