question_answer

Each of the circles of equal radii with centres A and B pas through the centre of one another circle they cut at C and D then\[\angle DBC\]is equal to

A) \[{{60}^{o}}\]                        

B) \[{{100}^{o}}\]

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

D)        \[{{140}^{o}}\]

Correct Answer: C

Solution :

In\[\Delta ABD\], \[AD=BD=AB=\]radius In \[\Delta ACB\], \[AC=CB=AB=\]radius \[\therefore \]      \[\angle DBC={{60}^{o}}+{{60}^{o}}={{120}^{o}}\]