Thread: Missing parts of triangle

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

////a//.\
////a/\.\\b
a//./____\
.........c

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

remember that

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

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?

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.

I just realized Angle B could also be 82 degrees so you have to figure out which angle it is talking about.