Physics Problem :(!

[mAdMaLuDaWg]

Hey,
I'm having a problem with this stupid problem. Could someone help me solve it? I know u gotta use the dot product but I could get only one equation in x and y, I can't seem to get the second, hence I can't solve it...
So here's the problem.. gotta use vectors:
An artist draws a line from the pixel location (10,20) to location (210,200). She wishes to draw a second line that starts at (10,20), is 250 pixels long, and measured at an angle of 30 degrees measured clockwise from the first line. At which pixel location should this second line end?
Thx in advance.

[caddmannq]

X= 254.5485 Y= 71.9233

I cheated, using AutoCAD.

[krohnjw]

You just need to use trig there. Use the initial given coordinates to set up a triangle. You can then grab the initial angle from the horizontal. With that you can get the end pixel coordinates using the X and Y coordinates 250 cos X and 250 sin X as your change in X and Y.

[mAdMaLuDaWg]

caddmanq, I dunno how u got that answer coz the answer is (87,258)

I could do it using trig, but I gotta do it using vectors and the dot product...

This is what I did:
A=(10,20) ; B=(210,200); C=(x,y)
vector AB=200i+180j
vector BC=(x-10)i+(y-20)j
AB.BC=200(x-10)+180(y-20)=|AB||BC|cos(30)
200(x-10)+180(y-20)=(250)(269)cos(30)

So I get an equation in x & y, I need one more equation to solve, and I got no idea how to get it !

[caddmannq]

That's what you get if you go 30 degrees counterclockwise.
exactly: X= 87.3073 Y= 257.7469

But your conventions might be different.

AutoCAD places 0,0 at the lower left. Angles are measured same as trig class: counterclockwise, but I went clockwise as per your first post.

I also set my precision to 4 decimals, but internal calculations are at 15 decimals & then rounded off.

<edit cause I can't type today...>

[mAdMaLuDaWg]

Damn this problem is killin me... ne1 gots any idea?