Sine rule and cosine ruleIB MYP Maths Extended: Revision notes

For a triangle $ABC$, the side opposite vertex $A$ is called $a$, the side opposite $B$ is $b$ and the side opposite $C$ is $c$. A side and the angle facing it are an opposite pair. The angle sum is $180^\circ$, so if you know two angles you can find the third straight away. SOH CAH TOA only works in a right-angled triangle. For other triangles you need the sine rule or the cosine rule.

For any triangle: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$ Use it when you know a complete opposite pair and one more side or angle. To find an angle, turn the rule over: $\frac{\sin A}{a}=\frac{\sin B}{b}$. Worked example. $BC=12$ cm, $B\hat AC=40^\circ$, $A\hat BC=65^\circ$. Then $AC=\frac{12\sin65^\circ}{\sin40^\circ}=16.9$ cm. The third angle is $75^\circ$, so $AB=\frac{12\sin75^\circ}{\sin40^\circ}=18.0$ cm.

Use the cosine rule when you know two sides and the included angle (the angle between them), or all three sides: $a^2=b^2+c^2-2bc\cos A\qquad\cos A=\frac{b^2+c^2-a^2}{2bc}.$ The side on the left must be opposite the angle $A$. Worked example. $PQ=6$, $PR=10$, $Q\hat PR=120^\circ$: $QR^2=36+100-120\cos120^\circ=136+60=196$, so $QR=14$ cm. Then $\cos P\hat QR=\frac{36+196-100}{2(6)(14)}=\frac{11}{14}$. A negative cosine means the angle is obtuse.

Look at what you are given:

• Right angle: SOH CAH TOA and Pythagoras.
• A side and its opposite angle, plus one more side or angle: sine rule.
• Two sides and the included angle: cosine rule for the third side.
• Three sides: cosine rule for an angle.
• Two angles and a side: find the third angle, then use the sine rule. When finding an angle, the cosine rule is safer, because the sine rule cannot tell an acute angle from an obtuse one with the same sine. If you use the sine rule, find the smaller angles first.

Many problems need two rules one after the other. Keep full calculator values between steps and round only at the end (3 significant figures for lengths and 1 decimal place for angles). Worked example. $A$ and $B$ are 800 m apart on a shore. $C\hat AB=52^\circ$ and $C\hat BA=71^\circ$. First $A\hat CB=57^\circ$, then $AC=\frac{800\sin71^\circ}{\sin57^\circ}=902$ m. A boat $D$ has $AD=500$ m and $C\hat AD=34^\circ$, so $CD^2=902^2+500^2-2(902)(500)\cos34^\circ$ and $CD=562$ m. Check that your calculator is in degrees mode.