In an isosceles triangle, if the unequal angle is twice the sum of the equal angles, then each equal angle is

Options

A. `120^@`
B. `60^@`
C. `30^@`
D. `90^@`

Answer

To solve the problem, we need to find the measure of each equal angle in an isosceles triangle where the unequal angle is twice the sum of the equal angles.

### Step-by-Step Solution:

1. **Define the Angles**:
Let the equal angles in the isosceles triangle be denoted as \(X\). Since there are two equal angles, we have:
\[ \text{Angle A} = X, \quad \text{Angle B} = X \]
Let the unequal angle be denoted as \(Y\).

2. **Set Up the Relationship**:
According to the problem, the unequal angle \(Y\) is twice the sum of the equal angles. The sum of the equal angles is \(X + X = 2X\). Therefore, we can express \(Y\) as:
\[ Y = 2 \times (X + X) = 2 \times 2X = 4X \]

3. **Use the Angle Sum Property**:
The sum of the angles in any triangle is \(180^\circ\). Thus, we can write:
\[ X + X + Y = 180^\circ \]
Substituting \(Y\) with \(4X\):
\[ X + X + 4X = 180^\circ \]
This simplifies to:
\[ 6X = 180^\circ \]

4. **Solve for \(X\)**:
To find \(X\), divide both sides of the equation by 6:
\[ X = \frac{180^\circ}{6} = 30^\circ \]

5. **Conclusion**:
Each equal angle in the isosceles triangle is:
\[ \boxed{30^\circ} \]