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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