= Solution
Take ordinary <basal slip> in <magnesium>, $(0001)\langle11\overline20\rangle$. In three-index hexagonal notation, $[101]$ denotes a vector proportional to $\mathbf a_1+\mathbf c$, not a cubic Cartesian $(1,0,1)$ with equal axis lengths. The <basal slip> direction closest to the tensile axis is along $\mathbf a_1$; its four-index representative is $[2\overline1\overline10]$.
For $r=c/a$,
$$
\cos\phi=\frac{r}{\sqrt{1+r^2}},\qquad
\cos\lambda=\frac1{\sqrt{1+r^2}},\qquad
m=\frac{r}{1+r^2}.
$$
Thus the algebraic answer is
$$
\boxed{\sigma_y=0.8\frac{1+r^2}{r}\,\mathrm{MPa}.}
$$
An external crystallographic reference gives $a=0.320927\,\mathrm{nm}$ and $c=0.521033\,\mathrm{nm}$, hence $r\simeq1.624$ and \b[$\sigma_y\simeq1.79\,\mathrm{MPa}$]. This is a conditional external-data estimate, not a fabricated value attributed to the supplied Data Book.<Crystal Structures by R. W. G. Wyckoff>[Volume1, magnesium entry reproduced in COD9008506]{ref}
The tensile axis rotates towards the selected <basal slip> direction during ideal single slip. As the axis approaches the basal plane, the resolved basal shear eventually becomes small; other modes, hardening and instability can interrupt the single-slip description. The boxed expression also permits direct substitution of any lattice ratio used in the examination Data Book; the numerical estimate above is explicitly tied to the cited crystallographic data. Taking a $45^\circ$ angle solely from the indices would incorrectly use a cubic metric.
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