Sine Rule

Statement

In any triangle (not just right-angled), the ratio of a side to the sine of its opposite angle is constant:

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

Why it’s true

• From the area of a triangle formula: \( \tfrac{1}{2}ab\sin C = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}ca\sin B \).
• Rearranging these equalities gives the sine rule.
• It works for both acute and obtuse angles.

Recipe (how to use it)

1. Label sides and opposite angles clearly.
2. Set up the proportion using known values.
3. Solve for the missing side or angle.

Spotting it

Use the sine rule when you know:

• two angles and one side (AAS or ASA), or
• two sides and a non-included angle (SSA).

Common pairings

• Cosine rule (when sine rule doesn’t apply directly).
• Ambiguous case (two possible triangles for SSA).
• Area formula \(\tfrac{1}{2}ab\sin C\).

Mini examples

1. In triangle ABC, \(A=40^\circ\), \(B=70^\circ\), \(a=8\). Find \(b\). → Use sine rule: \(b/\sin70 = 8/\sin40\). → \(b = (8×\sin70)/\sin40 ≈ 12.0\).
2. In triangle PQR, \(p=6\), \(q=9\), \(P=50^\circ\). Find \(Q\). → \(6/\sin50 = 9/\sin Q\). → \(\sin Q = (9×\sin50)/6 ≈ 1.15\). Not possible, so no triangle exists.
3. In triangle XYZ, \(X=60^\circ\), \(Y=80^\circ\), \(x=10\). Find \(y\). → \(y/\sin80 = 10/\sin60\). → \(y ≈ 11.3\).

Pitfalls

• Forgetting to match sides with opposite angles.
• Using sine rule when cosine rule is needed.
• Ambiguous case (SSA) — sine of an angle may correspond to two possible angles.

Exam strategy

• Check if sine rule applies (look for opposite side–angle pairs).
• Always check if second triangle is possible in SSA cases.
• Round answers sensibly (degrees to 1 d.p. or 2 s.f. unless stated).

Summary

The sine rule links sides and angles in any triangle. It is most useful when working with opposite side–angle pairs and is complementary to the cosine rule.