Question

Please answer Who Answer it correct with full explanation I will mark him as brainliest​

Answer 1

In ΔABC, ∠ABC = ∠ACB. (ΔABC is isosceles)

Thus, ∠ACB = 42°.

If we look at ΔACD, we see that ∠ACB is an exterior angle for it.

Thus, ∠ACB = ∠ACD + ∠CDA (Exterior Angle Theorem)

∠ACD = ∠CDA (ΔACD is isosceles)

Thus, ∠ACB = 2∠CDA

2∠CDA = 42°

∠CDA = 21°

In ΔABD, x is an exterior angle.

Thus, x = ∠ABC + ∠ADB.

x = 21° + 42°

x = 63°