Given : In quadrilateral ABCD ; C = 64^ , D = C – 8^ ; A = 5(a+2)… - Vidyadip

Question

Given : In quadrilateral $ABCD ;\angle C = 64^\circ , \angle D = \angle C – 8^\circ ; \angle A = 5(a+2)^\circ$ and $\angle B=2(2a+7)^\circ $.Calculate $\angle A.$

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Answer

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

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