Why is this a right angle in the mach cone?

This and similar depictions are used when deriving the equation for the Mach cone angle $$\theta$$.

It is usually stated that using algebra we can conclude that $$\sin\theta=\frac{ct}{vt},$$ and thus $$θ=\arcsin \left(\frac{c}{v}\right).$$ However, I noticed that that assumes a right angle between $$ct$$ and the wavefront, which is never mathematically proven.

So, can someone explain why there is a right angle between $$ct$$ and the wavefront?

• Because a tangent to the circle is at right angles to a radius intersecting at the tangent point. Jul 3 at 14:14
• Ok that makes sense. Is there a mathematical proof for "Because a tangent to the circle is at right angles to a radius intersecting at the tangent point"? Jul 3 at 14:26
• You can find some proofs by searching for "why is radius perpendicular to tangent". Jul 3 at 14:31
• Just to make sure the angle I get from the equation is only the angle for half a cone right? Jul 3 at 14:41
• yes, only half ...... Jul 3 at 14:42

Given: a circle $$C$$, a line tangent to the circle $$T$$, and a radius, $$R$$, from the center of the circle to the point of tangency.
Assume: $$R$$ is NOT perpendicular to $$T$$
Then there is some other line segment, $$S$$, from the center of $$C$$ which is perpendicular to $$T$$. Since $$S$$ is perpendicular to $$T$$ that makes $$R$$ the hypotenuse of a right triangle. Therefore the length of $$R$$ is greater than the length of $$S$$, so $$|R|>|S|$$.
Now, except for the point of tangency all other points on $$T$$ are outside of $$C$$. Since $$C$$ is defined as all points a distance of $$|R|$$ away from the center, all points outside of $$C$$ are a distance $$x>|R|$$ away from the center. Since $$S$$ connects the center to such a point we have $$|S|>|R|$$.
But $$|S|>|R|$$ contradicts $$|R|>|S|$$ so our assumption is false. Therefore $$R$$ is perpendicular to $$T$$. QED