How in the world do I find the missing parts of the triangle ABC if a = 12, b = 19, and A = 82 deg's?

Thanks very much

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- Jul 12th 2005, 04:17 PMJustardmsMissing parts of triangle help
How in the world do I find the missing parts of the triangle ABC if a = 12, b = 19, and A = 82 deg's?

Thanks very much - Jul 12th 2005, 09:07 PMMath HelpLaw of sines

////a//.\

////a/\.\\b

a//./____\

.........c

a = 12, b = 19, and A (opposite a) = 82 deg's

remember that $\displaystyle \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

now you can easily solve for B.

Once you have B you know that A+B+C = 180 so you can get C and finally use C and the law of sines to get c

let me know if you need more help - Jul 13th 2005, 06:22 AMJustardmsPlease Help!!!!
For angle B I get SinB = 19sin82deg / 12 which gives me 1.5679 and when I do arcsin 1.5679 I get an Error why is this?

- Jul 13th 2005, 10:58 AMMathGuruAmbiguity in the stated question
The reason is that the problem is not clearly stated. There is no way that the angle opposite a is 82 degrees. It can only be between a and b (I would have called that angle C as it is opposite c) Try it again this way and you will get the answer.

- Jul 13th 2005, 11:01 AMMathGuru
I just realized Angle B could also be 82 degrees so you have to figure out which angle it is talking about.