Question

For any triangle $ABC$, if $B=3C$, then $\sin \frac{A}{2}=$

• A. $\frac{b-c}{2c}$
• B. $\frac{b+c}{2c}$
• C. $\frac{b-c}{2b}$
• D. $\frac{b+c}{2b}$

Answer 1

The correct option is A $\frac{b-c}{2c}$

Given, $B=3C$

$\Rightarrow \sin B=\sin 3C$

$\Rightarrow \frac{\sin B}{\sin C}=3-4{\sin }^{2}C=\frac{b}{c}$
$\Rightarrow 3-4(1-{\cos }^{2}C)=\frac{b}{c}$

$\Rightarrow 4{\cos }^{2}C=\frac{b+c}{c}$

$\Rightarrow \cos C=\sqrt{\frac{b+c}{4c}}$
Now $A+B+C=\pi$

$\Rightarrow A=\pi -4C$

$\Rightarrow \frac{A}{2}=\frac{\pi }{2}-2C$
$\therefore \sin \frac{A}{2}=\cos 2C=2{\cos }^{2}C-1=\frac{2(b+c)}{4c}-1=\frac{b-c}{2c}$