A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15 m within water when sunlight is incident at an angle of 30° with the surface of water. If swimming pool is filled to a height of 1.5 m, then the height of the pole above the water surface in centimeters is $(n_{w} = 4 / 3)$ _______ (Round off to nearest integer)

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answer is 50.

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Detailed Solution

$\begin{array}{l} \tan 30^{\circ} = \frac{x}{2.15 - y} \\ 1 \sin 60^{\circ} = \frac{4}{3} \sin r \\ \sin r = \frac{3 \sqrt{3}}{8} \\ \tan r = \frac{3 \sqrt{3}}{\sqrt{37}} = \frac{y}{1.5} \\ \frac{3 \sqrt{3}}{\sqrt{37}} = \frac{2 y}{3} \Rightarrow y = \frac{9 \sqrt{3}}{2 \sqrt{37}} \\ x = \frac{1}{\sqrt{3}} (2.15 - \frac{9 \sqrt{3}}{2 \sqrt{37}}) = 0.5016 m = 50 cm \end{array}$

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