Angle at vertex of an isosceles triangle is `15^(@)` more than one of the at base angle at vertex is

Options

A. `35^(@)`
B. `55^(@)`
C. `65^(@)`
D. `70^(@)`

Answer

To solve the problem, we need to find the base angles of the isosceles triangle given that the angle at the vertex is 15 degrees more than one of the base angles.

### Step-by-Step Solution:

1. **Define the Angles**:
Let the measure of each base angle be \( x \). According to the problem, the angle at the vertex (the top angle) is \( x + 15^\circ \).

2. **Set Up the Equation**:
The sum of the angles in a triangle is always \( 180^\circ \). Therefore, we can write the equation:
\[ x + x + (x + 15^\circ) = 180^\circ \]

3. **Combine Like Terms**:
Simplifying the left side of the equation gives:
\[ 3x + 15^\circ = 180^\circ \]

4. **Isolate the Variable**:
To isolate \( 3x \), subtract \( 15^\circ \) from both sides:
\[ 3x = 180^\circ - 15^\circ \]
\[ 3x = 165^\circ \]

5. **Solve for \( x \)**:
Now, divide both sides by 3 to find \( x \):
\[ x = \frac{165^\circ}{3} \]
\[ x = 55^\circ \]

6. **Conclusion**:
The measure of each base angle is \( 55^\circ \).

### Final Answer:
The base angles of the isosceles triangle are \( 55^\circ \).

---