Dot product from magnitudes and cross product Given $|{\vec{a}}|=2, |{\vec{b}}|=5$ and $|{\vec{a}}\times{\vec{b}}|=8$

Find ${\vec{a}}.{\vec{b}}$

Books says answer should be $6$, I think answer should $\pm6$ since $|sin \theta| = \frac{8}{10}$ and thus $ cos \theta =\pm \frac{3}{5}$

So Basically, why is $|{\vec{a}}\times{\vec{b}}|=|{\vec{a}}||{\vec{b}}|Sin \theta$ and mod isn't used around $Sin \theta?$

[Update]

I found an example for which ${\vec{a}}.{\vec{b}}=-6$

${\vec{a}}=2\hat{i}$

${\vec{b}}=-3\hat{i}+4\hat{j}$
Just exploit Lagrange's identity
$$|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-\langle\vec a,\vec b\rangle^2,$$
hence $6$ and $-6$ are the solutions.
the equation $$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin(\theta)$$ is the absolute value of the cross product and the length of the vector $\vec{a}\times \vec{b}$
see also here https://en.wikipedia.org/wiki/Cross_product