In a Delta ABC,if sin A + sin B + sin C = 1 + | vedclass

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(D) Given: $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$.For a right-angled isosceles triangle,the angles are $9

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Right angled isosceles

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(D) Given: $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$.For a right-angled isosceles triangle,the angles are $90^{\circ}, 45^{\circ}, 45^{\circ}$.Let $A = 90^{\circ}, B = 45^{\circ}, C = 45^{\circ}$.Then $\sin 90^{\circ} + \sin 45^{\circ} + \sin 45^{\circ} = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 1 + \frac{2}{\sqrt{2}} = 1 + \sqrt{2}$.Also,$\cos 90^{\circ} + \cos 45^{\circ} + \cos 45^{\circ} = 0 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Since these values satisfy the given equations,the triangle is a right-angled isosceles triangle.

In a $\Delta ABC$,if $\sin A + \sin B + \sin C = 1 + \sqrt{2}$ and $\cos A + \cos B + \cos C = \sqrt{2}$,then the triangle is:

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