Question

(This is the law of cosines)
How do I solve for angle C, when side a=9 side b=12 side c=15.

Answer 1

c^2 = a^2 + b^2 - 2ab(cosC) would be the formula if I remember correctly.

Answer 2

So we just need to plug in values and then be careful with our algebra from there.

Answer 3

So try plugging in a, b, and c, and see what you're left with. Can check your work after that and help you finish theproblem :3

Answer 4

15^2 = 9^2 + 12^2 - 2(9)(12) cosC
Then 225 = 81 + 144 - 216 cosC
then 225 = 9cosC
then 25 = cosC
Now I'm stuck.

Answer 5

They kind of gave you a trick question, but no, not quite. Notice that in this particular problem, c^2 is the same as a^2 + b^2. If you add 81 and 144 on the right side of your equation, youll get 225, which would cancel out the other side of the equation, leaving you with 0 = -216cosC

Answer 6

So the -2(a)(b) is attached to the cosC???

Answer 7

Yes, a complete multiplication of.

Answer 8

Okay.
So what do I do about the 0=-216cosC

Answer 9

Sorry, the site went verrrrrrry laggy on me.

Answer 10

Well, essentially it's saying that cosC = 0. So you could do inverse cosine or if you happen to just know where cos is at 0, you'll have your answer for C.

Answer 11

It's one

Answer 12

The angle measure for a triangle is 1 degree?

Answer 13

Nope, it'd be pi/2, or 90 degrees.

Answer 14

Wait, so this is right triangle?????

Answer 15

Its a right triangle :P

Answer 16

Thats why I said its kind of a trick question. But yes, its 90 degrees. So you can now get the other two angles using the law of sines.

Answer 17

Thanks XDDD