Question

In a $\Delta ABC$, if $\angle A={30}^0$ and $b:c=2:\sqrt{3}$, then $\angle B=?$

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

Let $b=2k$ and $c=\sqrt{3}k$. Then,

$cosA=\frac{(b^2+c^2-a^2)}{2bc}\to cos{30}^0=\frac{4k^2+3k^2-a^2}{4\sqrt{3}{k}^2}$

$\therefore 7k^2-a^2=4\sqrt{3}{k}^2\times \frac{\sqrt{3}}{2}[\because cos{30}^0=\frac{\sqrt{3}}{2}]$$\Rightarrow k^2=a^2\Rightarrow a=k\Rightarrow b=2a,c=\sqrt{3}a$.

Now, $\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow \frac{a}{sin{30}^0}=\frac{2a}{sinB}\Rightarrow sinB=1\Rightarrow B={90}^0$.