posted by Melody .
In a triangle PQR, angle Q=50 degrees,PR=5CM, QR=4CM. find angle P?
Looks like the "ambiguous case" , or sometimes called the SSA setup, where A is NOT the contained angle
sinP/4 = sin 50°/5
sin P = 4sin50/5 = .61284
angle P = 37.8° or 180-37.8 ° which is 142.2°
but the second case would result in (angleR + 142.2+50) > 180, thus that case is not admissable.
so angle P is 37.8°