Question

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

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. None of these

Answer 1

The correct option is A ${15}^{\circ }$

Given

$A={30}^{\circ },b=2,c=\sqrt{3}+1$

So $CosA=\frac{b^2+c^2-a^2}{2bc}$

$cos30=\frac{4+(\sqrt{3}+1{)}^2-a^2}{2\left(2\right)(\sqrt{3}+1)}$

$⟹\sqrt{3}\left(2\right)(\sqrt{3}+1)=4+3+1+2\sqrt{3}-a^2$

$⟹a^2=8+2\sqrt{3}-6-2\sqrt{3}$

$⟹a^2=2⟹a=\sqrt{2}$

Using sine formula,

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

$\frac{\sqrt{2}}{sin30}=\frac{2}{sinB}=\frac{\sqrt{3}+1}{sinC}$

On solving we get

$B={45}^{\circ },C={75}^{\circ }$

$\therefore \frac{C-B}{2}=\frac{75-45}{2}={15}^{\circ }$