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Here is my suggestion for an issue that doesn't seem to be handled well in any online notes that I have seen. Can anyone give a counter-example?

If you are given $a,b,$ and $B$ in $\triangle ABC$ (using the standard trig naming conventions) and you are using the Sine Law to solve for $A$ in $\frac{a}{\sin A}=\frac{b}{\sin B}$, then you would have a so-called Ambiguous Case if and only if $b<a$.

In that case, calculate the first value $A_1=\sin^{-1}(\frac{a \cdot \sin B}{b})$. Calculate the second value $A_2=180^\circ - A_1$.


Follow-up

Based on discussions here and elsewhere, here is my corrected version of the criterion:

If you are given $a,b,$ and $B$ in $\triangle ABC$ (using the standard trig naming conventions) and you are using the Sine Law to solve for $A$ in $\frac{a}{\sin A}=\frac{b}{\sin B}$, then you would obtain at least one solution $A_1=\sin^{-1}(\frac{a \cdot \sin B}{b})$. And you would have a so-called Ambiguous Case if and only if $A_1\ne 90^\circ$ and $b\lt a$.

In that case, you would have a second solution $A_2=180^\circ - A_1$.

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(2) is not unnecessary because if you don't have (2) then you can have that (3) is immpossible (if you have A > $90^\circ$ then $a$ will be largest side) and you have to have (4) because if you don't you can have A = $90^\circ$ witch by (2) is immposible.

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  • $\begingroup$ Using the diagram at link, if $A>90^\circ$ then you will not have an ambiguous case. $\endgroup$ – Dan Christensen Nov 28 '14 at 18:50
  • $\begingroup$ Can you give a counter-example? $\endgroup$ – Dan Christensen Nov 28 '14 at 18:51
  • $\begingroup$ Yes and the criteria you posted have to be fulfilled for case to be ambiguous, and if you neglect (2) you can have non ambiguous cases. $\endgroup$ – Lale221 Nov 28 '14 at 18:56
  • $\begingroup$ Not sure I understand. Can you give an example of a triangle (as labelled at link) such that $a\geq c$ and we have an ambiguous case for $C$? $\endgroup$ – Dan Christensen Nov 28 '14 at 19:04
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    $\begingroup$ It's ambiguous if and only if a < c witch is equivalent with A < C, so it is not regardless of A. $\endgroup$ – Lale221 Nov 28 '14 at 19:25

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