In a ΔLMN, ∠ L = 118° and LM = LN. Find ∠ N.

Correct — A. LM = LN, so the triangle is isosceles about vertex L and the two base angles are equal.

The angles opposite the equal sides are ∠N and ∠M, so ∠M = ∠N
∠M + ∠N = 180° − 118° = 62°
∠N = 62° ÷ 2 = 31° → option (a)

Why the others are wrong

• (b)
  62° — 62° is the whole remainder 180° − 118°, which the two equal base angles share. Halving is the step being tested, and 118° + 62° + 62° comes to 242°.
• (c)
  42° — 42° fits an apex of 180° − 2(42°) = 96°, not the 118° given. With 42° at both M and N the three angles total 202°.
• (d)
  65° — 65° would put 65° at M as well, totalling 248°. It belongs to an apex of 50°, which is not this triangle.

In an isosceles triangle the angles opposite the equal sides are equal. The word opposite is what stops the wrong pair being matched.

LM = LN names the two sides that meet at L. The angles facing them are ∠N and ∠M, so those two are the equal pair and L is the apex.

Once the apex is fixed, each base angle is (180° − apex) ÷ 2. An obtuse apex of 118° forces both base angles to be acute, and 31° sits well under 90°.

Key facts

• Equal sides face equal angles, and the equality is read across the triangle rather than along it.
• Each base angle of an isosceles triangle equals (180° − apex angle) ÷ 2.
• A triangle holds at most one angle of 90° or more, so an apex of 118° forces two acute base angles.

Common traps

• Pairing ∠L with ∠M because LM is one of the named equal sides.
• Stopping at 62°, the sum of the two base angles, instead of halving it.
• Assuming an isosceles triangle must be acute, and so rejecting an apex of 118°.

No directly related past PYQ was found.