Ambiguous Case for Triangle Side-Side-Angle Congruence

Theorem

Let $\triangle ABC$ be a triangle.

Let the sides $a, b, c$ of $\triangle ABC$ be opposite $A, B, C$ respectively.

Let the sides $a$ and $b$ be known.

Let the angle $\angle B$ also be known.

Then it may not be possible to know the value of $\angle A$.

This is known as the .

From the Law of Sines, we have:

$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$

from which:

$\sin A = \dfrac {\sin a \sin B} {\sin b}$

We find that $0 < \sin A \le 1$.

We have that:

$\sin A = \map \sin {\pi - A}$

and so unless $\sin A = 1$ and so $A = \dfrac \pi 2$, it is not possible to tell which of $A$ or $\pi - A$ provides the correct solution.

$\blacksquare$

Ambiguous Case for Triangle Side-Side-Angle Congruence/Proof 2

1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $6$. The sine-formula.
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): solution of triangles
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): solution of triangles