# Algebra II

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How do you solve a triangle with two solutions? The problem says that A=40 degrees, b=10, and a=14. So I did this:

14>(10)(sin 40)
And this means that there are two solutions according to the equations my teacher gave me.

What do I do now?

• Algebra II -

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.

• Algebra II -

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!

• Algebra II -

You got it !!

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