Question

Given : In quadrilateral ABCD ; $\angle C={64}^o,\angle D=\angle C-8^o;\angle A=5(a+2{)}^{o}and\angle B=2(2a+7{)}^o$.

Answer 1

$\because \angle C={64}^o\left(Given\right)\therefore \angle D=\angle C=8^o={64}^o-8^o={56}^o\angle A=5(a+2{)}^o\angle B=2(2a+7{)}^{o}Now\angle A+\angle B+\angle C+\angle D={360}^{o}5(a+2{)}^o+2(2a+7{)}^o+{64}^o+{56}^o={360}^{o}5a+10+4a+{14}^o+{64}^o+{56}^o={360}^{o}9a+{144}^o={360}^{o}9a={360}^o-{144}^{o}9a={216}^{o}a={24}^o\therefore \angle A=5(a+2)=5(24+2)={130}^o$.