How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 42^{\circ}, a = 5, b = 6$ $A=42^circ, a=5, b=6$ ?

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Answer 1 · 2016-12-15T19:55:52

Check whether $a \geq h$ $a >= h$ :

$h = b sin (A)$ $h = bsin(A)$

$h = 6 sin (42^{\circ})$ $h = 6sin(42^@)$

$h \approx 4$ $h ~~ 4$

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

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