If the sides of a right angled triangle are in A.P. then the sines of the acute angles are

Options

A. `(3)/(5), (4)/(5)`
B. `(1)/(sqrt(3)), sqrt((2)/(3))`
C. `(1)/(2), (sqrt(3))/(2)`
D. none

Answer

To solve the problem, we need to find the sines of the acute angles in a right-angled triangle where the sides are in arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Define the sides of the triangle**:
Let the sides of the triangle be \( a - d \), \( a \), and \( a + d \), where \( a \) is the middle term and \( d \) is the common difference. Since it's a right-angled triangle, we can assume that \( a + d \) is the hypotenuse.

2. **Apply the Pythagorean theorem**:
According to the Pythagorean theorem, we have:
\[ (a - d)^2 + a^2 = (a + d)^2 \]

3. **Expand the equation**:
Expanding both sides gives:
\[ (a^2 - 2ad + d^2) + a^2 = (a^2 + 2ad + d^2) \]
Simplifying this, we get:
\[ 2a^2 - 2ad + d^2 = a^2 + 2ad + d^2 \]

4. **Rearranging the equation**:
Subtract \( a^2 + d^2 \) from both sides:
\[ 2a^2 - a^2 - 2ad - 2ad = 0 \]
This simplifies to:
\[ a^2 - 4ad = 0 \]

5. **Factor the equation**:
Factoring out \( a \):
\[ a(a - 4d) = 0 \]
Since \( a \) cannot be zero (as it represents a side length), we have:
\[ a = 4d \]

6. **Determine the sides**:
Now substituting \( a = 4d \) back into the sides:
- First side: \( a - d = 4d - d = 3d \)
- Second side: \( a = 4d \)
- Third side: \( a + d = 4d + d = 5d \)

Therefore, the sides of the triangle are \( 3d, 4d, 5d \).

7. **Calculate the sines of the acute angles**:
Let \( \theta_1 \) be the angle opposite the side \( 3d \) and \( \theta_2 \) be the angle opposite the side \( 4d \).

- For \( \theta_1 \):
\[ \sin(\theta_1) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3d}{5d} = \frac{3}{5} \]

- For \( \theta_2 \):
\[ \sin(\theta_2) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4d}{5d} = \frac{4}{5} \]

8. **Final Result**:
Thus, the sines of the acute angles are:
\[ \sin(\theta_1) = \frac{3}{5}, \quad \sin(\theta_2) = \frac{4}{5} \]

### Conclusion:
The sines of the acute angles in the right-angled triangle whose sides are in A.P. are \( \frac{3}{5} \) and \( \frac{4}{5} \).

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