Question

In the figure, given alongside, AOB is a diameter of the circle and ∠AOC = 110°. Find ∠BDC.

Answer 1

Join AD.

Here, `∠ADC = 1/2 ∠AOC`

= `1/2xx 110^circ`

= 55°

(Angle at the centre is double the angle at the circumference subtended by the same chord)

Also, ∠ADB = 90°

(Angle in a semicircle is a right angle)

∴ ∠BDC = 90° – ∠ADC

= 90° – 55°

= 35°