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

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(A) Let the angles be $4x, x,$ and $x.$ Since the sum of angles in a triangle is $180^\circ,$ we have $4x + x + x = 180^\circ.$ $6x = 180^\circ \Right

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the angles be $4x, x,$ and $x.$ Since the sum of angles in a triangle is $180^\circ,$ we have $4x + x + x = 180^\circ.$ $6x = 180^\circ \Rightarrow x = 30^\circ.$ The angles are $120^\circ, 30^\circ,$ and $30^\circ.$ Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k.$ Thus,$a = k \sin 120^\circ, b = k \sin 30^\circ, c = k \sin 30^\circ.$ The longest side is $a$ (opposite to $120^\circ$). The ratio of the longest side to the perimeter is $\frac{a}{a+b+c} = \frac{\sin 120^\circ}{\sin 120^\circ + \sin 30^\circ + \sin 30^\circ}.$ $= \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3} + 2}{2}} = \frac{\sqrt{3}}{2 + \sqrt{3}}.$

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

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