Question

Hey guys, I need serious help with this one. For the law of cosines (trying to find side "b") the problem looks like:
b^2= 5^2+ 3^2 -2*5*3 cos 90 degrees.
I keep getting an odd number, I need help solving this problem and finding b.

Answer 1

if you have a 90 degree angle, you have a right triangle.
so the problem should reduce to pythagoras.
It almost looks like a 3,4,5 right triangle, but the 5 would be the longest side.

Answer 2

OH! See, this is my problem and example, I'm supposed to make all of it. Would it help if I changed the 5 to 4 so the hypotenuse can be 5?

Answer 3

it would make the numbers nicer.

as you (should!) know, cos 90º is 0
so your equation up above simplifies to
b^2 = 5^2 + 3^2
b = sqrt(34)

if you use 4 instead of 5, you would get
b^2 = 4^2+9^2 = 25
b= sqr(25) = 5

Answer 4

So the -2ac*cos(90) just disappears?

Answer 5

yes, because cos 90 = 0

Answer 6

Thank you SO MUCH!!