In parallelogram $ABCD$, $\angle A = 70^{\circ}$. Then $\angle B$ is
In parallelogram $ABCD$, $\angle A = 70^{\circ}$. Then $\angle B$ is
- $110^{\circ}$
- $70^{\circ}$
- $90^{\circ}$
- $180^{\circ}$
Solution
Adjacent angles supplementary: $180 - 70 = 110^{\circ}$.
Asked in: MH-SSC-9
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