Answer:
Example A : 2.83 i + 3.94 j
Example B : -35.7 i + 6.5 j
Example C : 56.85 N, 50.71 °
Step-by-step explanation:
since all vectors are given in terms of magnitude and angle polar coordinates are the obvious choice.
A vector r with magnitude r and angle θ can be represented in polar coordinates as
r = rcosθ i + rsinθ j
where i and j are unit vectors in x , y direction
let's look at the question now ,
<u>Example A</u>
two vectors are given lets call the a and b
a = 23 cos27° i + 23 sin27° j = 20.5 i + 10.4j
b = 30 cos75° i + 30sin75° j = 7.8 i + 29.0 j
resultant vector,a + b =( 20.5 + 7.8 )i + (10.4 + 29.0)j
= 28.3 i + 39.4 j
if we give the scale 10 N = 1 cm vector will become
r = 2.83 i + 3.94 j
<u>Example B</u>
two vectors are given lets call the a and b
a = 27 cos224° i + 27 sin224° j = -19.4 i + -18.7j
b = 30 cos123° i + 30sin123° j = -16.3 i + 25.2 j
resultant vector,a + b =( -19.4 + -16.3 )i + (-18.7 + 25.2)j
=-35.7 i + 6.5 j
<u>Example C</u>
two vectors are given lets call the a and b
a = 16 cos80° i + 16sin80° j = 2.8 i + 15.8j
b = 48 cos144° i + 48sin144° j = -38.8 i + 28.2 j
resultant vector, a + b =( 2.8 + -38.8)i + (15.8 + 28.2)j
= 36.0 i + 44.0 j
magnitude, | a + b | = ![\sqrt{x^{2}+y^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%20%20%7D)
= ![\sqrt{36^{2}+44^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B36%5E%7B2%7D%2B44%5E%7B2%7D%20%20%7D)
= 56.85 N
angle = ![tan^{-1} (\frac{y}{x} )](https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20%28%5Cfrac%7By%7D%7Bx%7D%20%29)
= ![tan^{-1\frac{44}{36} }](https://tex.z-dn.net/?f=tan%5E%7B-1%5Cfrac%7B44%7D%7B36%7D%20%7D)
= 50.71 °