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(B) Let the height of the plane be $h = 5 	ext{ km}$.Let the positions of the plane be $A$ and $B$ at angles $15^{\circ}$ and $30^{\circ}$ respective

Vedclass · Accepted Answer

$720$

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(B) Let the height of the plane be $h = 5 	ext{ km}$.Let the positions of the plane be $A$ and $B$ at angles $15^{\circ}$ and $30^{\circ}$ respectively.Let $O$ be the observer on the ground.In $	riangle ODA$ (where $D$ is the point directly below $A$),$	an(15^{\circ}) = \frac{h}{OD} \implies OD = \frac{5}{	an(15^{\circ})} = 5 \cot(15^{\circ})$.Using $\cot(15^{\circ}) = 2 + \sqrt{3}$,$OD = 5(2 + \sqrt{3}) = 10 + 5\sqrt{3} 	ext{ km}$.In $	riangle OEB$ (where $E$ is the point directly below $B$),$	an(30^{\circ}) = \frac{h}{OE} \implies OE = \frac{5}{	an(30^{\circ})} = 5\sqrt{3} 	ext{ km}$.The distance traveled by the plane is $d = OD - OE = 10 + 5\sqrt{3} - 5\sqrt{3} = 10 	ext{ km}$.The time taken is $t = 50 	ext{ seconds} = \frac{50}{3600} 	ext{ hours} = \frac{1}{72} 	ext{ hours}$.Speed $v = \frac{d}{t} = \frac{10}{1/72} = 720 	ext{ kmph}$.

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