MCQ
If in $\triangle A B C$ with usual notations $a=18, b=24, c=30$ then $\sin \frac{A}{2}=$
- A$\frac{1}{\sqrt{5}}$
- B$\frac{1}{\sqrt{10}}$
- C$\frac{1}{\sqrt{15}}$
- D$\frac{1}{2 \sqrt{5}}$
(B) $\frac{1}{\sqrt{10}}$
$s=\frac{a+b+c}{2}=\frac{18+24+30}{2}=36$
$\sin \left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{(36-24)(36-30)}{24 \times 30}}=\sqrt{\frac{12 \times 6}{24 \times 30}}=\frac{1}{\sqrt{10}}$
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