Question

In the given figure, AC is the diameter of the circle with centre O. CD and BE are parallel. Angle ∠AOB = 80° and ∠ACE = 10°.

Calculate:

1. Angle BEC,
2. Angle BCD,
3. Angle CED.

Answer 1

i. ∠BOC = 180° – 80° = 100°  (Straight line)

And ∠BOC = 2∠BEC

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

`=> ∠BEC = (100^circ)/2 = 50^circ`

ii. DC || EB

∴ DCE = ∠BEC = 50°  (Alternate angles)

∴ ∠AOB = 80°

`=> ∠ACB = 1/2 ∠AOB = 40^circ`

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

We have,

∠BCD = ∠ACB + ∠ACE + ∠DCE

= 40° + 10° + 50°

= 100°

iii. ∠BED = 180° – ∠BCD

= 180° – 100°

= 80°

(Pair of opposite angles in a cyclic quadrilateral are supplementary)

`=>` ∠CED + 50° = 80°

`=>` ∠CED = 30°