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