# The Sine Law: A Simplified Criterion for the Ambiguous Case?

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 $A_2=180^\circ - A_1$.

(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.
• Using the diagram at link, if $A>90^\circ$ then you will not have an ambiguous case. – Dan Christensen Nov 28 '14 at 18:50
• 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 $C$? – Dan Christensen Nov 28 '14 at 19:04