# Vector from angle and length

• July 23rd 2012, 08:38 AM
mariusg
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:
$Cos(220)*5=-cos(40)*5$ (I have not been able to simplify this more)
$Sin(220)*5=-sin(40)*5$

and written it as this:
$-(cos(40)5)i-(sin(40)5)j$

Now i controlled this by checking this:
$\sqrt{((sin(220)5)^2+(cos(220)5)^2} =5$

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

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

My question is:
What am I doing wrong? I am completely stuck.
• July 23rd 2012, 08:48 AM
Plato
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:
$Cos(220)*5=-cos(40)*5$ (I have not been able to simplify this more)
$Sin(220)*5=-sin(40)*5$
and written it as this:
$-(cos(40)5)i-(sin(40)5)j$
Now i controlled this by checking this:
$\sqrt{((sin(220)5)^2+(cos(220)5)^2} =5$
But the answer provided in the back of the textbook is this:
$-\frac{5\sqrt{3}}{2}i-\frac{5}{2}j$
My question is:
What am I doing wrong? I am completely stuck.

You have done nothing wrong.
The answer $-\frac{5\sqrt{3}}{2}i-\frac{5}{2}j$ goes with $240^o$ and not $220^o$.