Question

If \( (b - c)\cos \frac{A}{2} = k \sin \frac{B - C}{2} \), then \( \frac{k}{\sin A} = \) :

• A. \( b \)
• B. \( 2R \)
• C. \( \sqrt{2R} \)
• D. \( a + c \)

Hint:

Replace sides using \(a=2R\sin A\) for quick simplification in triangle identities.

Answer 1

The Correct Option is B

Concept: Use \(b = 2R\sin B\), \(c = 2R\sin C\) and sum-to-product identities.

Step 1: Rewrite \(b-c\).
\[ b - c = 2R(\sin B - \sin C) \] \[ = 4R \cos\frac{B+C}{2}\sin\frac{B-C}{2} \]

Step 2: Use triangle identity.
\[ \frac{B+C}{2} = \frac{\pi - A}{2} = \frac{\pi}{2} - \frac{A}{2} \Rightarrow \cos\frac{B+C}{2} = \sin\frac{A}{2} \] \[ b - c = 4R \sin\frac{A}{2}\sin\frac{B-C}{2} \]

Step 3: Substitute into given equation.
\[ (b-c)\cos\frac{A}{2} = k\sin\frac{B-C}{2} \] \[ 4R \sin\frac{A}{2}\cos\frac{A}{2}\sin\frac{B-C}{2} = k\sin\frac{B-C}{2} \] \[ 2R\sin A = k \Rightarrow \frac{k}{\sin A} = 2R \] \[ \boxed{2R} \]