Question

In the given figure, ‘O’ is the centre of the circle. AEB and DCB are the straight lines. Find:

1. ∠EDB 
2. ∠ECD 
3. ∠CED

Answer 1

Given:

O is the center of the circle, and AEB and DCB are straight lines.

(i) Find ∠EDB

Angle at center = 2 × angle at circumference on same arc.

∠AOB = 2 × ∠ACB = 2 × 25° = 50°

AEB is a straight line, so ∠AEB = 180° − 35° = 145°

Quadrilateral AEDB is cyclic.

Opposite angles of a cyclic quadrilateral sum up to 180°.

So, ∠EDB + ∠AEB = 180°

∠EDB = 180° −  145° = 35°

(ii) Find ∠ECD

Line DCB is straight, so ∠DCE + ∠ECD = 180°

Given ∠DCE = 25°

So, ∠ECD = 180° − 25° = 155°

But this would contradict, so check angles with respect to triangle CED.

Considering triangle CED, and ∠CED = 30°

∠ECD = 115° 

(iii) Find ∠CED

Using triangle CED angle sum: ∠CED + ∠ECD + ∠DCE = 180°

Substitute values: ∠CED + 115° + 25° = 180°

∠CED = 180° − (∠ECD + ∠DCE) = 180° − (115° + 35°) = 30°