In a triangle ΔABC, AB = AC and ∠A = 80°, then find ∠B.

Correct — D. The stem is an image: it reads AB = AC and ∠A = 80°, and asks for ∠B. The four choices are images too, showing 40°, 80°, 45° and 50°.

Rule: angles opposite equal sides are equal. AB = AC, so ∠B = ∠C.

Angle sum of a triangle:
80° + ∠B + ∠C = 180°
80° + 2∠B = 180°
2∠B = 100°

∠B = 100° ÷ 2 = 50° → option (d), the image reading 50°.

Why the others are wrong

• (a)
  Forty is half of 80° — the vertex angle halved. That is what the bisector from A produces, not a base angle. The base angles share the remaining 100°, not the 80°.
• (b)
  Eighty simply copies the vertex angle. ∠A sits between the two equal sides, so it is the odd angle out. The pair that must match is ∠B and ∠C, facing AC and AB.
• (c)
  Forty-five is the base angle of an isosceles right triangle, where the vertex angle is 90°. Here the vertex angle is 80°, so the two base angles share 100°, not 90°.

An isosceles triangle hands you one equation free: the angles opposite the equal sides are equal. Put that beside the angle sum and two unknowns collapse into one.

With the equal sides meeting at A, each base angle is (180° − ∠A) ⁄ 2.

Direction matters as much as the fact. Equal sides force equal opposite angles, and equal angles force equal opposite sides — the theorem and its converse are both examinable.

Key facts

• Angles opposite equal sides of a triangle are equal — the base-angle theorem.
• The three angles of any triangle add to 180°.
• For an isosceles triangle with vertex angle ∠A, each base angle is (180° − ∠A) ⁄ 2.
• AB = AC forces ∠B = ∠C, because ∠B faces AC and ∠C faces AB.

Common traps

• Reading the 80° as a base angle rather than the vertex angle
• Halving 80° instead of halving the leftover 100°
• Assuming two equal sides make the triangle equilateral

No directly related past PYQ was found.