The shadow of a tower is 15m, when the Sun's elevation is `30^@`. What is the length of the shadow, when the Sun's elevation is `60^@`?

Options

A. 3m
B. 4 m
C. 5 m
D. 6m

Answer

To solve the problem, we will use the concept of trigonometry, specifically the tangent function, which relates the angle of elevation to the height of the tower and the length of its shadow.

### Step-by-Step Solution:

1. **Identify the given information:**
- Length of the shadow when the sun's elevation is \(30^\circ\) = 15 m.
- Angle of elevation = \(30^\circ\).

2. **Use the tangent function:**
The tangent of an angle in a right triangle is defined as the ratio of the opposite side (height of the tower) to the adjacent side (length of the shadow).
\[ \tan(30^\circ) = \frac{\text{Height of the tower}}{\text{Length of the shadow}} \]
Let the height of the tower be \(h\). Therefore,
\[ \tan(30^\circ) = \frac{h}{15} \]

3. **Calculate the height of the tower:**
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
\[ \frac{1}{\sqrt{3}} = \frac{h}{15} \]
Cross-multiplying gives:
\[ h = \frac{15}{\sqrt{3}} = 5\sqrt{3} \text{ m} \]

4. **Find the length of the shadow when the sun's elevation is \(60^\circ\):**
Now, we will use the same concept for \(60^\circ\):
\[ \tan(60^\circ) = \frac{\text{Height of the tower}}{\text{Length of the shadow at } 60^\circ} \]
Let the length of the shadow at \(60^\circ\) be \(x\). Thus,
\[ \tan(60^\circ) = \frac{h}{x} \]
We know that \(\tan(60^\circ) = \sqrt{3}\).
\[ \sqrt{3} = \frac{5\sqrt{3}}{x} \]

5. **Solve for \(x\):**
Cross-multiplying gives:
\[ x \cdot \sqrt{3} = 5\sqrt{3} \]
Dividing both sides by \(\sqrt{3}\):
\[ x = 5 \text{ m} \]

### Final Answer:
The length of the shadow when the sun's elevation is \(60^\circ\) is **5 meters**.

---