Trig in Sides?

By the law of cosines, $(AB)^2+(BC)^2-2(AB)(BC)\cos(θ)=(AC)^2$ which means $(21\cos(θ)+2)^2+(56\cos(θ)-1)^2-2(21\cos(θ)+2)(56\cos(θ)-1)\cos(θ)=(63\cos(θ)-1)^2$ . By simplification, $-2352\cos(θ)^3-574\cos(θ)^2+102\cos(θ)+4=0$ . Let $A=\cos(θ)$ : The equation becomes $-2352A^3-574A^2+102A+4=0$ . The roots of the equation in terms of $A$ are factors of $-\frac{1}{588}$ . Since the reciprocal of $\cos(θ)$ is $\sec(θ)$ , and $\sec(θ)$ is prime, $\cos(θ)$ is the reciprocal of a prime. Because $\cos(θ)>0$ , $\sec(θ)$ has to be a positive prime factor of $588$ . The prime factorization of $588$ is $2*2*3*7*7$ . Since $\sec(θ)$ is prime, $\sec(θ)$ is either $2$ , $3$ , or $7$ . Testing the reciprocals of these, $\frac{1}{7}$ is the only one that works. Thus, $5\sec(θ)=5*7=35$ which is the final answer.