One of the angles of a parallelogram is 420. What is the sum of half the smallest angle and twice the largest angle of the parallelogram ?
We know that consecutive angles of a parallelogram are supplementary (ie $$180^{o}$$).
Given $$\angle$$ ABC is $$42^{o}$$.
Hence $$\angle$$ DAB must be $$138^{o}$$.
$$ \frac{\angle ABC}{2} + 2\times \angle DAB$$ = $$21^{o} + 276^{o} = 297^{o}.$$
Hence Option C is the correct answer.
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