Sine and Cosine Rules Revision Notes | GCSE AQA Higher Further Maths

Worked example: Sine Rule — Finding a Missing Side

In triangle $ABC$, angle $A = 48^\circ$, angle $B = 73^\circ$, and side $b = 9.4\text{ cm}$. Find side $a$.

1. $$\frac{a}{\sin A} = \frac{b}{\sin B}$$ — We know $b$ and $B$, and we want $a$. Pick the two fractions that involve these four quantities. [M1]
2. $$\frac{a}{\sin 48^\circ} = \frac{9.4}{\sin 73^\circ}$$ — Substitute $A = 48^\circ$, $B = 73^\circ$, $b = 9.4$. [M1]
3. $$a = \frac{9.4 \times \sin 48^\circ}{\sin 73^\circ}$$ — Rearrange to isolate $a$. Keep the full decimal values on your calculator — do not round yet.
4. $$a = \frac{9.4 \times 0.7431\ldots}{0.9563\ldots} = \frac{6.985\ldots}{0.9563\ldots} \approx 7.31\text{ cm}$$ — Round only at the final step. Here $\sin 48^\circ \approx 0.7431$ and $\sin 73^\circ \approx 0.9563$. [A1]

Answer: $a \approx 7.31\text{ cm}$ (to 3 significant figures)

Worked example: Cosine Rule — Finding a Missing Side

In triangle $PQR$, $p = 7\text{ cm}$, $q = 5\text{ cm}$, and angle $R = 110^\circ$. Find side $r$.

1. $$r^2 = p^2 + q^2 - 2pq\cos R$$ — We know two sides ($p$ and $q$) and the included angle $R$ between them. Side $r$ is opposite $R$. [M1]
2. $$r^2 = 7^2 + 5^2 - 2 \times 7 \times 5 \times \cos 110^\circ$$ — Substitute $p = 7$, $q = 5$, $R = 110^\circ$. [M1]
3. $$7^2 = 7 \times 7 = 49, \quad 5^2 = 5 \times 5 = 25, \quad 2 \times 7 \times 5 = 70$$ — Expand each term separately to avoid arithmetic errors.
4. $$\cos 110^\circ \approx -0.3420$$ $$r^2 = 49 + 25 - 70 \times (-0.3420)$$ $$r^2 = 74 + 23.94 = 97.94$$ — Cosine of an obtuse angle is negative. Subtracting a negative number adds — so the side opposite an obtuse angle is the longest side. [M1]
5. $$r = \sqrt{97.94} \approx 9.90\text{ cm}$$ — Round only at this final step. [A1]

Answer: $r \approx 9.90\text{ cm}$ (to 3 significant figures)

Worked example: Cosine Rule — Finding a Missing Angle

A triangle has sides $a = 6\text{ cm}$, $b = 8\text{ cm}$, $c = 10\text{ cm}$. Find angle $C$.

1. $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ — All three sides are known. Angle $C$ is opposite side $c = 10$. [M1]
2. $$a^2 = 6 \times 6 = 36, \quad b^2 = 8 \times 8 = 64, \quad c^2 = 10 \times 10 = 100$$ $$a^2 + b^2 - c^2 = 36 + 64 - 100 = 0$$ — The numerator equals zero — this is a useful sanity check. [M1]
3. $$2ab = 2 \times 6 \times 8 = 96$$ — Compute the denominator separately to keep the working clear.
4. $$\cos C = \frac{0}{96} = 0$$ $$C = \cos^{-1}(0) = 90^\circ$$ — A cosine of zero means a right angle. This confirms the 6-8-10 triangle is a right-angled triangle (a multiple of the 3-4-5 Pythagorean triple). [A1]

Answer: $C = 90^\circ$

Worked example: Area Using the Sine Formula

Triangle $ABC$ has $a = 5\text{ cm}$, $b = 7\text{ cm}$, and angle $C = 38^\circ$. Calculate the area.

1. $$\text{Area} = \frac{1}{2}ab\sin C$$ — Two sides and the included angle are given — this formula applies directly. [M1]
2. $$\text{Area} = \frac{1}{2} \times 5 \times 7 \times \sin 38^\circ$$ — Substitute $a = 5$, $b = 7$, $C = 38^\circ$.
3. $$\frac{1}{2} \times 5 \times 7 = 17.5$$ $$\sin 38^\circ \approx 0.6157$$ $$\text{Area} = 17.5 \times 0.6157 \approx 10.77\text{ cm}^2$$ — Compute $\frac{1}{2} \times 5 \times 7$ first, then multiply by $\sin 38^\circ$. [A1]

Answer: $\text{Area} \approx 10.8\text{ cm}^2$ (to 3 significant figures)