Vector from angle and length

Hello, I need a little help again, I have been working on this for a couple hours, and I am stuck.

In my textbook I have this problem:

Find the vector that has a length of 5 and an angle of 220 with positive x axis on the form ai+bj

My best guess so far has been this:

$\displaystyle Cos(220)*5=-cos(40)*5$ (I have not been able to simplify this more)

$\displaystyle Sin(220)*5=-sin(40)*5$

and written it as this:

$\displaystyle -(cos(40)5)i-(sin(40)5)j$

Now i controlled this by checking this:

$\displaystyle \sqrt{((sin(220)5)^2+(cos(220)5)^2} =5$

But the answer provided in the back of the textbook is this:

$\displaystyle -\frac{5\sqrt{3}}{2}i-\frac{5}{2}j$

My question is:

What am I doing wrong? I am completely stuck.

Re: Vector from angle and length

Quote:

Originally Posted by

**mariusg** In my textbook I have this problem:

Find the vector that has a length of 5 and an angle of 220 with positive x axis on the form ai+bj

My best guess so far has been this:

$\displaystyle Cos(220)*5=-cos(40)*5$ (I have not been able to simplify this more)

$\displaystyle Sin(220)*5=-sin(40)*5$

and written it as this:

$\displaystyle -(cos(40)5)i-(sin(40)5)j$

Now i controlled this by checking this:

$\displaystyle \sqrt{((sin(220)5)^2+(cos(220)5)^2} =5$

But the answer provided in the back of the textbook is this:

$\displaystyle -\frac{5\sqrt{3}}{2}i-\frac{5}{2}j$

My question is:

What am I doing wrong? I am completely stuck.

You have done nothing wrong.

The answer $\displaystyle -\frac{5\sqrt{3}}{2}i-\frac{5}{2}j$ goes with $\displaystyle 240^o$ and not $\displaystyle 220^o$.