Imageshack - 87978225.jpg thats the problem please help i have a test tomorrow and if i fail i will loose one year please help
Imageshack - 87978225.jpg thats the problem please help i have a test tomorrow and if i fail i will loose one year please help
Alpha=x.. k?
cos x=5/13
x is between 270,360 so x is negative of the form (360-x).
cos x= cos -x
therefore, we can take cos x as acute.
but sin x is negative
sin360-x = - sin x
sin(60-x)=sin60*cosX -sin(-X)cos60
=root3 *5/13 + 12/13*1/2
=5root3 + 12
------------
26
cos(45+x)=cos45cosX-sin45sin(-X)
=1/root2*5/13 + 1/root2*12/13
= 17
------
13root2
Your addition of angles formula is the right way to go about this. The restriction on x just tells you the sign of sin(x), where you use Pythagorean theorem to find sin(x) without actually needing to solve for x. I would double-check your trig there, especially the sin(60).