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 of these

Explanation

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). Let's denote the sides of the triangle as follows:

1. 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.

2. Since it is a right-angled triangle, we can assume that the side \( a + d \) is the hypotenuse (the longest side), and the other two sides are the legs of the triangle.

### Step 1: Apply the Pythagorean Theorem
According to the Pythagorean theorem:
\[ \text{(Hypotenuse)}^2 = \text{(Side 1)}^2 + \text{(Side 2)}^2 \]
Thus, we have:
\[ (a + d)^2 = (a - d)^2 + a^2 \]

### Step 2: Expand the equation
Now, we expand both sides:
\[ a^2 + 2ad + d^2 = (a^2 - 2ad + d^2) + a^2 \]
Simplifying the right side:
\[ a^2 + 2ad + d^2 = 2a^2 - 2ad + d^2 \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation:
\[ a^2 + 2ad + d^2 - 2a^2 + 2ad - d^2 = 0 \]
This simplifies to:
\[ -a^2 + 4ad = 0 \]

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

### Step 5: Determine the side lengths
Now substituting \( a \) back into the side lengths:
- The sides are:
- \( a - d = 4d - d = 3d \)
- \( a = 4d \)
- \( a + d = 4d + d = 5d \)

### Step 6: Calculate the sines of the acute angles
Now we can find the sine of the acute angles \( A \) and \( B \):
1. **For angle \( A \)**:
\[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{3d}{5d} = \frac{3}{5} \]

2. **For angle \( B \)**:
\[ \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{4d}{5d} = \frac{4}{5} \]

### Final Result
Thus, the sines of the acute angles are:
\[ \sin A = \frac{3}{5}, \quad \sin B = \frac{4}{5} \]