In Delta ABC,angle A : angle B : angle C = 5 | vedclass

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(A) In $\Delta ABC$,the sum of the interior angles is $\angle A + \angle B + \angle C = 180^{\circ}$.Given the ratio $\angle A : \angle B : \angle C =

Vedclass · Accepted Answer

$45^{\circ}, 63^{\circ}, 72^{\circ}$

Vedclass · Answer

(A) In $\Delta ABC$,the sum of the interior angles is $\angle A + \angle B + \angle C = 180^{\circ}$.Given the ratio $\angle A : \angle B : \angle C = 5 : 7 : 8$,let the angles be $5x, 7x,$ and $8x$.Sum of the angles: $5x + 7x + 8x = 180^{\circ}$.$20x = 180^{\circ}$.$x = \frac{180^{\circ}}{20} = 9^{\circ}$.Therefore,the angles are:$\angle A = 5 	imes 9^{\circ} = 45^{\circ}$.$\angle B = 7 	imes 9^{\circ} = 63^{\circ}$.$\angle C = 8 	imes 9^{\circ} = 72^{\circ}$.

In $\Delta ABC$,$\angle A : \angle B : \angle C = 5 : 7 : 8$. Find the measure of each angle of $\Delta ABC$.

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