POP

G7.13 The Cosine Rule

Image · 914 × 270

01 — Explanation

Why do I need a different rule from the sine rule here?

The sine rule needs a complete opposite pair (a side and its directly-opposite angle). When you're given two sides and the included angle (SAS), or all three sides (SSS), no such pair exists — that's exactly when you switch to the cosine rule.

02 — Explanation

What's the formula, and how do I find a missing side with it?

a^2=b^2+c^2-2bc\cos A

Worked example (SAS): $PQ=11.2$ , $PR=9.4$ , $\angle P=78°$ . Find $QR$ .

QR^2=9.4^2+11.2^2-2(9.4)(11.2)\cos78° = 213.8-210.56\cos78°\approx170.0 \Rightarrow QR\approx13.0

Exam Tip: Treat $2bc\cos A$ as a single term subtracted from $b^2+c^2$ — don't bracket $(b^2+c^2-2bc)$ and multiply by $\cos A$ afterwards, which is a common order-of-operations slip.

03 — Explanation

How do I rearrange it to find a missing angle (SSS)?

\cos A=\frac{b^2+c^2-a^2}{2bc}

where $a$ is the side opposite the angle you want.

Worked example: sides $8, 10, 14$ . Find the largest angle.

The largest angle is always opposite the longest side, so find $\angle B$ opposite the side of length $14$ :

\cos B=\frac{10^2+8^2-14^2}{2(10)(8)}=\frac{-32}{160}=-0.2 \Rightarrow B=\cos^{-1}(-0.2)\approx101.5°

04 — Explanation

Why doesn't the cosine rule have an "ambiguous case" like the sine rule does?

$\cos^{-1}$ always returns a unique value between $0°$ and $180°$ — a negative cosine automatically gives an obtuse angle, a positive cosine an acute one. There's no second possible solution to check, unlike $\sin^{-1}$ .

Exam Tip: A negative value partway through your working (e.g. $\cos B=-0.2$ ) isn't a mistake — it's the formula telling you the angle is obtuse. Keep the sign.

05 — Explanation

How do I solve a multi-step problem combining both rules (e.g. a quadrilateral)?

Split the shape into two triangles sharing one side. Solve whichever triangle has enough information first, use that result as the "bridge" into the second triangle, then apply whichever rule fits the second triangle's new information.

Worked exam-style question: Quadrilateral $ABCD$ : $AB=9.8$ , $BC=7.4$ , $\angle ABC=100°$ , $CD=11.5$ , $\angle CAD=35°$ . Find $\angle ADC$ .

$\triangle ABC$ is SAS, so use the cosine rule for the bridge side $AC$ :

AC^2=9.8^2+7.4^2-2(9.8)(7.4)\cos100°\approx176.0 \Rightarrow AC\approx10.40

In $\triangle ADC$ : side $CD=11.5$ is opposite $\angle CAD=35°$ — a complete pair — so the sine rule finds $\angle ADC$ (store the unrounded $AC$ for this step):

\frac{\sin(ADC)}{AC}=\frac{\sin35°}{11.5} \Rightarrow \sin(ADC)\approx0.661 \Rightarrow \angle ADC\approx41.4°

06 — Explanation

How do bearings problems use the cosine rule?

Draw a north line at every point. Use parallel-line angle facts (alternate/co-interior) and "angles around a point sum to $360°$ " to find the triangle's included angle, then apply the cosine rule as normal (SAS → missing side).

Worked example: A ship sails from $A$ on a bearing of $075°$ for $50$ km to $B$ , then on a bearing of $160°$ for $65$ km to $C$ . Find $AC$ .

The back bearing from $B$ to $A$ is $075°+180°=255°$ . The included angle $\angle ABC$ is the difference between this and the bearing to $C$ : $255°-160°=95°$ . Then:

AC^2=65^2+50^2-2(65)(50)\cos95°\approx7291.5 \Rightarrow AC\approx85.4\text{ km}

07 — Common mistakes

Common mistakes to avoid

Mixing up the rearranged formula's signs: it's $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ , with the side opposite the target angle subtracted on top.
Applying $2bc\cos A$ outside the correct order of operations.
Forgetting the cosine rule never needs an ambiguous-case check — don't invent a second solution.
Rounding a "bridge" value before carrying it into the second triangle.