Question

In ΔABC, AC = BC. DE is drawn parallel to BC through A. ∠EAC = 42°. Find angle x.

Answer 1

Given:

Triangle ABC with AC = BC isosceles triangle

DE is drawn parallel to BC through A

∠EAC = 42°

Find angle x which is ∠DAB

1. Since DE is parallel to BC and AC = BC, triangle ABC is isosceles.

2. Because DE || BC, ∠EAC and ∠ACB are corresponding angles.

3. Thus, ∠ACB = ∠EAC = 42°.

4. In triangle ABC, angles at A and B are equal because AC = BC.

5. So, ∠BAC = ∠ABC = x (say).

6. Sum of angles in triangle ABC is 180°.

So, x + x + 42° = 180°.

2x = 180° – 42°

2x = 138°

x = 69°

7. Angle x = ∠DAB = 69° since ∠DAB = ∠ABC by alternate interior angles due to DE || BC.

Angle x = 69°