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

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(A) Let the angles of the triangle be $2x, 3x,$ and $7x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ}$,w

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$\sqrt{2} : 2 : (\sqrt{3} + 1)$

Vedclass · Answer

(A) Let the angles of the triangle be $2x, 3x,$ and $7x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.The angles are $A = 2(15^{\circ}) = 30^{\circ}$,$B = 3(15^{\circ}) = 45^{\circ}$,and $C = 7(15^{\circ}) = 105^{\circ}$.By the Sine Rule,the ratio of the sides $a : b : c$ is $\sin A : \sin B : \sin C$.$a : b : c = \sin 30^{\circ} : \sin 45^{\circ} : \sin 105^{\circ}$.We know $\sin 30^{\circ} = \frac{1}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,and $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.Thus,$a : b : c = \frac{1}{2} : \frac{1}{\sqrt{2}} : \frac{\sqrt{3} + 1}{2\sqrt{2}}$.Multiplying by $2\sqrt{2}$,we get $a : b : c = \sqrt{2} : 2 : (\sqrt{3} + 1)$.

If the angles of a triangle are in the ratio of $2 : 3 : 7$,then the sides are in the ratio of

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