Ambiguous Case

[Captn Cook]

Hey Guys,
If I am given a triangle requiring me to use the sine rule, how do I know when the ambiguous case is present?

Thanks!!

[jazzycab]

Hey Guys,
If I am given a triangle requiring me to use the sine rule, how do I know when the ambiguous case is present?

Thanks!!

The ambiguous case of the sine rule comes about when you know two sides \(a\) and \(b\) where \(a>b\) and the angle opposite the shorter side, \(\angle{\left(B\right)}\), where you're trying to find the angle opposite the longer known side, \(\angle{\left(A\right)}\).
One additional condition that also must be met is \(b>a\sin{\left(B\right)}\).
I will give a more complete explanation when I'm at my computer.
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Consider a pendulum that overhangs a table (as shown in the diagram below), where there is a beam fixed at an angle 'hanging' the pendulum over the table.

For the set of conditions where the length of the pendulum is shorter than the length of the beam, but long enough to overhang the table, we can create two triangles with the same set of information (the beam length, the pendulum length and the angle that the beam makes with the table top).
These triangles in the image above include the beam, the pendulum when it's in the green position and the line segment joining the bottom of the pendulum to the bottom of the beam.
The height that the top of the beam is above the table can be found using right-angled trig \(h=a\sin{\left(\angle{\left(B\right)}\right)}\).
Thus, \(a>b>a\sin{\left(\angle{\left(B\right)}\right)}\).