Sine Rule

Video lesson4 worked examples

Theory

The Sine Rule is used for non-right-angled triangles when you know one corresponding side-and-angle pair, plus one additional piece of information.

The formula provided on the exam sheet is:

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

The Golden Rule: use the Sine Rule when you have a matching pair — a side and its opposite angle — plus one more piece of information. Put the unknown on top: use the fractions the “normal” way up to find a side, and flip them (angles on top) to find an angle.

⚠️ Common Examiner Traps

Side opposite angle: each side pairs with the angle opposite it, not the one next to it.
Third angle: if you are given two angles, the third is $180^\circ$ minus their sum — find it if you need its opposite side.
Application problems: a “height from two angles of elevation” question needs the Sine Rule in one triangle first, then basic trig for the height.
Rounding: keep full accuracy from the calculator until the very end.

Worked examples

Example 1

In $\Delta PQR$ , angle $P = 45^\circ$ , angle $Q = 60^\circ$ , and side $p = 10\text{ cm}$ . Find side $q$ .

Step 1: Set up the ratio:

\frac{q}{\sin(60^\circ)} = \frac{10}{\sin(45^\circ)}

Step 2: Multiply both sides by $\sin(60^\circ)$ :

q = \frac{10\sin(60^\circ)}{\sin(45^\circ)}

Answer: $q \approx 12.2\text{ cm}$ .

Example 2

In $\Delta XYZ$ , side $x = 8\text{ cm}$ , side $y = 12\text{ cm}$ , and angle $X = 30^\circ$ . Find angle $Y$ .

Step 1: Set up the ratio with angles on top for easier solving:

\frac{\sin Y}{12} = \frac{\sin(30^\circ)}{8}

Step 2: Rearrange to isolate $\sin Y$ :

\sin Y = \frac{12\sin(30^\circ)}{8} = \frac{6}{8} = 0.75

Answer: $Y = \sin^{-1}(0.75) \approx 48.6^\circ$ .

Example 3

In $\Delta ABC$ , angle $A = 40^\circ$ , angle $B = 80^\circ$ , and side $c = 15\text{ m}$ . Find side $a$ .

Step 1: First find angle $C$ .

C = 180^\circ - (40^\circ + 80^\circ) = 60^\circ

Step 2: Set up the sine rule ratio:

\frac{a}{\sin(40^\circ)} = \frac{15}{\sin(60^\circ)}

Step 3: Solve for $a$ :

a = \frac{15\sin(40^\circ)}{\sin(60^\circ)}

Answer: $a \approx 11.1\text{ m}$ .

Example 4

🎯 Exam-style (height from two angles)

Two observers A and B stand 40 m apart on level ground, in line with the point directly below a balloon. The angle of elevation of the balloon is 30° from A and 45° from B (B is nearer). Calculate the height of the balloon.

Step 1: Work in the triangle ABD, where D is the balloon. The angle at A is 30°. Because A is on the far side of B, the angle DBA is the supplement of the 45° elevation: $180^\circ - 45^\circ = 135^\circ$ .

Step 2: The third angle is $\angle ADB = 180^\circ - 30^\circ - 135^\circ = 15^\circ$ . Use the Sine Rule to find BD:

\frac{BD}{\sin 30^\circ} = \frac{40}{\sin 15^\circ} \implies BD = \frac{40\sin 30^\circ}{\sin 15^\circ} \approx 77.3 \text{ m}

Step 3: Now drop to the right-angled triangle under B. The height is $BD \sin 45^\circ$ :

\text{height} = 77.3 \times \sin 45^\circ \approx 54.7 \text{ m}

Answer: approximately 54.7 m.

Ready to practise?

10 Sine Rule questions with answers and video solutions