If the lengths of the sides of a right- angled triangle ABC, right angled at C, are in arithmetic progression, then the value of `5(sinA+sinB)` is

Answer

To solve the problem, we need to find the value of \( 5(\sin A + \sin B) \) for a right-angled triangle ABC, where the lengths of the sides are in arithmetic progression (AP) and the right angle is at C.

### Step-by-Step Solution:

1. **Define the sides of the triangle**:
Since the sides are in arithmetic progression, we can denote the lengths of the sides as:
- \( a - d \) (one leg)
- \( a \) (the other leg)
- \( a + d \) (the hypotenuse)

2. **Apply Pythagorean theorem**:
According to the Pythagorean theorem:
\[ (a + d)^2 = (a - d)^2 + a^2 \]
Expanding both sides:
\[ a^2 + 2ad + d^2 = (a^2 - 2ad + d^2) + a^2 \]
Simplifying the right side:
\[ a^2 + 2ad + d^2 = 2a^2 - 2ad + d^2 \]
Canceling \( d^2 \) from both sides gives:
\[ a^2 + 2ad = 2a^2 - 2ad \]
Rearranging terms:
\[ 4ad = a^2 \]

3. **Solve for \( a \)**:
From the equation \( 4ad = a^2 \), we can rearrange it to:
\[ a^2 - 4ad = 0 \]
Factoring out \( a \):
\[ a(a - 4d) = 0 \]
Since \( a \) cannot be zero (as it represents a length), we have:
\[ a = 4d \]

4. **Determine the lengths of the sides**:
Substituting \( a = 4d \):
- First leg: \( 4d - d = 3d \)
- Second leg: \( 4d \)
- Hypotenuse: \( 4d + d = 5d \)

5. **Calculate \( \sin A \) and \( \sin B \)**:
- For angle \( A \):
\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3d}{5d} = \frac{3}{5} \]
- For angle \( B \):
\[ \sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4d}{5d} = \frac{4}{5} \]

6. **Calculate \( 5(\sin A + \sin B) \)**:
\[ \sin A + \sin B = \frac{3}{5} + \frac{4}{5} = \frac{7}{5} \]
Therefore:
\[ 5(\sin A + \sin B) = 5 \times \frac{7}{5} = 7 \]

### Final Answer:
The value of \( 5(\sin A + \sin B) \) is \( \boxed{7} \).