In the given figure, AB = AD. ∠ACB = 95° + ∠BAC and ∠BAD = 150°. Find ∠ACB.

130°

In the given figure, AB = AD. ∠ACB = 95° + ∠BAC and ∠BAD = 150°.

∠ACB = 95° + ∠BAC

Also in triangle ABC, ∠ACB + ∠BAC + ∠ABC = 180

95° + 2∠BAC + ∠ABC = 180

2∠BAC + ∠ABC = 180° - 95° = 85°

Also, in triangle ABD,   ∠ABD + ∠BAD + ∠ADB = 180°

AD = AB, thus ∠ABD = ∠ADB and ∠BAD = 150°

2∠ABD + 150° = 180°

∠ABD = ∠ABC = 15°

2∠BAC + 15° = 85°

∠BAC = 35°

∠ACB + 35 + 15 = 180

∠ACB = 130

The correct answer is Option (4) → 130°