= Solution
A numerical answer needs an assumption about fault orientation. For a cohesionless fault optimally oriented for <Mohr-Coulomb failure criterion>, with vertical $\sigma_1$, write $\varphi$ for the <internal friction angle> and $\delta$ for the <fault dip>. The angle between the plane and $\sigma_1$ is $90^\circ-\delta$, so the <optimal normal-fault dip relation> gives
$$
90^\circ-\delta=45^\circ-\frac\varphi2,\qquad
\varphi=2\delta-90^\circ=40^\circ.
$$
Hence \b[$\mu_f=\tan40^\circ=0.839\simeq0.84$] under that model.
For the <Mohr circle>, plot effective compression $\sigma_n'$ horizontally and <shear stress> $\tau$ vertically. Draw a circle with diameter $[\sigma_3',\sigma_1']$ tangent to $\tau=\mu_f\sigma_n'$. Its centre and radius obey $R/C=\sin\varphi$. One illustrative normalization is $\sigma_3'=1$, $\sigma_1'=4.59891$, with tangency at the resolved stresses on the $65^\circ$ plane. <Pore fluid pressure> shifts the total-stress circle horizontally; the friction law uses <effective stress>.
\Image[/past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2022/paper-1-fault-friction.png]
{title=Original normalized <Mohr circles> for optimally oriented normal faults: different dips imply different tangent slopes only under the stated failure model}
\b[The dip does not determine the actual sliding <coefficient of friction> without the optimal-orientation assumption.] A reactivated inherited plane may be nonoptimal, <cohesion> may matter, and the stress ratio and <pore fluid pressure> are not measured. For an arbitrary compression-positive stress state,
$$
\mu_f=\frac{(\sigma_1'-\sigma_3')\sin\delta\cos\delta}
{\sigma_1'\cos^2\delta+\sigma_3'\sin^2\delta}
$$
at cohesionless sliding. Thus the requested single value is a conditional inference, not a friction measurement.
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