Question

In a triangle \(ABC\), if \(s=\dfrac{15}{2},\; a=3,\; b=5\), then \[ \frac{\sin \dfrac{B}{2}}{\sin \dfrac{A}{2}}= \]

• A. \(\cot\dfrac{C}{2}\)
• B. \(2\sin\dfrac{C}{2}\)
• C. \(2\cosec\dfrac{C}{2}\)
• D. \(\sin\dfrac{C}{2}\)

Hint:

Remember the half-angle formulas in a triangle: \[ \sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}, \] \[ \sin\frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}, \] \[ \sin\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}. \]

Answer 1

The Correct Option is B

Step 1: Use the half-angle formula in a triangle. We know that \[ \sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}, \] and \[ \sin\frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}. \] Therefore, \[ \frac{\sin\frac{B}{2}}{\sin\frac{A}{2}} = \sqrt{\frac{(s-a)b}{(s-b)a}}. \]

Step 2: Substitute the given values. Since \[ s=\frac{15}{2},\qquad a=3,\qquad b=5, \] we have \[ s-a=\frac{15}{2}-3=\frac92, \] and \[ s-b=\frac{15}{2}-5=\frac52. \] Hence, \[ \frac{\sin\frac{B}{2}}{\sin\frac{A}{2}} = \sqrt{\frac{\frac92\cdot5}{\frac52\cdot3}} = \sqrt3. \]

Step 3: Find \(\sin\dfrac{C}{2}\). First, \[ c=2s-a-b =15-3-5=7. \] Using \[ \sin\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}, \] we obtain \[ \sin\frac{C}{2} = \sqrt{\frac{\frac92\cdot\frac52}{3\cdot5}} = \sqrt{\frac38} =\frac{\sqrt3}{2\sqrt2}. \] Also, \[ \sqrt3 = 2\sin\frac{C}{2}. \] Therefore, \[ \boxed{ \frac{\sin\frac{B}{2}}{\sin\frac{A}{2}} = 2\sin\frac{C}{2} }. \] Hence, the correct option is \(\boxed{(B)}\).