Question

A pole 6 m high casts a shadow $2\sqrt{3}$ m long on the ground. The sun's altitude is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 1

(c) 60^o

Let AB be the pole and AC be its shadow.
$$\begin{gathered} \mathrm{Then},\mathrm{AB}=6\mathrm{m}\mathrm{and}\mathrm{AC}=2\sqrt{3}\mathrm{m} \\ \mathrm{Let}\angle \mathrm{ACB}={\theta }^{\mathrm{o}} \\ \mathrm{Then},\tan \theta =\frac{AB}{AC}=\frac{6}{2\sqrt{3}}=\sqrt{3}=\tan {60}^{\mathrm{o}} \\ \Rightarrow \theta ={60}^{\mathrm{o}} \end{gathered}$$