IB Maths AA SL Topic 3 — Geometry & Trig Paper 1 & 2 🎯 Skill ~3 min practice

AA SL Sine Rule skills

For triangles that aren’t right-angled. The sine rule pairs each side with the angle opposite it — so if you can spot one matching pair, you can always find a missing piece. Use it when you have an angle and the side opposite, plus one more bit of info.

The Method

asin A  =  bsin B  =  csin C side ÷ sin(opposite angle) is the same for all three pairs · ✓ in formula booklet
  1. Label sides and angles correctly. Side a is opposite angle A, side b is opposite angle B, side c is opposite angle C.
  2. Pick two pairs — one with both pieces known, one with the unknown. Set them equal.
  3. Cross-multiply and solve. If you’re finding an angle, take sin⁻¹ at the end.

The labelling rule

A B C c a b

Side a is opposite angle A — and same for b/B, c/C. Always label the triangle this way before plugging into the formula.

When to use the sine rule

✓ USE sine rule when

You have an angle and its opposite side, plus one more side or angle.

Triangle types: AAS, ASA, SSA

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Ambiguous case (SSA)

If the given angle is acute and you’re solving sin θ = ..., there can be two valid answers: θ and (180° − θ). Always check whether both fit the triangle’s angle sum (180°). If the question says “obtuse triangle”, pick the obtuse one.

Worked examples

WE 1 EASY

In triangle ABC, A = 40°, B = 75°, side a = 8 cm. Find side b.

step 1 — pick the right pairs have a/sin A pair, want b → use b/sin Bstep 2 — set up sine rule b / sin 75° = 8 / sin 40°step 3 — solve b = 8 × sin 75° / sin 40° b = 8 × 0.9659 / 0.6428b ≈ 12.0 cm (3 sf) always isolate what you want first — multiply both sides by sin(opposite)!
WE 2 MEDIUM

In triangle PQR, p = 12, q = 9, P = 65°. Find angle Q.

step 1 — set up sine rule for angle sin Q / 9 = sin 65° / 12step 2 — solve for sin Q sin Q = 9 × sin 65° / 12 sin Q = 9 × 0.9063 / 12 = 0.6797step 3 — take inverse sin Q = sin⁻¹(0.6797)Q ≈ 42.8° (3 sf) when finding an angle, flip the formula — sin θ on top of the side it’s opposite.
WE 3 HARD

In triangle ABC, A = 35°, a = 7, b = 11. Triangle is obtuse. Find angle B.

step 1 — sine rule for sin B sin B / 11 = sin 35° / 7 sin B = 11 × sin 35° / 7 sin B = 0.9013step 2 — both possible angles B = sin⁻¹(0.9013) ≈ 64.3° or B = 180° − 64.3° ≈ 115.7°step 3 — pick the obtuse one obtuse means > 90° → 115.7° check: 35° + 115.7° = 150.7° < 180° ✓B ≈ 115.7° (1 dp) SSA always check both 180° − θ — the question hints which to pick!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Sketch the triangle if it helps — labels matter.
Q1 EASY In triangle ABC, A = 50°, B = 60°, a = 10. Find b. Show answer ▼Hide answer ▲
b / sin 60° = 10 / sin 50° b = 10 × sin 60° / sin 50° b ≈ 11.3 cm (3 sf)
Q2 EASY In triangle PQR, p = 14, P = 80°, Q = 45°. Find q. Show answer ▼Hide answer ▲
q / sin 45° = 14 / sin 80° q = 14 × sin 45° / sin 80° q ≈ 10.1 (3 sf)
Q3 MEDIUM In triangle ABC, a = 6, b = 9, A = 30°. Find B (acute). Show answer ▼Hide answer ▲
sin B / 9 = sin 30° / 6 sin B = 9 × 0.5 / 6 = 0.75 B = sin⁻¹(0.75) B ≈ 48.6° (3 sf)
Q4 MEDIUM In triangle ABC, A = 55°, C = 70°, a = 12. Find c. Show answer ▼Hide answer ▲
c / sin 70° = 12 / sin 55° c = 12 × sin 70° / sin 55° c ≈ 13.8 (3 sf)
Q5 HARD In triangle ABC, a = 8, b = 13, A = 28°. The triangle is obtuse — find B. Show answer ▼Hide answer ▲
sin B / 13 = sin 28° / 8 sin B = 13 × sin 28° / 8 = 0.7627 two possible angles B = sin⁻¹(0.7627) ≈ 49.7° or B = 180° − 49.7° = 130.3° obtuse → pick 130.3° B ≈ 130° (3 sf) check: 28° + 130.3° = 158.3° < 180° ✓

⚠ Common mistakes

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Want the theory?

Read the full Sine & Cosine Rules notes for the proof, the link to area = ½ ab sin C, and worked problems involving bearings.

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