# Help calculating a point in 2d space on a rotating right angle triangle

#### Gummby8

I am trying to solve for a point on a triangle that rotates around a fixed point. For ease of calculation this point is at X: 0 Y: 0

The length and width of the triangle are always fixed, the only thing that changes is the rotation angle. How can I solve for X: ??? and Y: ??? taking into account the rotation angle?

I am trying to place a 2d rendered object in Java so I need to break the formula down to it's base elements so I can put it into code and have the formula spit out X and Y coordinates

The known variables are all listed in the pictures. The angle of the triangle, the length & width. Anything else that is needed to find the ??? must be calculated first I guess.

Please don't just link wiki pages. I have read them. I don't understand them. Hence why I am asking for help walking through the solution.

Thank you!

#### ebaines

Start by determining the angle of the hypotenuse of the original triangle:

$$\displaystyle \theta = \tan^{-1} \frac W L$$

and the length of the hypotenuse of the triangle, which I will call C:

$$\displaystyle C = \sqrt{L^2 + W^2}$$

The initial point (x,y) is then:

$$\displaystyle (x_1,y_1) = (C \cos \theta, C \sin \theta)$$

Note: $$\displaystyle (x_1, y_1)$$ also equals $$\displaystyle (W,L)$$

Now let's assume you rotate the triangle by the angle alpha. The new (x,y) coordinates are:

$$\displaystyle (x_2, y_2) = (C \cos (\theta + \alpha), C \sin (\therta + \alpha))$$

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#### skeeter

MHF Helper
$\displaystyle \left[\begin{array}{cc}x'\\y'\end{array}\right] = \left[\begin{array}{cc}\cos(35)&-\sin(35)\\\sin(35)&\cos(35)\end{array}\right] \cdot \left[\begin{array}{cc}3\\-5\end{array}\right]$

$x' = 3\cos(35) + 5\sin(35) \approx 5.33$

$y' = 3\sin(35) - 5\cos(35) \approx -2.38$

#### DenisB

I'm a bit confused...

1st, why not (as your example) use a 3-4-5 right triangle (hypotenuse = 5)
instead of the 3-5-SQRT(34) that you used?
(not important...just seems strange)

2nd, why bring in a right triangle at all?
Isn't your problem simply a given straight line (who cares if it's an hypotenuse!)
with one end at (0,0) moving such that a circle with radius = line length is formed?
And you want to calculate the coordinates of other end of line, given an angle?

Like with Skeeter's solution, 5.33^2 + 2.38^2 = ~SQRT(34).

Or am I just too dumb to understand "rotation angle"?

#### Gummby8

DennisB

I have an L shaped figure in my program that rotates around the top of the L. I need to place another object at the end of the horizontal part of the L in 2d space

So it is a right angle triangle. You just don't see the hypotenuse.

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#### DenisB

Yes, I understand that. I follow everything except:
Skeeter's 35 degrees...where, oh where, praytell,
does that come from?
I'll probably kick myself when you tell me !

Btw, I have a Phoenix Suns jacket with Steve Nash on the back...so be nice

#### Gummby8

The 35 degrees comes from teh second image. I need the point at the end of the L if it is rotated 35 degrees. it is a bit small font

#### DenisB

OK; gotcha! Had to get my binoculars

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#### ebaines

I see I made two errors in my original post. First I used a coordinate stystem where positive y is down, whereas down should be negative y, and secondly I confused sines and cosines (doh!). Hence this part of my post:

The initial point (x,y) is then:

$$\displaystyle (x_1,y_1) = (C \cos \theta, C \sin \theta)$$
should be:

$$\displaystyle (x_1,y_1) = (C \sin \theta, -C \cos \theta)$$

And this part:

The new (x,y) coordinates are:

$$\displaystyle (x_2, y_2) = (C \cos (\theta + \alpha), C \sin (\therta + \alpha))$$
should be:

$$\displaystyle (x_2, y_2) = (C \sin (\theta + \alpha), -C \cos(\therta + \alpha))$$

Sorry for any confusion. With these changes and using alpha = 35 degrees I get $$\displaystyle (x_2,y_2) = (5.325,-2.375)$$. By the way, it appears that the figure was actually drawn with alpha = 30 degrees, not 35 degrees.

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1 person

#### Gummby8

Yeah I went back and looked at my photoshop, it was 30
Actualy it was -30 (>.<) but I got the formula and I can confirm it works in the program.
Thank you all so much.