#### Answer

$\theta=6.08^{\circ}$
$I_{max}=1.74\times 10^3in^4$
and $I_{min}=435in^4$

#### Work Step by Step

We can find the orientation of the principal axes and the required principle moments of inertia as follows:
$tan2\theta=\frac{-2I}{I_x-I_y}$
$\implies tan2\theta=\frac{-2(188)}{450-1730}$
$\implies \theta=6.08^{\circ}$
Now $I_{max/min}=\frac{I_x+I_y}{2}\pm \sqrt{(\frac{I_x-I_y}{2})^2+I_{xy}^2}$
We plug in the known values to obtain:
$I_{max/min}=\frac{450+1730}{2}\pm \sqrt{(\frac{450-1730}{2})^2++(138)^2}$
This simplifies to:
$I_{max}=1.74\times 10^3in^4$
and $I_{min}=435in^4$