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(B) Let the angles of the triangle be $x, 2x$,and $3x$. Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + 3x = 180^{\circ}$ $\R

Vedclass · Accepted Answer

$1: \sqrt{3}: 2$

Vedclass · Answer

(B) Let the angles of the triangle be $x, 2x$,and $3x$. Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$. Thus,the angles are $30^{\circ}, 60^{\circ}$,and $90^{\circ}$. According to the Sine Rule,the sides $a, b, c$ are proportional to the sines of their opposite angles: $a: b: c = \sin A: \sin B: \sin C$. Substituting the angles: $a: b: c = \sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}$. $a: b: c = \frac{1}{2}: \frac{\sqrt{3}}{2}: 1$. Multiplying by $2$,we get the ratio $1: \sqrt{3}: 2$.

In any triangle,if the angles are in the ratio $1: 2: 3$,then their corresponding sides are in the ratio.

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