ABCD is a trapezium in which BC ∥ AD and AC = CD. If ∠ABC = 69° and ∠BAC = 23° , then what is the measure of ∠ACD (in degree) .?

Question

Question

This question was previously asked in

RRB NTPC Graduate Level CBT-I Official Paper (Held On: 06 Jun, 2025 Shift 1)

Answer 1

Given:

ABCD is a trapezium (trapezoid) with BC parallel to AD (BC || AD).

AC = CD (This means triangle ACD is an isosceles triangle).

Angle ABC (∠ABC) = 69°

Angle BAC (∠BAC) = 23°

Find: The measure of Angle ACD (∠ACD).

Calculation:

Find Angle ACB in Triangle ABC.

The sum of angles in any triangle is 180°.

In Triangle ABC:

∠ACB = 180° - (∠ABC + ∠BAC)

∠ACB = 180° - (69° + 23°)

∠ACB = 180° - 92°

∠ACB = 88°

Use the property of parallel lines to find Angle CAD.

Since BC is parallel to AD (BC || AD) and AC is a transversal line, the alternate interior angles are equal.

∠CAD = ∠ACB

Since ∠ACB = 88° (from Step 1), then ∠CAD = 88°

Find Angle ACD in Triangle ACD.

We are given that AC = CD. This means Triangle ACD is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal.

The angle opposite side CD is ∠CAD.

The angle opposite side AC is ∠CDA.

Therefore, ∠CDA = ∠CAD = 88°.

Now, apply the sum of angles property to Triangle ACD:

∠ACD + ∠CAD + ∠CDA = 180°

∠ACD + 88° + 88° = 180°

∠ACD + 176° = 180°

∠ACD = 180° - 176°

∠ACD = 4°

The measure of ∠ACD is 4 degrees.

Download Solution PDF