Angles of a triangle are in the ratio 4:1:1. | vedclass

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(B) Let the angles of the triangle be $4x, x,$ and $x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $4x + x + x = 180^{\circ}$,which

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

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(B) Let the angles of the triangle be $4x, x,$ and $x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $4x + x + x = 180^{\circ}$,which implies $6x = 180^{\circ}$,so $x = 30^{\circ}$.The angles are $120^{\circ}, 30^{\circ},$ and $30^{\circ}$.By the sine rule,$\frac{a}{\sin 120^{\circ}} = \frac{b}{\sin 30^{\circ}} = \frac{c}{\sin 30^{\circ}} = k$.Thus,$a = k \sin 120^{\circ} = k \frac{\sqrt{3}}{2}$,$b = k \sin 30^{\circ} = k \frac{1}{2}$,and $c = k \sin 30^{\circ} = k \frac{1}{2}$.The greatest side is $a$ (opposite to $120^{\circ}$).The perimeter is $a + b + c = k(\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{2}) = k(\frac{\sqrt{3}+2}{2})$.The ratio of the greatest side to the perimeter is $\frac{a}{a+b+c} = \frac{k \frac{\sqrt{3}}{2}}{k \frac{\sqrt{3}+2}{2}} = \frac{\sqrt{3}}{2+\sqrt{3}}$.

Angles of a triangle are in the ratio $4:1:1$. Then the ratio of its greatest side to its perimeter is

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