In the given figure, AB = AD. ∠ACB = 95° + ∠BAC and ∠BAD = 150°. Find ∠ACB.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 130°
In an isosceles triangle the angles opposite the equal sides are equal, and the three angles of any triangle add up to 180°.
Triangle ABD (AB = AD, so ∠ABD = ∠ADB):
The two base angles share what is left of 180° after the angle at A: \(\angle ABD+\angle ADB=180^\circ-150^\circ=30^\circ\)
Being equal, each base angle is half of it: \(\angle ABD=\angle ADB=15^\circ\)
Triangle ABC (taking \(\angle BAC=x\), so \(\angle ACB=95^\circ+x\) as given):
Angle sum of triangle ABC: \(\angle ABC+\angle BAC+\angle ACB=180^\circ\)
\(\angle ABC\) is the same angle as \(\angle ABD\), 15°: \(15^\circ+x+(95^\circ+x)=180^\circ\)
Solving: \(2x=70^\circ\), so \(\angle BAC=x=35^\circ\)
The angle asked for: \(\angle ACB=95^\circ+35^\circ=130^\circ\)
