Solving Triangles with the Cosine Rule Revision Notes | GCSE AQA Higher Maths

Worked example: Finding a Missing Side Using the Cosine Rule

In triangle $ABC$, $b = 7$ cm, $c = 5$ cm, and angle $A = 60^\circ$. Calculate the length of side $a$. Give your answer to 1 decimal place. [3 marks]

1. $a^2 = b^2 + c^2 - 2bc\cos A$ — State the cosine rule before substituting — this earns the method mark [M1].
2. $a^2 = 7^2 + 5^2 - 2 \times 7 \times 5 \times \cos 60^\circ$ — Replace $b = 7$, $c = 5$, $A = 60^\circ$. Keep $\cos 60^\circ$ as a symbol until the next step [M1].
3. $7^2 = 7 \times 7 = 49$ $5^2 = 5 \times 5 = 25$ $2 \times 7 \times 5 = 70$ $\cos 60^\circ = 0.5$ $70 \times 0.5 = 35$ — Expand every power longhand to avoid arithmetic errors. $\cos 60^\circ = 0.5$ is an exact value worth memorising.
4. $a^2 = 49 + 25 - 35 = 39$ — Add the first two terms, then subtract the correction term.
5. $a = \sqrt{39} \approx 6.244...$ — Do not round until the very last step. Keep the full calculator display.
6. $a \approx 6.2$ cm (1 d.p.) — Round only now, and always include units [A1].

Answer: $a \approx 6.2$ cm

Worked example: Finding a Missing Angle Using the Cosine Rule

Triangle $PQR$ has sides $p = 8$ cm, $q = 11$ cm, $r = 6$ cm. Find angle $P$, giving your answer to 1 decimal place. [3 marks]

1. Angle $P$ is opposite side $p = 8$ cm. The other two sides are $q = 11$ cm and $r = 6$ cm. — Always confirm which side is opposite the angle you want before substituting.
2. $\cos P = \frac{q^2 + r^2 - p^2}{2qr}$ — State the rearranged cosine rule — this earns the method mark [M1].
3. $q^2 = 11 \times 11 = 121$ $r^2 = 6 \times 6 = 36$ $p^2 = 8 \times 8 = 64$ $2 \times 11 \times 6 = 132$ — Expand every square longhand. This prevents the common error of misremembering squared values [M1].
4. Numerator: $121 + 36 - 64 = 93$ Denominator: $132$ $\cos P = \frac{93}{132}$ — Keep the fraction exact at this stage — do not round the decimal yet.
5. $\cos P = \frac{93}{132} \approx 0.70454...$ — Keep all decimal places on your calculator display.
6. $P = \cos^{-1}(0.70454...) \approx 45.2^\circ$ — Use the $\cos^{-1}$ (or SHIFT COS) button on your calculator. Round only here [A1].

Answer: $P \approx 45.2^\circ$

Worked example: Cosine Rule in Context — Two-Stage Problem

Two ships leave a port. Ship $A$ travels 15 km on a bearing of $040^\circ$ and Ship $B$ travels 20 km on a bearing of $160^\circ$. Calculate the distance between the two ships. Give your answer to 3 significant figures. [4 marks]

1. Bearing of $A$: $040^\circ$ Bearing of $B$: $160^\circ$ Angle between paths $= 160^\circ - 40^\circ = 120^\circ$ — The angle at the port (between the two routes) is the included angle. This is the $A$ in the cosine rule [M1].
2. $b = 15$ km (distance of Ship $A$) $c = 20$ km (distance of Ship $B$) $A = 120^\circ$ (included angle at port) — Side $a$ is the unknown distance between the two ships — the side opposite the $120^\circ$ angle.
3. $a^2 = b^2 + c^2 - 2bc\cos A$ $a^2 = 15^2 + 20^2 - 2 \times 15 \times 20 \times \cos 120^\circ$ — State the formula before substituting [M1].
4. $15^2 = 15 \times 15 = 225$ $20^2 = 20 \times 20 = 400$ $2 \times 15 \times 20 = 600$ $\cos 120^\circ = -0.5$ $600 \times (-0.5) = -300$ — $\cos 120^\circ$ is negative — this is a key result. Subtracting a negative means adding, so the triangle is larger than Pythagoras alone would suggest.
5. $a^2 = 225 + 400 - (-300) = 225 + 400 + 300 = 925$ — Subtracting $-300$ gives $+300$. This is where many students lose a mark — watch the double negative carefully [M1].
6. $a = \sqrt{925} \approx 30.413...$ $a \approx 30.4$ km (3 s.f.) — Round only at the final step. State units [A1].

Answer: The distance between the two ships is approximately $30.4$ km.