question_answer

In a circle with centre O, AOC is a diameter of the circle, BD is a chord and OB and CD are joined. If \[\angle AOB=130{}^\circ ,\] then \[\angle BDC\] is equal to

A) \[30{}^\circ \]                          

B) \[25{}^\circ \]  

C) \[50{}^\circ \]                        

D) \[60{}^\circ \]

Correct Answer: B

Solution :

\[\because \]       \[AOB=130{}^\circ \]

and       \[\angle AOB+\angle BOC=180{}^\circ \]

\[\therefore \]      \[\angle BOC=180{}^\circ -130{}^\circ =50{}^\circ \]

Again,   \[\angle BDC=\frac{1}{2}\,\,\angle BOC\]

[angle at circle is half the angle at centre]

\[\Rightarrow \]   \[\angle BOC=\frac{1}{2}\times 50{}^\circ =25{}^\circ \]