If a pole of 24m height casts a shdow of 8√3 m long on the ground then the sun's angle of elevation at that instant will be how much?

Options

A. a.`60^(@)`
B. b.`30^(@)`
C. c.`75^(@)`
D. d.`45^(@)`

Answer

To find the sun's angle of elevation when a pole of height 24 meters casts a shadow of length \(8\sqrt{3}\) meters, we can use the concept of trigonometry, specifically the tangent function.

### Step-by-Step Solution:

1. **Identify the components**:
- Height of the pole (perpendicular) = 24 m
- Length of the shadow (base) = \(8\sqrt{3}\) m

2. **Set up the tangent function**:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height of the pole}}{\text{length of the shadow}} = \frac{24}{8\sqrt{3}} \]

3. **Simplify the fraction**:
\[ \tan(\theta) = \frac{24}{8\sqrt{3}} = \frac{24 \div 8}{8\sqrt{3} \div 8} = \frac{3}{\sqrt{3}} \]

4. **Rationalize the denominator**:
\[ \tan(\theta) = \frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

5. **Find the angle**:
- We know that \(\tan(60^\circ) = \sqrt{3}\).
- Therefore, \(\theta = 60^\circ\).

6. **Conclusion**:
The sun's angle of elevation at that instant is \(60^\circ\).