In a triangle `ABC, sin A-cosB=cosC,` then angle B, is

Options

A. `pi//2`
B. `pi//3`
C. `pi//4`
D. `pi//6`

Answer

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

\[ \sin A - \cos B = \cos C \]

### Step-by-Step Solution:

1. **Rearranging the Equation:**
Start by rearranging the given equation to isolate \(\sin A\):
\[ \sin A = \cos B + \cos C \]

**Hint:** Remember that you can move terms from one side of the equation to the other by performing the same operation on both sides.

2. **Using the Cosine Rule:**
We know that in any triangle, the angles satisfy the equation \(A + B + C = 180^\circ\) or \(\pi\) radians. Therefore, we can express \(C\) in terms of \(A\) and \(B\):
\[ C = 180^\circ - A - B \]

**Hint:** Use the property of triangle angles summing up to \(180^\circ\).

3. **Substituting for \(\cos C\):**
We can use the cosine of the angle sum identity:
\[ \cos C = \cos(180^\circ - A - B) = -\cos(A + B) \]
However, for simplicity, we will use the relationship directly later.

4. **Using the Sine Rule:**
We can also apply the sine rule which states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
But for this problem, we will focus on the relationship derived from the rearrangement.

5. **Using the Identity for \(\cos B + \cos C\):**
We can apply the cosine addition formula:
\[ \cos B + \cos C = 2 \cos\left(\frac{B + C}{2}\right) \cos\left(\frac{B - C}{2}\right) \]
However, we will keep it simple and proceed with the values we have.

6. **Equating the Sine Values:**
From our rearrangement, we have:
\[ \sin A = \cos B + \cos C \]
Substitute \(\cos C\) with \(\sin A\):
\[ \sin A = \cos B + \cos(180^\circ - A - B) \]
This leads us to the conclusion that:
\[ \sin A = \cos B - \sin A \]

7. **Finding Angle B:**
From the derived equations, we can find that:
\[ 2\sin A = \cos B \]
Using the identity for angles in a triangle, we can derive:
\[ B = 90^\circ \]

**Hint:** Recall that in a right triangle, the sine and cosine functions have specific relationships that can simplify the equations.

8. **Conclusion:**
Therefore, the angle \(B\) in triangle \(ABC\) is:
\[ B = 90^\circ \]

### Final Answer:
The angle \(B\) is \(90^\circ\).