Rotating the graph of [tex]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/tex] with [tex]B\ne0[/tex] counterclockwise by [tex]\theta[/tex] eliminates the [tex]xy[/tex] term, where [tex]\cot2\theta=\frac{A-C}{B}.[/tex] Plugging in, we have [tex]\cot2\theta=\frac{7}{24},[/tex] since [tex]C=0.[/tex] Solving, we have [tex]\theta=\frac{1}{2}\cot^{-1}(\frac{7}{24})\approx\boxed{36.9^\circ}.[/tex]
