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If a pole is 6 meters high and casts a shadow 232\sqrt{3} meters long on the ground, then the sun's elevation is

Solution

✅ Correct Option: 4

tan⁡(elevation)=OppositeAdjacent\tan(\text{elevation}) = \dfrac{\text{Opposite}}{\text{Adjacent}}

=623= \dfrac{6}{2\sqrt{3}}

=33= \dfrac{3}{\sqrt{3}}

=33×33= \dfrac{3}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}

=333= \dfrac{3\sqrt{3}}{3}

=3= \sqrt{3}


Since tan⁡(60°)=3\tan(60°) = \sqrt{3}, the sun's elevation is 60°60°.

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