MCQ
If A, B and C are interior angles of a triangle ABC, then $\sin\Big(\frac{\text{B}+\text{C}}{2}\Big)=$
- A$\sin\frac{\text{A}}{2}$
- ✓$\cos\frac{\text{A}}{2}$
- C$-\sin\frac{\text{A}}{2}$
- D$-\cos\frac{\text{A}}{2}$
✓
Answer
Correct option: B.
$\cos\frac{\text{A}}{2}$
We know tht in triangle ABC
$\text{A+B+C}=180^\circ$
$\Rightarrow\text{B+C}=180^\circ-\text{A}$
$\Rightarrow\frac{\text{B+C}}{2}=\frac{90^\circ}{2}-\frac{\text{A}}{2}$
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
Since $\sin(90^\circ-\text{A})=\cos{\text{A}}$
So,
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Hence the correct option is (b)
$\text{A+B+C}=180^\circ$
$\Rightarrow\text{B+C}=180^\circ-\text{A}$
$\Rightarrow\frac{\text{B+C}}{2}=\frac{90^\circ}{2}-\frac{\text{A}}{2}$
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
Since $\sin(90^\circ-\text{A})=\cos{\text{A}}$
So,
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Hence the correct option is (b)
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