Answer:
Magnitude of vector A is 51.4 m
The angle of vector A with positive x-axis is θ= [tex]127.1^o}[/tex]
Explanation:
Given is
[tex]A_{x}[/tex]=-31m
[tex]A_{y\\[/tex]=41m
Let [tex]A_{x}[/tex] and [tex]A_{y\\[/tex] are components of a vector A,
The magnitude of vector A will be:
[tex]A=\sqrt{A_{x}^2 +A_{y}^2}[/tex]
= [tex]\sqrt{(-31)^{2} +41^2}[/tex]
A = 51.4 m
we know
θ1= [tex]tan^{-1} \frac{A_{y} }{A_{x} }[/tex]
= [tex]tan^{-1}(-\frac{41}{31} )[/tex]
=[tex]-52.9^{o[/tex]
Vector A is present in 2nd quadrant as x-component of vector A is negative and y-component is positive.
For finding the angle of vector with positive x-axis,
θ=180-52.9
θ= [tex]127.1^o}[/tex]
