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?