If the angles of a triangle are in the ratio | vedclass

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(D) Given the ratio of the angles of a triangle is $1: 1: 4$. Let the angles be $A, B$,and $C$.$	herefore A: B: C = 1: 1: 4$.Let $A = x, B = x$,and $

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

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(D) Given the ratio of the angles of a triangle is $1: 1: 4$. Let the angles be $A, B$,and $C$.$	herefore A: B: C = 1: 1: 4$.Let $A = x, B = x$,and $C = 4x$.Since $A + B + C = 180^{\circ}$,we have $x + x + 4x = 180^{\circ}$,which implies $6x = 180^{\circ}$,so $x = 30^{\circ}$.Thus,$A = 30^{\circ}, B = 30^{\circ}$,and $C = 120^{\circ}$.The largest angle is $120^{\circ}$,so the largest side is $c$.The ratio of the perimeter to the largest side is $(a + b + c) : c$.Using the sine rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.Ratio $= (2R \sin 30^{\circ} + 2R \sin 30^{\circ} + 2R \sin 120^{\circ}) : 2R \sin 120^{\circ}$.Ratio $= (\sin 30^{\circ} + \sin 30^{\circ} + \sin 120^{\circ}) : \sin 120^{\circ}$.Ratio $= (\frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2} = (1 + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2} = (2 + \sqrt{3}) : \sqrt{3}$.

If the angles of a triangle are in the ratio $1: 1: 4$,then the ratio of the perimeter of the triangle to its largest side is

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