Practice Maths

The Sine Rule

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Key Terms

Sine rule
a/sin A = b/sin B = c/sin C — links two sides and their opposite angles in any triangle.
When to use
Use when two angles + one side are known (AAS/ASA), or two sides + a non-included angle (SSA — check for ambiguous case).
Ambiguous case (SSA)
Given two sides and a non-included angle, there may be 0, 1, or 2 valid triangles; always check with a diagram.
Labelling
Side a is opposite angle A; side b is opposite angle B; side c is opposite angle C.
Finding an angle
Rearrange to sin A / a = sin B / b; then A = sin²(a sin B / b) — always check which quadrant is valid.
Non-right triangle
The sine rule applies to ANY triangle, not just right-angled ones.

Standard Triangle Labelling

In any triangle, we label sides and angles using corresponding lower and upper case letters: side a is opposite angle A, side b is opposite angle B, side c is opposite angle C.

A B C a b c
Labelling rule: Each angle is at a vertex. The side directly opposite it carries the same letter in lower case. Always set up your labelling before writing any formula.

The Sine Rule

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]

QCAA Formula Sheet

When to Use the Sine Rule

  • AAS (two angles and any side known): find any remaining side
  • SSA (two sides and an angle opposite one of them known): find the missing angle — but watch for the ambiguous case

Finding a Missing Side

Choose the ratio that connects two known quantities plus the unknown:

\[\frac{a}{\sin A} = \frac{b}{\sin B} \implies a = \frac{b \sin A}{\sin B}\]

Finding a Missing Angle

\[\frac{\sin A}{a} = \frac{\sin B}{b} \implies \sin A = \frac{a \sin B}{b} \implies A = \sin^{-1}\!\left(\frac{a \sin B}{b}\right)\]

The Ambiguous Case (SSA)

When using SSA, the inverse sin can give two results: A and 180° − A. Both may be valid if each produces a triangle where all angles are positive and sum to 180°.

Ambiguous case check: Given angle B, side b and side a (with A opposite a):
Compute sin A = a sin B / b.
• If sin A > 1: no solution exists.
• If sin A = 1: one right-angle solution (A = 90°).
• If sin A < 1: A1 = sin&sup8;&sup9;(sin A). Also try A2 = 180° − A1.
For A2 to be valid: B + A2 < 180°.

Worked Example 1 — Finding a Side (AAS)

Problem: In triangle ABC, A = 42°, B = 73°, a = 14. Find b.

Step 1: Check: C = 180° − 42° − 73° = 65°. All angles valid.

Step 2: Use sine rule with the pair (a, A) and (b, B):

\[\frac{b}{\sin B} = \frac{a}{\sin A}\] \[b = \frac{a \sin B}{\sin A} = \frac{14 \times \sin 73°}{\sin 42°} = \frac{14 \times 0.9563}{0.6691} \approx \mathbf{20.01}\]

Worked Example 2 — Ambiguous Case (SSA)

Problem: In triangle ABC, a = 8, b = 11, B = 60°. Find angle A.

Step 1: Use the sine rule to find sin A:

\[\sin A = \frac{a \sin B}{b} = \frac{8 \times \sin 60°}{11} = \frac{8 \times 0.8660}{11} = \frac{6.928}{11} \approx 0.6298\]

Step 2: First solution: A1 = sin&sup8;&sup9;(0.6298) ≈ 39.1°

Step 3: Second solution: A2 = 180° − 39.1° = 140.9°

Check A2: B + A2 = 60° + 140.9° = 200.9° > 180°. Invalid!

Answer: Only one solution: A ≈ 39.1°

Hot Tip: Always verify the ambiguous case by checking whether B + A2 < 180°. If it is not, the second solution is geometrically impossible and must be discarded.

Why Does the Sine Rule Work?

Drop a perpendicular from C to side c (base AB). Call its height h. Then:

  • In the left triangle: h = b sin A
  • In the right triangle: h = a sin B

Since both equal h: b sin A = a sin B  ⇒  a/sin A = b/sin B. This applies to all three pairs, giving the full sine rule.

Step-by-Step Approach

  1. Label the triangle: identify which sides and angles are given and which are unknown.
  2. Decide AAS or SSA. Choose the appropriate pair of (side, opposite angle) to use.
  3. Write the sine rule equation with only one unknown. Solve for it.
  4. If finding an angle (SSA), check the ambiguous case.
  5. If all three angles are now known, find remaining sides as needed.

Guide Example

Problem: In triangle PQR, P = 52°, Q = 71°, p = 30 cm. Find q.

R = 180° − 52° − 71° = 57°

\[\frac{q}{\sin Q} = \frac{p}{\sin P} \implies q = \frac{30 \sin 71°}{\sin 52°} = \frac{30 \times 0.9455}{0.7880} \approx 35.99 \text{ cm}\]
Common Mistake: Writing the sine rule upside down when finding a side (writing sin A / a instead of a / sin A). Keep the pattern consistent: side over sine when finding a side; sine over side when finding an angle.

Mastery Practice

See Answers ➔
  1. Fluency Find the missing side using the sine rule:
    1. A = 55°, B = 75°, a = 12. Find b.
    2. B = 40°, C = 65°, b = 18. Find c.
    3. A = 36°, C = 82°, c = 25. Find a.
    View Solution
  2. Fluency Find the missing angle using the sine rule:
    1. a = 9, b = 13, A = 38°. Find B.
    2. a = 18, c = 22, C = 65°. Find A.
    3. b = 14, c = 11, B = 82°. Find C.
    View Solution
  3. Fluency In a triangle, A = 48°, B = 67°, b = 21. Find the third angle C, then find all remaining sides a and c. View Solution
  4. Understanding Ambiguous case: a = 10, b = 14, A = 44°. Find all possible values of angle B and describe the corresponding triangles. State how many triangles are possible. View Solution
  5. Understanding A surveyor stands at point C and sights two points A and B. Angle ACB = 74°, angle CAB = 56°, and AB = 85 m. Find:
    1. Angle ABC.
    2. Distance CA.
    3. Distance CB.
    View Solution
  6. Understanding Two ships leave the same port. Ship 1 sails on bearing 040° and Ship 2 on bearing 110°. After some time, Ship 1 is 80 km from port. The angle at the port between the two ships is 70° (i.e., 110° − 040° = 70°). The distance between the two ships is 95 km. Find the distance of Ship 2 from port. View Solution
  7. Understanding An architect designs a triangular roof truss. The base is 14 m wide. The angles at the two base corners are 52° and 63°. Find:
    1. The apex angle.
    2. The lengths of the two sloping sides.
    3. Which sloping side is longer and by how much?
    View Solution
  8. Understanding Point P is observed from two points A and B, where B is due east of A and AB = 600 m. From A, the bearing of P is N 25° E. From B, the bearing of P is N 60° W. Find the distances AP and BP. View Solution
  9. Problem Solving A triangular paddock has one side of 320 m. The angles at each end of this side are 61° and 74°. Find:
    1. The remaining angle and the other two sides.
    2. The total perimeter of the paddock.
    3. The area of the paddock (using Area = ½ base × height, where height is calculated using trigonometry).
    View Solution
  10. Problem Solving In triangle ABC, AB = 8 cm, BC = 6 cm, and angle BAC = 35°.
    1. Use the sine rule to find angle BCA.
    2. Check whether the ambiguous case applies.
    3. For each valid triangle, find the remaining side AC.
    4. Find the area of each valid triangle using Area = ½ab sin C.
    View Solution