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September 3, 2014

September 3, 2014

Posted by **Samantha** on Thursday, May 15, 2008 at 7:50pm.

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 -
**Reiny**, Thursday, May 15, 2008 at 8:02pmby 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 -
**Samantha**, Thursday, May 15, 2008 at 8:07pmOh, 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 -
**Reiny**, Thursday, May 15, 2008 at 8:18pmYou got it !!

- Algebra II -

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