If the three angles of a triangle are:
`(x + 15 ^(@) ), (( 6x)/(5) + 6 ^(@)) and ((2x)/( 3) + 30 ^(@)),` then the triangle is :
यदि एक त्रिभुज के तीन कोण हैं
`(x + 15 ^(@) ), (( 6x)/(5) + 6 ^(@)) and ((2x)/( 3) + 30 ^(@)),` तो त्रिभुज है

Options

A. समद्विबाहु
B. समकोणीय
C. समबाहु
D. विषमबाहु

Explanation

To solve the problem, we need to find the angles of the triangle and determine the type of triangle formed based on those angles.

### Step-by-step Solution:

1. **Set up the equation for the sum of angles in a triangle:**
The sum of the angles in a triangle is always 180 degrees. Therefore, we can write the equation:
\[ (x + 15) + \left(\frac{6x}{5} + 6\right) + \left(\frac{2x}{3} + 30\right) = 180 \]

2. **Combine the terms:**
First, we will combine the constant terms and the terms with \(x\):
\[ x + \frac{6x}{5} + \frac{2x}{3} + 15 + 6 + 30 = 180 \]
This simplifies to:
\[ x + \frac{6x}{5} + \frac{2x}{3} + 51 = 180 \]

3. **Isolate the \(x\) terms:**
Subtract 51 from both sides:
\[ x + \frac{6x}{5} + \frac{2x}{3} = 180 - 51 \]
\[ x + \frac{6x}{5} + \frac{2x}{3} = 129 \]

4. **Find a common denominator:**
The common denominator for 1, 5, and 3 is 15. Rewrite each term:
\[ \frac{15x}{15} + \frac{18x}{15} + \frac{10x}{15} = 129 \]
Combine the fractions:
\[ \frac{15x + 18x + 10x}{15} = 129 \]
\[ \frac{43x}{15} = 129 \]

5. **Solve for \(x\):**
Multiply both sides by 15:
\[ 43x = 129 \times 15 \]
Calculate \(129 \times 15\):
\[ 129 \times 15 = 1935 \]
Now divide by 43:
\[ x = \frac{1935}{43} = 45 \]

6. **Calculate the angles:**
Now substitute \(x = 45\) back into the expressions for the angles:
- First angle:
\[ x + 15 = 45 + 15 = 60^\circ \]
- Second angle:
\[ \frac{6x}{5} + 6 = \frac{6 \times 45}{5} + 6 = 54 + 6 = 60^\circ \]
- Third angle:
\[ \frac{2x}{3} + 30 = \frac{2 \times 45}{3} + 30 = 30 + 30 = 60^\circ \]

7. **Determine the type of triangle:**
Since all three angles are equal to \(60^\circ\), the triangle is an equilateral triangle.

### Conclusion:
The triangle is an **equilateral triangle**.

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