Question

If in a $△ABC,\angle A=\frac{\pi }{6}$ and $b:c=2:\sqrt{3}$, find $\angle B$

Answer 1

$\frac{b}{c}=\frac{2}{\sqrt{3}}...\left(1\right)$ $\angle A=\frac{\pi }{6}...$ [Given]

Using cosine rule

$cos\angle A=\frac{b^2+c^2-a^2}{2bc}$

$cos\frac{\pi }{6}=\frac{b^2+c^2-a^2}{2bc}$

$\frac{\sqrt{3}}{2}=\frac{b^2+c^2-a^2}{2abc}$

$\sqrt{3bc}=b^2+c^2-a^2$

Dividing throughout by $b^2$

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

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

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

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

${\left(\frac{a}{b}\right)}^2=\frac{1}{4}$

$\frac{a}{b}=\frac{1}{2}...\left(2\right)$

Multiplying equations (1) and (2)

$\frac{b}{c}\times \frac{a}{b}=\frac{2}{\sqrt{3}}\times \frac{1}{2}$

$\frac{a}{c}=\frac{1}{\sqrt{3}}$

$cos\angle B=\frac{a^2+c^2-b^2}{2ac}$ - [using cosine rule]

$cos\angle B=\frac{a}{2c}+\frac{c}{2a}-\frac{1}{2}\times \frac{b}{a}\times \frac{b}{c}$

$=\frac{1}{2\times \sqrt{3}}+\frac{\sqrt{3}}{2}-\frac{1}{2}\times 2\times \frac{2}{\sqrt{3}}$

$=\frac{1}{2\sqrt{3}}+\frac{\sqrt{3}}{2}-\frac{2}{\sqrt{3}}$

$\therefore =\frac{\sqrt{3}}{2}-\frac{3}{2\sqrt{3}}$

$\therefore cos\angle B=0$

$cos\angle B={90}^{\circ }$