= Solution
Use compression-positive <principal stresses>, $\sigma_1\geq\sigma_2\geq\sigma_3$. For the intended <thrust fault> regime during approximately north–south <tectonic shortening>, the likely orientations are
$$
\boxed{\sigma_1:\ \text{horizontal N--S};\quad\sigma_2:\ \text{horizontal E--W};\quad\sigma_3:\ \text{vertical}.}
$$
A locally horizontal traction-free surface motivates taking one principal direction as vertical, while the strong north–south compression selects the maximum horizontal <principal stress>. The expected <fault strike> is then approximately east–west.
This ordering assumes the two horizontal compressions exceed the vertical <principal stress>. North–south shortening by itself does not establish the magnitudes of all three: if the vertical <principal stress> were intermediate, <strike-slip faulting> would instead be possible. The answer above is the simple compressional end member used in the following parts.
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