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February 27, 2015

February 27, 2015

Posted by **Laurianne** on Tuesday, January 19, 2010 at 6:25pm.

- Trigonometry -
**bobpursley**, Tuesday, January 19, 2010 at 6:34pma^2=b^2+c^2-2bcCosA

CosA= 1/2 * (b^2+c^2-a^2)/bc

A= 139 deg, you are right on that.

Lets find angle B, by the law of sines

sinB/b=SinA/a or sinB= .3269 or B=19 deg

Lets find angle C by the law of sines

SinC/c=SinA/a or SinC=.3718 C=21.8 deg

check: do the sum of angles = 180 deg?

139+21.8+19=180 rounded.

All is well.

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