1. ## direction angles

Vector f has a direction angle a= 45 (degrees) it is known that cos b is twive as large as cos y find angles b and y (a= alpha b=beta, y= gama)

i want to know if i did this equation right

i know that cosb = 2cosy

and i have the formula cos a^2 + cos b^2 + cos y^2=1
so i plug in cosb=2cosy in for cosb^2

and i get : cos45^2 + 2cosy^2 + cosy^2 = 1

then: .5+ 3cosy^2=1

3cosy^2= .5
y= 65.9 deg

then plug it back in to original

so: cos45^2+cosb^2+cos 65.9^2=1

then you get cosb^2=.333
cosb=sqrt(.333)
b=Cos^-1(.333)
b=54.74deg

thanks for the help

2. Originally Posted by valvan
Vector f has a direction angle a= 45 (degrees) it is known that cos b is twive as large as cos y find angles b and y (a= alpha b=beta, y= gama)
??? It would be nice to know what "b" and "y" have to do with vector f!

i want to know if i did this equation right

i know that cosb = 2cosy

and i have the formula cos a^2 + cos b^2 + cos y^2=1
So a, b, and y ( $\alpha$, $\beta$, and $\gamma$) are the direction angles, the angles vector f makes with the x, y, and z axes? It would have been nice if you had said that to begin with!
(I see now that you titled this thread "direction angles" but I still think you should have been more explicit.)

so i plug in cosb=2cosy in for cosb^2

and i get : cos45^2 + 2cosy^2 + cosy^2 = 1
But you didn't do what you just said! If cos b= 2 cos y then $cos^2 b= (2 cos y)^2= 4 cos^2 y$.

then: .5+ 3cosy^2=1=
No. $.5+ 5cos^2 y= 1$ (and be careful: Write either cos^2 y or (cos y)^2. "cosy^2" means cos(y^2).)

3cosy^2= .5
y= 65.9 deg
5 cos^2 y= .5 so cos^2 y= .1.

then plug it back in to original

so: cos45^2+cosb^2+cos 65.9^2=1

then you get cosb^2=.333
cosb=sqrt(.333)
b=Cos^-1(.333)
b=54.74deg

thanks for the help