question_answer

The diagonals of a parallelogram ABCD intersect at O. If \[{{240}^{o}}\], then \[{{60}^{o}}\] is

A)  \[{{120}^{o}}\]                              

B) \[{{180}^{o}}\]

C)  \[\angle R={{138}^{o}}\]                          

D)  \[{{90}^{o}}\]

Correct Answer: D

Solution :

\[\Rightarrow \] or \[\angle 1+\angle 2+\angle 3={{180}^{o}}\] Alternate Method: In triangle, OAB, since sum of angles is\[\Rightarrow \], we have \[BD=CD\]