Question

in the figure angles ABC =40 then find angles ABC​

Answer 1

Answer:

In ΔOAB, we have

OA=OB .....(radii of the same circle).

∴ΔOAB is isosceles.

∴∠OAB=∠OBA⇒∠OAB+∠OBA=2∠OAB 

So, ∠AOB+∠OAB+∠OBA=∠AOB+2∠OAB=180o ....(angle sum property of triangles).

∴∠OAB=21(180o−∠AOB) ..........(i)

Now ∠AOB=2∠ACB=2×40o=80o ....(The angle subtended by a chord of a circle at the centre is double of that at the cicumference)

From (i), we have

∠OAB=21(180o−80o)=50o