Triangle Inequality works here right?

$SinB=\dfrac{45/4} 6 *8/15=1. \ \ \therefore\ \angle\ B=90^o. \\ c = \sqrt{6^2-(45/4)^2}=\dfrac{3\sqrt{161}} 4.\ \$

Checking for the maximal possible length of $b$ , we see that $\dfrac{8}{15} = \dfrac{6}{b} \implies b = \dfrac{45}{4}$

From this we know that $b$ has the maximal possible length so $\triangle ABC$ must be a right triangle, with $b$ being the hypotenuse.

$\therefore c = \sqrt{\left(\dfrac{45}{4}\right)^2 - 36} = \sqrt{\dfrac{45^2}{16} - \dfrac{576}{16}} \\ \text{So } c = \sqrt{\dfrac{1449}{16}} \implies \dfrac{3\sqrt{161}}{4}$

So the correct answer choice is $\triangle ABC$ is a right triangle and $c = \dfrac{3\sqrt{161}}{4}$ units.