If the angles of a triangle are in A.P.with common difference equal `1//3` of the least angle ,the sides are in the ratio

Options

A. `sqrt2:2sqrt3:sqrt6+sqrt2`
B. `2sqrt2:sqrt3:sqrt6-sqrt2`
C. `2sqrt2:2sqrt3:sqrt6-sqrt2`
D. `2sqrt2:2sqrt3:sqrt6+sqrt2`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step-by-Step Solution:

1. **Assume the least angle**:
Let the least angle of the triangle be \( a \).

2. **Determine the angles in A.P.**:
Since the angles are in arithmetic progression (A.P.) with a common difference of \( \frac{1}{3} \) of the least angle, the angles can be expressed as:
- First angle: \( a \)
- Second angle: \( a + \frac{a}{3} = \frac{4a}{3} \)
- Third angle: \( a + 2 \cdot \frac{a}{3} = \frac{5a}{3} \)

3. **Sum of angles in a triangle**:
The sum of the angles in a triangle is \( 180^\circ \). Therefore, we can write the equation:
\[ a + \frac{4a}{3} + \frac{5a}{3} = 180^\circ \]

4. **Combine the angles**:
To combine the angles, we find a common denominator:
\[ a + \frac{4a + 5a}{3} = 180^\circ \]
This simplifies to:
\[ a + \frac{9a}{3} = 180^\circ \]
\[ a + 3a = 180^\circ \]
\[ 4a = 180^\circ \]

5. **Solve for \( a \)**:
Dividing both sides by 4 gives:
\[ a = \frac{180^\circ}{4} = 45^\circ \]

6. **Calculate the angles**:
Now substituting \( a \) back to find the other angles:
- First angle: \( a = 45^\circ \)
- Second angle: \( \frac{4a}{3} = \frac{4 \times 45^\circ}{3} = 60^\circ \)
- Third angle: \( \frac{5a}{3} = \frac{5 \times 45^\circ}{3} = 75^\circ \)

7. **Use the Sine Rule to find the ratio of sides**:
By the sine rule, the ratio of the sides opposite to the angles is given by:
\[ \frac{a}{b} : \frac{b}{c} : \frac{c}{a} = \sin(45^\circ) : \sin(60^\circ) : \sin(75^\circ) \]

8. **Calculate the sine values**:
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

9. **Form the ratio**:
The ratio of the sides is:
\[ \frac{1/\sqrt{2}}{\sqrt{2}/2} : \frac{\sqrt{3}/2}{\sqrt{2}/2} : \frac{(\sqrt{3}+1)/2\sqrt{2}}{\sqrt{2}/2} \]
Simplifying gives:
\[ 1 : \sqrt{3} : \sqrt{3} + 1 \]

### Final Ratio of Sides:
The sides of the triangle are in the ratio \( 1 : \sqrt{3} : \sqrt{3} + 1 \).