Question

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 6, c = 11, B = 109°

Answer 1

law of cosines for this one

Answer 2

draw a picture to start

Answer 3

make that
\[b^2=a^2+c^2-2ac\cos(B)\]

Answer 4

fill in the numbers and you get \(b^2\) right away with a calculator
then use the law of sines to find the other angle

Answer 5

this is what i get, very close to 200 for \(b\)
http://www.wolframalpha.com/input/?i=6^2%2B11^2-2*6*11*cos%28109%29

Answer 6

i mean for \(b^2\)

Answer 7

These are my choices
A. b = 14.1, A = 24°, C = 47°
B. b = 19.9, A = 22°, C = 49°
C. b = 17, A = 26°, C = 45°
D. no triangle

Answer 8

So, did you draw a picture?

Answer 9

I'm not quite sure how to

Answer 10

Ok so draw a triangle that has a big 120 degree angle, one side that is 6, and another that is 11. Can you do that?