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

Options

A. Equilateral
B. Isosceles
C. Right angled
D. Obtuse triangle

Answer

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

**Step 1: Write down the given equation.**
We have:
\[ \cos A = \frac{\sin B}{2 \sin C} \]

**Step 2: Cross-multiply to eliminate the fraction.**
Multiplying both sides by \(2 \sin C\) gives:
\[ 2 \sin C \cos A = \sin B \]

**Step 3: Use the angle sum property of triangles.**
In triangle ABC, we know that:
\[ A + B + C = \pi \]
From this, we can express \(B\) in terms of \(A\) and \(C\):
\[ B = \pi - A - C \]

**Step 4: Substitute \(B\) into the equation.**
Now we substitute \(B\) into our equation:
\[ 2 \sin C \cos A = \sin(\pi - A - C) \]
Using the sine identity \(\sin(\pi - x) = \sin x\), we have:
\[ \sin(\pi - A - C) = \sin(A + C) \]

**Step 5: Apply the sine addition formula.**
Using the sine addition formula \(\sin(A + C) = \sin A \cos C + \cos A \sin C\), we rewrite the equation:
\[ 2 \sin C \cos A = \sin A \cos C + \cos A \sin C \]

**Step 6: Rearrange the equation.**
Rearranging gives:
\[ 2 \sin C \cos A - \cos A \sin C - \sin A \cos C = 0 \]
This simplifies to:
\[ \sin C \cos A - \sin A \cos C = 0 \]

**Step 7: Factor the equation.**
This can be factored using the sine difference identity:
\[ \sin(C - A) = 0 \]

**Step 8: Solve for the angles.**
The equation \(\sin(C - A) = 0\) implies:
\[ C - A = 0 \quad \text{or} \quad C = A \]

**Step 9: Conclude the type of triangle.**
Since \(C = A\), we have two angles that are equal. Therefore, triangle ABC is an isosceles triangle, where two angles are equal, and consequently, the sides opposite those angles are also equal.

**Final Answer:**
The triangle is an **isosceles triangle**.

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