Finding the magnitude (length) and direction (angle) of a vector

[Ineedhelpwithphysics]

Homework Statement

In the Picture

Relevant Equations

Pythagoras theorem, inverse tan function

So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?

 

[kuruman]

Homework Statement: In the Picture
Relevant Equations: Pythagoras theorem, inverse tan function

View attachment 333724
So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?

You are not right. Remember ##~\text{tan} =\dfrac{\text{opposite}}{\text{adjacent}}.##

 

[Tomy World]

Homework Statement: In the Picture
Relevant Equations: Pythagoras theorem, inverse tan function

View attachment 333724
So i found the magnitude which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?

Hi,

In your case, + or - π will give you the right answer ;)