= Solution
For a plane whose normal lies in the $\sigma_1$–$\sigma_3$ plane, rotate through $2\theta$ from $(\sigma_1,0)$ in the stated sign convention. The circle point gives
$$
\boxed{\sigma_n=C+R\cos2\theta=\sigma_1\cos^2\theta+\sigma_3\sin^2\theta,\qquad\tau=R\sin2\theta.}
$$
The normal-to-plane angle is essential: if the angle supplied is between the \b[plane itself] and $\sigma_1$, use its complement for $\theta$. The <principal stress planes> have zero <shear stress>, and planes at $45^\circ$ to those <principal stress planes> have maximum $|\tau|$.
For an arbitrary plane in a fully three-dimensional <stress> field, let $n_i$ be its direction cosines in the <principal stress> basis. Then
$$
\sigma_n=\sum_{i=1}^{3}\sigma_i n_i^2,\qquad |\tau|=\sqrt{\sum_{i=1}^{3}\sigma_i^2n_i^2-\sigma_n^2}.
$$
These follow by resolving the <traction> into its normal and tangential components; the compression-positive tensor is the negative of the usual tension-positive <Cauchy stress tensor>. The family of <Mohr circles> visualizes the possible combinations, but the two-dimensional $2\theta$ rule alone does not specify an arbitrary three-dimensional orientation.<Stress and strength lecture notes by MIT OpenCourseWare>[Lecture 10, stress tensor, traction resolution and Mohr construction; signs here are explicitly compression positive.]{ref}
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