question_answer

In a parallelogram ABCD, if \[\angle \mathbf{A}=\mathbf{4}\angle \mathbf{B}\], then \[\angle \mathbf{A}+\angle \mathbf{C}\]

A)  \[{{260}^{{}^\circ }}\]                      

B)  \[{{150}^{{}^\circ }}\]            

C)  \[{{140}^{{}^\circ }}\]                      

D)  \[{{288}^{{}^\circ }}\]

Correct Answer: D

Solution :

(d): By law of transversal, \[\angle A+\angle B={{180}^{{}^\circ }}\] \[\angle A=4\angle B\] \[\Rightarrow 5\angle B={{180}^{{}^\circ }}=\angle B={{36}^{{}^\circ }}\] \[\angle \,A=\angle C\]     \[\therefore \angle A+\angle C=2\angle A\] \[\Rightarrow 2\times 4\times \angle B=8\angle B\] \[\Rightarrow 8\times {{36}^{{}^\circ }}={{288}^{{}^\circ }}\]