In `triangleABC`, if `a^(2)+c^(2)-b^(2)=ac`, then `angleB=`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the measure of angle B in triangle ABC given the equation:

\[ a^2 + c^2 - b^2 = ac \]

We can use the cosine rule, which states:

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a^2 + c^2 - b^2 = ac \]

2. **Substitute the given equation into the cosine rule:**
According to the cosine rule:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
We can substitute \( a^2 + c^2 - b^2 \) with \( ac \):
\[ \cos B = \frac{ac}{2ac} \]

3. **Simplify the expression:**
\[ \cos B = \frac{1}{2} \]

4. **Find the angle B:**
The cosine of angle B is \( \frac{1}{2} \). The angle whose cosine is \( \frac{1}{2} \) is:
\[ B = 60^\circ \]

### Final Answer:
Thus, the measure of angle B is:
\[ \angle B = 60^\circ \]