Question

Use the law of cosines to find the value of 2*3*4 cos (theta).

Please explain. Thank you!

Answer 1

@jim_thompson5910

Answer 2

The law of cosines is

c^2 = a^2 + b^2 - 2ab*cos(C)

we'll replace C with theta. Also, we'll plug in a = 3, b = 4, c = 2 to get

c^2 = a^2 + b^2 - 2ab*cos(C)
2^2 = 3^2 + 4^2 - 2*3*4*cos(theta)

isolate the "2*3*4*cos(theta)"

Answer 3

@jim_thompson5910

For 2^2 = 3^2 + 4^2 - 2*3*4*cos(theta) I got 4 = 25-24cos(theta)

http://www.wolframalpha.com/input/?i=2%5E2+%3D+3%5E2+%2B+4%5E2+-+2*3*4*cos%28theta%29

I don't understand what isolate the "2*3*4*cos(theta)" means?

Answer 4

think of "2*3*4*cos(theta)" as its own variable

let z = "2*3*4*cos(theta)"

2^2 = 3^2 + 4^2 - 2*3*4*cos(theta)

2^2 = 3^2 + 4^2 - z

solve for z

Answer 5

21

Answer 6

so would that make the answer a. 21

Answer 7

yes