Finding Components of Vectors in the xy Plane

[ff4930]

[SOLVED] Vectors, finding components

Homework Statement

Determine the x and y components of the following three vectors in the xy plane
a)a 10-m displacement vector that makes and angle of 30degree clockwise from the +y axis.

Homework Equations

The Attempt at a Solution

The answer is x = 5.0 and y = 8.7
but I don't know how to get it.
I follow the textbook example, don't I do 10*cos 30 and 10*sin 30?
Thanks for your help.

 

[rock.freak667]

When you make the triangle(or parallelogram depending on what you draw). Use some simple trigs to find the side opposite to the angle and the the side adjacent to the angle.

 

[suspenc3]

Yeah, just draw the vector, remember that $$sin(\theta)=\frac{opp}{hyp}$$ & $$cos(\theta)=\frac{adj}{hyp}$$

 

[ff4930]

I looked through my textbook and is suppose to be
10*cos 30 and 10*sin 30 but why the textbook answer say component x = 5.0 and y = 8.7?

 

[chocokat]

You have to be sure to read the problem correctly:

Homework Statement

Determine the x and y components of the following three vectors in the xy plane
a)a 10-m displacement vector that makes and angle of 30degree clockwise from the +y axis.

I follow the textbook example, don't I do 10*cos 30 and 10*sin 30?
Thanks for your help.

The vector makes an angle 30 degrees clockwise from the +y axis. If you want to use the +x axis as your guide, and you drop a perpendicular line from the vector, what is the angle from the +x axis?

 

[ff4930]

Is it 60?

 

[chocokat]

Is it 60?

Yes.

Using the 60 degree angle, you can use the (what I'll call) traditional sin for the 'y' length and cos for the 'x' length. Now your calculations should work.

You can also use the 30 degree angle. But you'd have to use sin for the 'x' length, and cos for the 'y' length. It's always helpful to me to draw a picture to help get the components correct.

 

[ff4930]

But component of x the answer from the textbook says 5.0 and component y = 8.7
and when I do 10*cos60 is not 5.0 nor 10*sin60, why is that?

Edit:
my graphing calculator was on radians, and not degrees. Thanks.