If the angles A, B and C of a triangle ABC ar | vedclass

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(B) The angles of triangle $ABC$ are in the ratio $2:3:7$. Let the common multiple be $x$. $\angle A = 2x, \angle B = 3x, \angle C = 7x$. Since the su

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$\sqrt{2}: 2:(\sqrt{3}+1)$

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(B) The angles of triangle $ABC$ are in the ratio $2:3:7$. Let the common multiple be $x$. $\angle A = 2x, \angle B = 3x, \angle C = 7x$. Since the sum of the angles of a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ} \implies 12x = 180^{\circ} \implies x = 15^{\circ}$. Thus,$\angle A = 30^{\circ}, \angle B = 45^{\circ}, \angle C = 105^{\circ}$. By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Substituting the values: $\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 105^{\circ}}$. We know $\sin 30^{\circ} = \frac{1}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,and $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$. Therefore,$\frac{a}{1/2} = \frac{b}{1/\sqrt{2}} = \frac{c}{(\sqrt{3}+1)/(2\sqrt{2})}$. Multiplying by $\frac{1}{2}$,we get $\frac{a}{\sqrt{2}} = \frac{b}{2} = \frac{c}{\sqrt{3}+1}$. Thus,the ratio $a:b:c = \sqrt{2}: 2: (\sqrt{3}+1)$.

If the angles $A, B$ and $C$ of a triangle $ABC$ are in the ratio $2:3:7$ respectively,then the sides $a, b$ and $c$ are respectively in the ratio

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