If `cos A= (sinB)/(2 sinC), " then " Delta ABC` is

Options

A. rt, angled
B. isosceles
C. equilateral
D. none

Answer

To solve the problem where \( \cos A = \frac{\sin B}{2 \sin C} \), we will use the properties of triangles and trigonometric identities.

### Step-by-step Solution:

1. **Start with the given equation**:
\[ \cos A = \frac{\sin B}{2 \sin C} \]

2. **Use the cosine rule**:
The cosine rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Here, \( a, b, c \) are the lengths of the sides opposite to angles \( A, B, C \) respectively.

3. **Use the sine rule**:
The sine rule states:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \]
From this, we can express \( \sin B \) and \( \sin C \) in terms of \( k \):
\[ \sin B = kb \quad \text{and} \quad \sin C = kc \]

4. **Substitute \( \sin B \) and \( \sin C \) into the equation**:
Substitute these into the original equation:
\[ \cos A = \frac{kb}{2kc} = \frac{b}{2c} \]

5. **Set the two expressions for \( \cos A \) equal**:
Now we have:
\[ \frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{2c} \]

6. **Cross-multiply to eliminate the fractions**:
\[ (b^2 + c^2 - a^2) \cdot 2c = b \cdot 2bc \]
Simplifying gives:
\[ 2c(b^2 + c^2 - a^2) = 2b^2c \]

7. **Divide both sides by \( 2c \) (assuming \( c \neq 0 \))**:
\[ b^2 + c^2 - a^2 = b^2 \]
This simplifies to:
\[ c^2 - a^2 = 0 \]

8. **Rearranging gives**:
\[ c^2 = a^2 \]

9. **Taking the square root**:
\[ c = a \]

10. **Conclusion**:
Since \( c = a \), it implies that two sides of triangle \( ABC \) are equal. Therefore, triangle \( ABC \) is an **isosceles triangle**.

### Final Answer:
Triangle \( ABC \) is an **isosceles triangle**.