MCQ
Choose the correct answer from the given four options. A pole $6m$ high casts a shadow $2\sqrt{3}\text{m}$ long on the ground, then the Sun’s elevation is :
- ✓$60^\circ$
- B$45^\circ$
- C$30^\circ$
- D$90^\circ$
✓
Answer
Correct option: A.
$60^\circ$
Let $\text{BC}=6\text{m}$ be the height of the pole and $\text{AB}=2\sqrt{3}\text{m}$ be the lenth of the shadow on the ground.
let the Sun's makes an angle $\theta$ on the ground.
Now, in $\triangle\text{BAC},\ \tan\theta=\frac{\text{BC}}{\text{AB}}$
$\Rightarrow\ \tan\theta=\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\ \tan\theta=\frac{3\sqrt{3}}{3}=\sqrt{3}$
$=\tan60^\circ\ [\because\tan60^\circ=\sqrt{3}]$
$\therefore\ \theta=60^\circ$
Hence, the Sun's elevation is $60^\circ .$
let the Sun's makes an angle $\theta$ on the ground.
Now, in $\triangle\text{BAC},\ \tan\theta=\frac{\text{BC}}{\text{AB}}$
$\Rightarrow\ \tan\theta=\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\ \tan\theta=\frac{3\sqrt{3}}{3}=\sqrt{3}$
$=\tan60^\circ\ [\because\tan60^\circ=\sqrt{3}]$
$\therefore\ \theta=60^\circ$
Hence, the Sun's elevation is $60^\circ .$
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