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The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeysThe relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set $s=-\tau >0$. Its circle has center $(-Us,0)$ and radius $cs$. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse $Us$ and opposite side $cs$, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as $M$ increases: its derivative is $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.