Question

In $\Delta ABC$, if $A={30}^0$, b=2, $c=\sqrt{3}+1$, then $\frac{C-B}{2}$ is equal to

• A. ${15}^0$
• B. ${30}^0$
• C. ${45}^0$
• D. none of these

Answer 1

The correct option is A ${15}^0$

$cosA=\frac{b^2+c^2-a^2}{2bc}$
$\frac{\sqrt{3}}{2}=\frac{8+2\sqrt{3}-a^2}{4(\sqrt{3}+1)}$
$2\sqrt{3}(\sqrt{3}+1)=8+2\sqrt{3}-a^2$
$6=8-a^2$
$a=\sqrt{2}$.
Hence
$\frac{a}{sinA}=\frac{b}{sinB}$
$2\sqrt{2}=2\frac{1}{sinB}$
$sinB=\frac{1}{\sqrt{2}}={45}^0$ ...(i)
And
$\frac{a}{sinA}=\frac{c}{sinC}$
$2\sqrt{2}=\frac{\sqrt{3}+1}{sinC}$
$sinC=\frac{\sqrt{3}+1}{2\sqrt{2}}={75}^0$
Hence
$\frac{{75}^0-{45}^0}{2}={15}^0$.