Question

the altitude of the sun when the length of the shadow is $7\sqrt{3}m$.

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

Answer: (B)

${45}^o$

The sun is at a degree of ${30}^0$
And it is given that the length of the shadow of the pole is 21m.
Hence
$\tan {30}^0=\frac{poleheight}{shadowlenght}$
$\frac{1}{\sqrt{3}}=\frac{h}{21}$
$h=\frac{7\times 3}{\sqrt{3}}$
$h=7\sqrt{3}m$
Now the altitude of the sun changes however, height of the pole remains constant.
It is given that the length of shadow is equal to $7\sqrt{3}$
Let $\theta$ be the altitude.
Then
$\tan \theta =\frac{poleheight}{shadowlenght}$
$=\frac{7\sqrt{3}}{7\sqrt{3}}$
$=1$
$\tan \theta =1$
$\theta ={45}^0$