I don\'t understand how to know which angle is correct when you are finding an a
ID: 2232890 • Letter: I
Question
I don't understand how to know which angle is correct when you are finding an angle and direction of a vector sum. For example, when using arctan(x/-y). Say the answer is -37.5 degrees, the solution manual says to add 360 for an angle of 322.5. This is for a question asking to find angle relative to x axis in positive direction. My question is, why can't you just add 90 degrees? It would result in 52.5 degrees in the first quadrant, or you could also add 180 for a result of 142.5 degrees in third quadrant, and aren't all of these answers the same with respect to x axis?Explanation / Answer
folllow this Find the x and y components for each vector and simply add these together for the resultant vector. The vector A has no Y component, since it lies on the x axis opposite to its positive direction, so it has a magnitude of - 15.9 m on that axis. For B vector the y component will be sin(70.2 degrees)* 8.7 or in terms of radians sin (70.2/360* 6.28)= .94 thus the y component is .94*8.7= 8.18m. The x component will be cos(70.2/360*6.28)*8.7= 2.95m. Adding the x components 2.95-15.9= -12.95m. Thus the sum of these vectors has a negative x component of -12.95 and a y component of 8.18. Applying Pythagorean theorem the magnitude will be sq rt[-12.95^2+ 8.18^2]= 15.317 m Now to find the angle of the vector, since we have the X and Y component values, and we know that tan= y/x=8.18/-12.95= -.6316, we need to ask what angle has a tangent value of -.6316 where tan^-1(-.6316)= -.563 radians. Multiplying by 360 degrees/2pi radians gives -32.27 degrees which would be an answer in the fourth quadrant which in term of positive degrees would be 360- 32.27= 327.7 degrees. But we need to know the answer made in the second quadrant since the x component is negative but the y component is positive, so we simply subtract 180 degrees from that answer giving 147.7 degrees. Thus the answer is magnitude =15.317 m at 147.7 degrees from the positive x axis.
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