question_answer

In the figure given below, AO = CD where 0 is the centre of the circle. What is \[\angle APB\]equal to?

A) \[60{}^\circ \]              

B)  

C) 45°                   

D)  

Correct Answer: A

Solution :

AO = CD \[\Rightarrow \] OC = OD = CD [\[\therefore \] AO = OC = OD= radii] A COD is equilateral. \[\angle \]x + \[\angle \]y = 180 - 60 and \[\angle \]x = \[\angle \]y     \[\therefore \] \[\angle \]2x = 120° \[\Rightarrow \] \[\angle \]x= 60° and \[\Delta \]AOC is equilateral. \[\therefore \] \[\angle \]DCP = 180° - 120° = 60° and \[\angle \]CDP = 60° \[\therefore \]\[\angle \]APB = 360° - (60° + 120° + 120°) \[\angle \]APB = [60°]