Question

In the given figure, O is the centre of the circle. If
ADC = 140°,
find
BAC.
[Hint : ABC = 40° and ACB = 90°.]​

Answer 1

Answer:

50 °

Step-by-step explanation:

OA=OC⇒∠OAC=∠ACO=X

∠ADC+∠AMC=180  

∘

 (cyclic quadrilateral )

∠AMC=180  

∘

−140  

∘

=40  

∘

 

∠ADC=2∠AMC=2×40  

∘

=80  

∘

 

In △AOC  

∠ADC+∠OAC+∠OCA=180  

∘

 

x+80  

∘

+x=180  

∘

 

2x=100  

∘

 

x=50  

∘

Answer 2

OA=OC⇒∠OAC=∠ACO=X

ADC+∠AMC=180 (cyclic quadrilateral )

∠AMC=180° −140° =40 °

∠ADC=2∠AMC=2×40° =80°

In △AOC

∠ADC+∠OAC+∠OCA=180°

x+80°+x=180 °

2x=100°

x=50°