Here is the question: In the diagram below, AOB is a diameter of the center O. OC//AD and <OCA=48 degree.
Calculate <ABC.
Below is the diagram. The angle inside is 48 degree.

42...!

Me think is 132 degree

benbrains: 42...!

How did you get 42? I need the solution not just the answer.

seaman007: Me think is 132 degree

Show the working please cos that is what I need.

Triangle ABC is in the semi-circle making <BCA=90 ( or angle at d circumference is half d angle d centre, i.e 180/2)
hence <BCO +<OCA=90
<BCO + 48=90
.: <BCO =42.
|OB|=|OC| (radii)
<BCO=<ABC=42(base angle of isos. triangle)

I can't see any diagram

k2039: I can't see any diagram

prof K in d building

k2039: I can't see any diagram

The diagram is there pls.

@Timijo, you dont appear to be cool with the answer that I gave.

42

benbrains: Triangle ABC is in the semi-circle making <BCA=90 ( or angle at d circumference is half d angle d centre, i.e 180/2)
hence <BCO +<OCA=90
<BCO + 48=90
.: <BCO =42.
|OB|=|OC| (radii)
<BCO=<ABC=42(base angle of isos. triangle)

Thanks, your solution has given me a hint.

Timijo: Thanks, your solution has given me a hint.

Actually he is correct.

k2039: Actually he is correct.

Have you seen the diagram now?

Timijo: Have you seen the diagram now?

Yes, but do you understand what he did?.

k2039: Yes, but do you understand what he did?.

Yes, I understood it perfectly. I proved it and it is correct.