Use cosine rule to determine the length of a side of a triangle. Cannot replicate textbook solution Given a triangle of lengths $4$, $5$ and $x$, where the sides of length $4$ and $5$ meet at an angle of $60^{\circ}$, I must calculate the length of side $x$.

The cosine rule is: $$a^2 = b^2 + c^2 - 2bc\cos(\alpha)$$
So, applying to my problem:
$$\begin{align}
a^2&=b^2+c^2-2bc\cos(\alpha) \\
x^2&=4^2+5^2-2(4)(5)(\cos(60^{\circ})) \\
x^2&=16+25-40\cos(60^{\circ}) \\
x^2&=41-40\cos(60^{\circ}) \\
x^2&=41-40(0.9524) \\
x^2&=41-38.1 \\
x^2&=2.9 \\
x&=\sqrt{2.9}
\end{align}$$
The textbook solution says I should arrive at $x = \sqrt{21}$, not $x=\sqrt{2.9}$. The textbook does not provide working to the solution, only the final answer.
Where did I go wrong?

Hint: $\cos(60^o)=0.5$ maybe you mixed radians ($60^o=1/3\pi$) with degrees on your calculator?
