In triangle ABC, a = 4, b = 7, and c = 9. Determine CosB to the nearest ten-thousandth

(1) 0.7340
(2) 0.2857
(3) 0.6667
(4) -.02857

I got (4) but the answer is (3). Can you explain why?

Using the law of cosines we have:
b^2 = a^2 + c^2 - 2ac• cos(B)

Solve for cos(B) and we have:
(b^2 - a^2 - c^2) / -2ac = cos(B)

Plug in the numbers and you should get:
(7^2 - 4^2 - 9^2) / (-2(4)(9)) = cos(B)
2/3 = cos(B)

Or in decimal form:
0.66666... = cos(B), which is answer (3)