by Sine Law :
sinB/b = sinA/a
sinB/10 = sin40/14
sinB = .459134
angle B = 27.33º or 180-27.33º
so B is either 27.33 or 152.67º
draw the two possible triangles, and use the sine law again to find c.
Oh, I see. So there would be two solutions for both B and C. I would just solve it like a regular triangle, and B' and C' would be 180 minus the values of B and C. Thanks!
You got it !!
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