The angle of elevation of an aeroplane as obs | vedclass

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(A) Let $H$ be the height of the aeroplane from the water surface of the lake.Let the point of observation be $P$,which is $30 \ m$ above the water su

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$60$

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(A) Let $H$ be the height of the aeroplane from the water surface of the lake.Let the point of observation be $P$,which is $30 \ m$ above the water surface.The height of the aeroplane above the point $P$ is $(H - 30) \ m$.The depth of the image of the aeroplane in the lake is $H$ below the water surface.Therefore,the total depth of the image from the point $P$ is $(H + 30) \ m$.Let $x$ be the horizontal distance of the aeroplane from the point of observation.From the angle of elevation: $	an(30^{\circ}) = \frac{H - 30}{x} \implies x = \frac{H - 30}{	an(30^{\circ})} = (H - 30)\sqrt{3}$.From the angle of depression of the image: $	an(60^{\circ}) = \frac{H + 30}{x} \implies x = \frac{H + 30}{	an(60^{\circ})} = \frac{H + 30}{\sqrt{3}}$.Equating the two expressions for $x$: $(H - 30)\sqrt{3} = \frac{H + 30}{\sqrt{3}}$.$3(H - 30) = H + 30$.$3H - 90 = H + 30$.$2H = 120$.$H = 60 \ m$.

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