angle ACD is an exterior angle of Delta ABC. | vedclass

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(A) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its tw

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(A) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.Therefore,$\angle ACD = \angle A + \angle B$.Given that $\angle ACD = 122^{\circ}$ and $\angle A = 68^{\circ}$.Substituting the values,we get: $122^{\circ} = 68^{\circ} + \angle B$.Solving for $\angle B$: $\angle B = 122^{\circ} - 68^{\circ} = 54^{\circ}$.Thus,$\angle B = 54^{\circ}$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle ACD = 122^{\circ}$ and $\angle A = 68^{\circ}$,then $\angle B = \ldots$ (in $^{\circ}$)

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