- #1
th77
- 16
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I have a problem that includes an acceleration vector 'a' located in the 3 quadrant and it makes an angle (theta) 30 degrees to the Negative Y axis.
The solution manual shows the the vector resolved like
a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j
Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.
The solution manual shows the the vector resolved like
a = (12.0 sin 30 m/s^2) i - (12.0 cos 30 m/s^2) j
Why take the sin of 30 for the x direction and the cos of 30 for y? I've always done it cos of x and sin of y.