Chemistry - STEREO ISOMERISM Question with Solution | TestHub

Given:

* $a=1+\sqrt{3}$

* $b=2$

* $\angle C=60^{\circ }$

Using the cosine rule:

$$c^2=a^2+b^2-2ab\cos C$$

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

$$c^2=(4+2\sqrt{3})+4-(2+2\sqrt{3})=6$$

$$c=\sqrt{6}$$

Since $b=2<c=\sqrt{6}<a=1+\sqrt{3}$, the smallest side is $b$. Therefore, the smallest angle is $\angle B$.

Using the sine rule:

$$\frac{b}{\sin B}=\frac{c}{\sin 60^{\circ }}$$

$$\sin B=\frac{b\sin 60^{\circ }}{c}=\frac{2\cdot \frac{\sqrt{3}}{2}}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{6}}=\frac{1}{\sqrt{2}}$$

$$B=45^{\circ }=\frac{\pi }{4}$$

Answer: (A) $\boxed{\frac{\pi }{4}}$