Question

Length of the shadow of a $15$ meter high pole is $5\sqrt{3}$ meters at 7'o clock in the morning. Then what is the angle of elevation of the Sun rays with the ground at that time?

Answer 1

Length of pole is $XY=15m$

Shadow of pole is $=YZ=5\sqrt{3}$

angle of elevation of sun with ground is $\varphi$ as shown in figure.

Here in a right angled triangle,

$\tan ({90}^{\circ }-\varphi )=\frac{YZ}{XY}=\frac{5\sqrt{3}}{15}=\frac{1}{\sqrt{3}}\Rightarrow \tan ({90}^{\circ }-\varphi )=\frac{1}{\sqrt{3}}\Rightarrow {90}^{\circ }-\varphi ={60}^{\circ }\Rightarrow \varphi ={30}^{\circ }$

Elevation is ${30}^{\circ }$