In the sum , A+ B = C vector A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vecto

Question

In the sum , A+ B = C vector  A has a magnitude of 12.0 m
and is angled 40.0&deg; counterclockwise from the +x  direction, and vector  C
has a magnitude of 15.0 m and is angled 20.0&deg; counterclockwise
from the -x direction. What are (a) the magnitude and (b) the
 
angle (relative to +x  ) of  B ?

BoomSchroasted · Answer

When adding vectors, you split the directions into X and Y components by using trigonimetry.  
 
In this case, we know that the A vector has an angle of 40 degrees, and a magnitude of 12.  Draw this into a triangle on an axis, with the 40 degree incline coming up horizontally from the X plane, and the 12 meters is how long the hypotenuse of the triangle is.  
We can find the opposite side of this triangle by using the formula "sin(angle) = opposite/hypotenuse".  We know the hypotenuse is 12, and the angle is 40.  Use some algebra, and see that "hyp * sin(ang) = opp" or, 12 * sin(40) = opp.
 
From this, we see that the opposite side is 7.71.  This is our X component for this vector.  Now we can figure out the Y component using A^2 + B^2 = C^2 (because we know 2 sides of the triangle), and we get that the X component = 9.19
 
So, A(x) = 9.19
A(y) = 7.71
 
Now we can figure out C using the same methods, except this time they will be negative components, so you draw the triangle on the other side of the axis, relative to the X axis.  The components for this, are
 
C(x) = -5.13
C(y) = -14.095
 
Since we see that A + B = C (the vector addition), we add the components with respect to their direction.  In this case...
 
A(x) + B(x) = C(x)
9.19 + B(x) = -14.095  (solve using addition/substraction to get B(x) by itself)
B(x) = -23.29
 
A(y) + B(y) = C(y)
7.71 + B(y) = -5.13
B(y) = -12.84
 
So, B(x) = -23.29
B(y) = -12.84
 
Now, we can get the hypotenuse (magnitude) by using the A^2 + B^2 = C^2 again.
 
a)  In this case, the magnitude of the components is 26.60 (magnitude is always positive)
 
b)  Since both of the components are negative, the hypotenuse will also be negative.  The degree of the arcsin(opp/hyp) is 28.86 degrees, but since it is negative, you subtract it from ninety, making the angle -61.14 degrees.

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In the sum , A+ B = C vector A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vecto

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