I have a rather simple question, not sure if this is in the right SO.

I am attempting to find the angle of a side of a right angle triangle. And also attempting to find the mid point of a right angle triangle no matter the rotation(for a webapp I am making). enter image description here I am attempting to find the angle where the question mark is, then test that my algorithm/methodology to find the position of the mid point of a shape works(this is for my HTML5 Canvas Web App).

Is my maths correct?...

// This is half code/mathematics
? = tan-1( opposite/adjacent );  // inverse tan
? = tan-1(50/25);
? = 63.43; // is that correct??

My math algorithm below attempts to find the mid point when a shape is rotated. In my webapp I will only ever know the x,y pos & the width & height, but it will also be rotated around the x,y point which means the midpoint can be different. Is it correct?

// To find the x,y midpoint all I have to do is have   
// the hypotenuse of the triangle & the angle of one   
// side(the angle I found above). Is that correct?

// h stands for hypotenuse, a stands for the angle I found above
midX = cos(a)*h
midY = sin(a)*h

// so to take the example above, if I use those formulas I should get an mid x,y value of 25,12.5
midX = cos(63.4)*75;
     = 33.58; // shouldn't this be 25?
midY = sin(63.4)*75;
     = 67.06; // should be 12.5?
  • 2
    $\begingroup$ Are you interested in finding the "mid-point" of the triangle that is invariant under rotation or find mid points of sides? $\endgroup$
    – user21436
    Jan 26, 2012 at 1:46
  • 2
    $\begingroup$ $63.43^\circ$ is correct to two decimal places. As for the midpoint of a right angled triangle, you are going to be clearer about what you mean. A triangle has many centres. $\endgroup$
    – Henry
    Jan 26, 2012 at 1:51
  • $\begingroup$ @Henry thanks for the confirmation. I have added more info about the mid points, its really just x=half width, y=half height. $\endgroup$
    – sazr
    Jan 26, 2012 at 2:16
  • $\begingroup$ In order to provide an answer, can you tell us the the dimensions of the first triangle? Also, let me know if I understand, if a triangle have width=(0,0)--(0,1) and height=(0,0)--(0,-1), the midpoint that you want is (0.5,-0.5), do am I right? $\endgroup$
    – leo
    Jan 26, 2012 at 4:01

1 Answer 1


$63.43^\circ$ is correct to two decimal places.

Your midpoint seems to be halfway along the hypotenuse (the circumcentre of a right angled triangle), rather than being inside the triangle. The hypotenuse has length $\sqrt{50^2+25^2} \approx 55.90$ so half of that is about $27.95$.

So for a given angle $\theta$, the coordinates are about $(27.95 \sin \theta, 27.95 \cos \theta)$.

Plug in the angle you found earlier and you get $(25,12.5)$ within rounding errors


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