Ambiguous Case for Triangle Side-Side-Angle Congruence/Proof 1
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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 ambiguous case.
Proof
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.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): ambiguous case
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): ambiguous case
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): ambiguous case