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

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(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triang

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$1 : 2 : 3$

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(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 2k$.Let the angles opposite to sides $a, b, c$ be $A, B, C$ respectively.Then $C = 90^{\circ}$.Using the sine rule or trigonometric ratios in a right triangle:$\sin A = \frac{a}{c} = \frac{k}{2k} = \frac{1}{2} \implies A = 30^{\circ}$.$\sin B = \frac{b}{c} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \implies B = 60^{\circ}$.Thus,the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$.The ratio of the angles is $30^{\circ} : 60^{\circ} : 90^{\circ} = 1 : 2 : 3$.

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

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