Question

In the adjoining figure, ABCD is an isosceles trapezium in which AB || DC and AD = BС. If ∠ACD = 30° and ∠ADB = 25°, then find:

1. ∠AOD
2. ∠CBD

Answer 1

Given:

ABCD is an isosceles trapezium with AB || DC and AD = BC diagonals AC and BD meet at O.

∠ACD = 30° and ∠ADB = 25°.

Step-wise calculation:

1. Angle between the diagonals (∠AOD):

At C, the ray C. A makes 30° with CD given: ∠ACD = 30°. 

Thus, AC is inclined 30° above the direction CD to the left.

By symmetry of the isosceles trapezium, AD = BC and AB || DC, the ray D.

B is symmetrically inclined 30° above the direction DC to the right.

Hence, BD makes 30° with the base in the opposite sense.

Therefore, the acute angle between AC and BD

= 30° + 30°

= 60°

So, ∠AOD = 60°.

2. Angle ∠CBD:

Let α = ∠DAB = ∠ABC   ...(Base angles equal in an isosceles trapezium)

In triangle ADB:

α + 25° + ∠ABD = 180°

⇒ ∠ABD = 155° – α

At vertex B the ray BA, BD and BC occur in that order.

So, ∠ABD + ∠CBD = ∠ABC = α. 

Hence, ∠CBD = α – ∠ABD

= α – (155° – α)

= 2α – 155°

We now find α.

From the symmetry and the given ∠ACD = 30° one gets by angle-chasing / coordinate-symmetry that α = 125°.

This value satisfies the trapezium symmetry and the given ∠ADB = 25°.

Substitute α = 125°:

∠CBD = 2 × 125° – 155° 

= 250° – 155°

= 95°