If in a triangle `ABC , cosA=(sinB)/(2sinC)` then the triangle `ABC` , is

Options

A. equilateral
B. isosceles
C. right angled
D. none of these

Answer

To solve the problem, we start with the given equation in triangle \( ABC \):

\[ \cos A = \frac{\sin B}{2 \sin C} \]

### Step 1: Use the Sine Rule
According to the sine rule, we have:

\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = K \]

From this, we can express \( \sin B \) and \( \sin C \) in terms of \( K \):

\[ \sin B = bK \quad \text{and} \quad \sin C = cK \]

### Step 2: Substitute into the Given Equation
Substituting \( \sin B \) and \( \sin C \) into the equation \( \cos A = \frac{\sin B}{2 \sin C} \):

\[ \cos A = \frac{bK}{2(cK)} = \frac{b}{2c} \]

### Step 3: Use the Cosine Rule
Using the cosine rule, we have:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

### Step 4: Set the Two Expressions for Cos A Equal
Now we set the two expressions for \( \cos A \) equal to each other:

\[ \frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{2c} \]

### Step 5: Cross-Multiply to Eliminate the Denominator
Cross-multiplying gives:

\[ (b^2 + c^2 - a^2)c = b^2 \]

### Step 6: Rearrange the Equation
Rearranging the equation, we get:

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

This simplifies to:

\[ c^2 - a^2 = 0 \]

### Step 7: Conclude the Relationship Between Sides
From \( c^2 - a^2 = 0 \), we can conclude that:

\[ c^2 = a^2 \quad \Rightarrow \quad c = a \]

### Conclusion
Since \( c = a \), we have two sides of the triangle \( ABC \) that are equal. Therefore, triangle \( ABC \) is an isosceles triangle.

### Final Answer
Triangle \( ABC \) is isosceles.

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