In quadrilateral ABCD,angle A = 100^{circ},an | vedclass

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(B) The sum of the interior angles of a quadrilateral is always $360^{\circ}$.Given that $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.Sub

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$60$

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(B) The sum of the interior angles of a quadrilateral is always $360^{\circ}$.Given that $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.Substituting the given values: $100^{\circ} + 80^{\circ} + 120^{\circ} + \angle D = 360^{\circ}$.Summing the known angles: $300^{\circ} + \angle D = 360^{\circ}$.Therefore,$\angle D = 360^{\circ} - 300^{\circ} = 60^{\circ}$.

In quadrilateral $ABCD$,$\angle A = 100^{\circ}$,$\angle B = 80^{\circ}$,and $\angle C = 120^{\circ}$. Find $\angle D$. (in $^{\circ}$)

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