MCQ
If the elevation of the sun changes form $30^\circ$ to $60^\circ,$ then the difference between the lengths of shadows of a pole $15m$ high, is :
- A$7.5\text{m}$
- B$15\text{m}$
- ✓$10\sqrt{3}\text{m}$
- D$5\sqrt{3}\text{m}$
✓
Answer
Correct option: C.
$10\sqrt{3}\text{m}$
Let $AB$ be the pole and $AC$ and $AD$ be its shadows.
We have :
$\angle\text{ACB}=30^\circ,\angle\text{ADB}=60^\circ$ and $AB = 15m.$
In $\triangle\text{ACB},$
We have :
$\frac{\text{AC}}{\text{AB}}=\cot30^\circ=\sqrt{3}$
$\Rightarrow\frac{\text{AC}}{15}={\sqrt{3}}$
$\Rightarrow\text{AC}=15\sqrt{3}\text{m}$
Now, in $\triangle\text{ADB},$ we have:
$\frac{\text{AD}}{\text{AB}}=\cot60^\circ=\frac{1}{\sqrt{3}}$
$\Rightarrow\frac{\text{AD}}{15}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{AD}=\frac{15}{\sqrt{3}}=\frac{15\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\frac{15\sqrt{3}}{3}=5\sqrt{3}\text{m}$
$\therefore$ Difference between the lengths of the shadows
$=\text{AC}-\text{AD}=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\text{m}$
We have :
$\angle\text{ACB}=30^\circ,\angle\text{ADB}=60^\circ$ and $AB = 15m.$
In $\triangle\text{ACB},$
We have :
$\frac{\text{AC}}{\text{AB}}=\cot30^\circ=\sqrt{3}$
$\Rightarrow\frac{\text{AC}}{15}={\sqrt{3}}$
$\Rightarrow\text{AC}=15\sqrt{3}\text{m}$
Now, in $\triangle\text{ADB},$ we have:
$\frac{\text{AD}}{\text{AB}}=\cot60^\circ=\frac{1}{\sqrt{3}}$
$\Rightarrow\frac{\text{AD}}{15}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{AD}=\frac{15}{\sqrt{3}}=\frac{15\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\frac{15\sqrt{3}}{3}=5\sqrt{3}\text{m}$
$\therefore$ Difference between the lengths of the shadows
$=\text{AC}-\text{AD}=15\sqrt{3}-5\sqrt{3}=10\sqrt{3}\text{m}$
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