Question

In the adjoining figure, BC = CD. Find ∠ACB.

Answer 1

We are given:

• ∠DAB = 140°
• ∠ADC = 120°
• BC = CD   ...(So triangles ΔBCD is isosceles)

We are to find:

∠ACB

Step-wise calculation:

Step 1: Use the straight line property at point A.

∠DAB + ∠BAD = 180°   ...(Linear pair)

⇒ ∠BAD = 180° – 140°

⇒ ∠BAD = 40°

Step 2: In triangle ΔADC.

We are given:

∠ADC = 120°

∠DAB = 40°   ...(Just found)

So in triangle ΔADC, the third angle is:

∠ACD = 180° – (120° + 40°)

∠ACD = 20°

Step 3: Use an isosceles triangle ΔBCD.

Given:

BC = CD

So, ∠CBD = ∠CDB

Let x = ∠CBD = ∠CDB

Then in triangle ΔBCD:

x + x + ∠BCD = 180°

2x + 20° = 180°

2x = 160°

x = 80°

Step 4: Find ∠ACB.

From the diagram:

∠ACB = ∠DCB

∠ACB = 80°