In the given figure, O is the centre of the circle.


In the given figure, O is the centre of the circle. If ABD = 35° and ∠BAC = 70°, find ∠ACB.



It is clear that BD is the diameter of the circle.
Also, we know that the angle in a semicircle is a right angle.
i.e., BAD = 90°
Now, considering the ΔBAD, we have:
ADB + BAD + ABD = 180°  (Angle sum property of a triangle)
⇒ ADB + 90° + 35° = 180°
⇒ ADB = (180° - 125°) = 55°
Angles in the same segment of a circle are equal.
Hence, ACB = ADB = 55°


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