Question when solve $10r = |25 - 3r \cos \theta - 4r \sin \theta|$

Question

How can I know the sign of the term within the absolute value in: $10r = |25 - 3r \cos \theta - 4r \sin \theta|$

If we say that one vector is $\vec{A} = (r \cos \theta, r \sin \theta)$ and the other one is $\vec{B} = (3, 4)$, then the equation is equivalent to $10r = |25 - \vec{A}\cdot\vec{B}|$, but how can I prove that the term in the absolute value is always non-negative, so the original equation is in fact equivalent to $10r = 25 - \vec{A}\cdot\vec{B}$?

Thanks!

Answer 1

Note that

$$3r \cos \theta + 4r \sin \theta=5r\left(\frac35\cos\theta+\frac45\sin\theta \right)=5r\sin(\theta+\alpha)$$

where $\cos\alpha=\frac45, \sin\alpha=\frac35$, hence $$|25 - 3r \cos \theta - 4r \sin \theta|=5|5 - r\sin(\theta+\alpha)|$$

As long as $0\le r\le 5$, we can guarantee the expression inside the absolute value is non-negative.