Answer :

Let us consider the above problem, by the following diagram, where 'A' depicts the aero plane, 'C' and 'D' depict two ends of river.

Now,

Angle of depression from plane to 'C' = ∠XAC = 45°

∠XAC = ∠ ACD = 45° = θ_{1}(say) [Alternate angles]

Angle of depression from plane to 'D' = ∠YAD = 60°

∠YAD = ∠ ADC = 60° = θ_{2}(say) [Alternate angles]

AP = Height at which aero plane is flying = 300 m

Now, In ΔACP and ΔADP

⇒ CP = 300 m

⇒DP = 100√3 m

⇒DP = 100(1.732)

⇒DP = 173.2 m

Now, distance between two banks, CD = CP + DP

⇒ CD = 300 + 173.2 = 473.2 m

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