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Illustration regarding the Ambiguous Case for the Law of Sines
John Travis
Mississippi College
When one is given two sides of a triangle and an angle opposite one of the sides, the Law of Sines may produce two different but viable triangles. This ambiguity might occur due to the fact the sine function is positive for both actute angles (i.e. quadrant I) and obtuse angles (i.e. quadrant II) and becuase triangles can of course contain both actute and obtuse angles.
To see when this happens, consider the standard expression for the Law of Sines:
$\displaystyle\frac{\sin(B)}{b} = \displaystyle\frac{\sin(A)}{a}$
with B unknown. Solving for $\sin(B)$ yields
$\sin(B) = \frac{b \sin(A)}{a}$
By taking the inverse sine, one obtains a reference angle for B:
$B_r = \sin^{-1}( \frac{b \sin(A)}{a})$
Two options between $0$ and $\pi$ for $B$ are
$B_r$ or $\pi - B_r$.
Suppose $ B = \pi - B_r $.
If $A + B > \pi$, it is not possible to select a positive value for angle $C$ so that $A + B + C = \pi$.
So, there is no ambiguity and the only angle choice is
$B = B_r$
$C = \pi-A-B_r$
If $A + B < \pi$, then we get the following angle choices:
$B = B_r$
$C = \pi-A-B_r$
or
$B = \pi-B_r$
$C = B_r - A$
Notice that in each case the sum of all angles is $\pi$ or $180^o$.
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