The angles of a triangle are in the ratio 5:1 | vedclass

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(D) Let the angles of the triangle be $5x, x, 6x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $5x + x + 6x = 180^{\circ}$,which giv

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$\sqrt{3}-1: 2 \sqrt{2}$

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(D) Let the angles of the triangle be $5x, x, 6x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $5x + x + 6x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.The angles are $75^{\circ}, 15^{\circ}, 90^{\circ}$.Let the sides opposite to these angles be $a, b, c$ respectively. By the Sine Rule,$\frac{a}{\sin 75^{\circ}} = \frac{b}{\sin 15^{\circ}} = \frac{c}{\sin 90^{\circ}} = k$.The smallest side is $b$ (opposite to $15^{\circ}$) and the greatest side is $c$ (opposite to $90^{\circ}$).The ratio of the smallest side to the greatest side is $\frac{b}{c} = \frac{\sin 15^{\circ}}{\sin 90^{\circ}}$.We know $\sin 15^{\circ} = \sin(45^{\circ}-30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.Since $\sin 90^{\circ} = 1$,the ratio is $\frac{\sqrt{3}-1}{2\sqrt{2}} : 1 = \sqrt{3}-1 : 2\sqrt{2}$.

The angles of a triangle are in the ratio $5:1:6$. The ratio of the smallest side to the greatest side is:

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