simple trig problem making me go nuts

[ultrus]

Here it is:

http://www.ultran.info/gamePlan1.gif

 

How do I solve for x2 and y2? I remember seeing this in high school, but I can't recall it.

 

Thanks much for any sanity you can provide.

[.josh]

x2 = 12

y2 = 16

 

I'm not a math whiz so I don't know how to officially figure it out, but here's what I did: 

 

c2 = 50 / 20 = 2.5

x2 = 40 / 2.5 = 16

y2 = 30 / 2.5 = 12

 

And to check:

 

x2^2 + y2^2 = c2^2

 

16^2 + 12^2 = 20^2

256 + 144 = 400

 

In essence, I just like, scaled it, lol. 

[akitchin]

CV's right, because congruent triangles have the same trig ratios.  therefore:

 

x2 = (x1/c1) * c2
y2 = (y1/c1) * c2

 

where x1/c1 is the cosine and y1/c1 is the sine (assuming you consider the angle from the x-axis).

 

just figured i'd offer a mathematical reason for it .

[.josh]

And that's what makes me so great.  I don't need no stinking math.

[ultrus]

Heh thanks much! I was using the math to move xyz coordinates of stars in our sky to coordinates around a sphere. It worked! You can see the working result at www.christopherstevens.cc/fun . Click and drag your mouse to look up, down, left, right. Hold apple/ctrl+drag to zoom in and out. Hold option/alt+drag to move around horizontally. Now I just need to make it look pretty, and see if my stars are inverted and/or backwards.

[Maq]

Heh, that looks pretty cool, gj!