How do you determine whether #triangle ABC# has no, one, or two solutions given #A=42^circ, a=5, b=6#?

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Answer 1 · 2016-12-15T19:55:52

Check whether #a >= h#:

#h = bsin(A)#

#h = 6sin(42^@)#

#h ~~ 4#

A is acute and #h < a < b#, therefore, two possible triangles exist.