Cosine Rule

The Cosine Rule is $c^2=a^2+b^2-2ab\cdot \cos (C)$, where $a$ and $b$ are the two sides that enclose angle $C$, and $c$ is the side opposite $C$.

Two things must be right. The sign in front of $2ab\cdot \cos (C)$ is a minus, not a plus. And the subtraction applies only to that whole term: you work out $2ab\cdot \cos (C)$ first and then subtract it, rather than working out $a^2+b^2-2ab$ and multiplying that by $\cos (C)$.

The sine rule, $\frac{a}{\sin (A)}=\frac{b}{\sin (B)}=\frac{c}{\sin (C)}$, is a different formula and cannot be used when you know two sides and the angle between them.

The Cosine Rule links all three sides to one angle: $a^2=b^2+c^2-2bc\cos (A)$.

Use it to find a missing side when you know two sides and the angle between them (the included angle), and to find a missing angle when you know all three sides.

If instead you know two angles and a side, the Sine Rule is the tool to reach for. The Cosine Rule does not give area or perimeter directly.

The $60^{\circ }$ angle sits between the two known sides, so use the Cosine Rule $x^2=a^2+b^2-2ab\cos (C)$ with $a=4$, $b=5$ and $C=60^{\circ }$.

$x^2=4^2+5^2-2\times 4\times 5\times \cos (60^{\circ })$

Since $\cos (60^{\circ })=0.5$: $x^2=16+25-40\times 0.5=41-20=21$.

Only the $2ab\cos (C)$ term is multiplied by the cosine, and the final step is to square root: $x=\sqrt{21}\text{ cm}\approx 4.58\text{ cm}$.

The $60^{\circ }$ angle is the included angle, so use $x^2=a^2+b^2-2ab\cos (C)$ with $a=5$, $b=12$ and $C=60^{\circ }$.

$x^2=5^2+12^2-2\times 5\times 12\times \cos (60^{\circ })$

Since $\cos (60^{\circ })=0.5$: $x^2=25+144-120\times 0.5=169-60=109$.

Square root at the end: $x=\sqrt{109}\text{ cm}\approx 10.44\text{ cm}$. It is not $13\text{cm}$. That would only be right if the angle were $90^{\circ }$.

Use $x^2=a^2+b^2-2ab\cos (C)$ with $a=6$, $b=10$ and $C=120^{\circ }$.

$x^2=6^2+10^2-2\times 6\times 10\times \cos (120^{\circ })$

The angle is obtuse, so its cosine is negative: $\cos (120^{\circ })=-0.5$. Subtracting a negative adds on:

$x^2=136-120\times (-0.5)=136+60=196$

So $x=\sqrt{196}=14\text{ cm}$. Sensibly, the side opposite the obtuse angle is longer than both $6\text{cm}$ and $10\text{cm}$.

The angle between a and b is the angle C, opposite the side $c=7$. Rearranging the Cosine Rule gives

$\cos (C)=\frac{a^2+b^2-c^2}{2ab}=\frac{3^2+5^2-7^2}{2\times 3\times 5}=\frac{9+25-49}{30}=\frac{-15}{30}=-0.5$

A negative cosine means the angle is obtuse, so $C={\cos }^{-1}(-0.5)=120^{\circ }$. Keep the minus sign: dropping it gives the acute angle $60^{\circ }$ instead.