0 1 or 2 triangles

Question

how many possible triangles are there

a=20

c=24.5

C=32.3 degrees

Answer 1

You'll notice that this is a SSA situation and, if you remember back in Geometry, there are 5 ways of proving triangle congruence (essentially meaning that there are 5 ways of defining a specific triangle) and SSA is not one of them. This means that you cannot for sure (in all cases) define a triangle given this SSA condition. But sometimes you can. Hence the question.

There are lots of textbooks that give a table with a list of criteria and listing and results (if one side is bigger than another . . .) but I prefer the sure fire method (works every time) of using the Law of Cosines.

The Law of Cosines says:

c² = a² + b² - 2abcos(C)

Plugging in the values we get:

(24.5)² = (20)² + b² - 2(20)bcos(32.2°)

600.25 = 400+ b² - 33.8105b

b² - 33.8105b - 200.25 = 0

Using the quadratic formula you get b = -5.141 or b = 38.952

Since it is impossible for a triangle to have a negative length on a side, there can be only 1 triangle associated with these measurements.