With usual notations,if the angles of a trian | vedclass

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(D) Let the angles be $x, 2x, 3x$. Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\

Vedclass · Accepted Answer

$1: \sqrt{3}: 2$

Vedclass · Answer

(D) Let the angles be $x, 2x, 3x$. 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 of the triangle are $A = 30^{\circ}, B = 60^{\circ}, C = 90^{\circ}$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Substituting the values,$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$. $\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$. Multiplying by $1/2$,we get $a : b : c = \sin 30^{\circ} : \sin 60^{\circ} : \sin 90^{\circ} = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$. Multiplying throughout by $2$,we get $a : b : c = 1 : \sqrt{3} : 2$.

With usual notations,if the angles of a triangle are in the ratio $1: 2: 3$,then their corresponding sides are in the ratio.

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