Let a, b and c denote the lengths of sides BC | vedclass

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(B) Given $BC = a, CA = b$ and $AB = c$.To find $a: b: c$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$\angle ACB = 180^{\circ} - (

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$1: \sqrt{3}: 2$

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(B) Given $BC = a, CA = b$ and $AB = c$.To find $a: b: c$.Since the sum of angles in a triangle is $180^{\circ}$,we have:$\angle ACB = 180^{\circ} - (\angle BAC + \angle ABC)$$\angle ACB = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.By using the sine rule,we have:$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$Substituting the values of the sine functions:$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$Multiplying by $1/2$,we get:$a : b : c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$Multiplying the ratio by $2$,we get:$a : b : c = 1 : \sqrt{3} : 2$.

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

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