Vectors in 3D, angle formed with axes

Find the vector in i-j-k form.

Magnitude - 10
Angle with x-axis - 130
Angle with y-axis - 80
Angle with z-axis - 41.75

I can't visualise what these angles are... Help is appreciated, once I can do this, I'll be able to do similar questions.

This is called "Director cosine " , the projection of the vector on each axis .-it is a slant angle , formed on the plane that contains the respective axis and the vector .- The projectionis
V dot Axis .- On X axis , you a unit vector i , so
V dot i = IVI I1I cos X ( X the slant angle )

We are looking for IVI cos X = V dot i in the same way
IVIcosY = V dot j
IVIcosZ= Vdot k
cosX, cosY, cosZ are the "director cosines "

In your example , the projection on X axis is vx= IVI cos X = 10 cos130
vy= IVI cos Y= 10 cos80
vz=IVIcos Z= 10 cos41.75

So the vector is V= 10 ( cos 130 i +cos 80 j+ cos41.75 k )