I solve these two equations and I got x=51.65degree but in my book x=128.35

so Iam confused

the equations are:

54.684=F Cos x...........(1)

69.313=F Sin x.............(2)

Printable View

- May 31st 2013, 10:29 AMkakashiIm I right??
I solve these two equations and I got x=51.65degree but in my book x=128.35

so Iam confused

the equations are:

54.684=F Cos x...........(1)

69.313=F Sin x.............(2) - May 31st 2013, 11:08 AMHallsofIvyRe: Im I right??
IF the question is as you stated it, "128.35" is impossible because if F is positive F cos x would be negative and if F is negative, F sin x would be negative.

- May 31st 2013, 11:24 AMkakashiRe: Im I right??
sorry

the first equation is negative - May 31st 2013, 12:44 PMHallsofIvyRe: Im I right??
So your equations are F cos(x)= -54.684 and F sin(x)= 69.313?

Then clearly 51.65 degrees cannot be correct. I presume neglecting the negative was your error in doing the calculation and you see now that

128.35 degrees is correct? (Actually, I get 128.27 degrees.) - Jun 1st 2013, 12:27 AMkakashiRe: Im I right??
why??

I got tan(x)=-1.264 and when you take the enverse of tan the answer will be x=-51.65

so can explain how did you solve it,please - Jun 1st 2013, 12:53 AMProve ItRe: Im I right??
Surely you know that if sine is positive and cosine is negative then your angle is in the second quadrant.

The arctangent function is defined to only give angles in the fourth and first quadrants to make it a one to one function.