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Given the following measurements, A=28, a=19cm, and c=41cm how many triangles are possible?
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using law of sines \[\frac{\sin C}{41} = \frac{\sin 28}{19}\] \[C = \sin^{-1} (\frac{41 \sin 28}{19})\] If 41sin28/19 > 1, Then triangle doesn't exist If 41sin28/19 = 1, Then only 1 triangle exists, a right triangle as C=90 If 41sin28/19 < 1, Then 2 triangles exist, C could have 2 values
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