If a tower 9m high casts a shadow of 3 √(3 m long on the ground, then the suns elevation)

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Question 1 5 A pole 6 m high casts a shadow 2 √3 m long on the ground, then the Sun's elevation isA 60∘B 45∘C 30∘D 90∘

$\tan \theta =\frac{BC}{AB}$

$\Rightarrow \tan \theta =\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{3\sqrt{3}}{3}=\sqrt{3}=\tan {60}^{\circ }$

$\therefore \theta ={60}^{\circ }$

Hence, the Sun’s elevation is ${60}^{\circ }$