My problem is as follows. I have two flat isosceles triangles (marked in yellow below) -- with sides of $L$ and base of width $w$ -- touching at a common point (marked in red), such that the angle between them is $\theta_i$. Note that $\theta_i$ is also the apex angle of another isosceles triangle, which I have depicted in lavender.

Now, I tilt these triangles (shown in deep blue) towards each other, by an angle $\alpha$. Then what is the equation relating the final angle $\theta_f$ to $\theta_i$ and $\alpha$? I know from measurements that $\theta_f \leq \theta_i$.

I realise that I only need to figure out how the lateral sides of the yellow triangles behave under the tilt $\alpha$, but I am unable to do so. Can someone please help?

enter image description here enter image description here

Note: In the picture, the entire deep blue configuration is tilted off the horizontal plane, but this is just for viewing convenience.

  • $\begingroup$ How exactly does your "tilting" work? $\theta$ shouldn't change if your "tilting" is like folding a sheet of paper along the side. Anyway, this doesn't look like mathematics at the moment. $\endgroup$ Mar 9, 2021 at 6:24
  • $\begingroup$ Good point. By 'tilt', I believe I mean a rotation of angle $\alpha$ about the red point. More precisely, the axis of rotation lies in the horizontal plane along the plane of mirror symmetry. $\endgroup$
    – ap21
    Mar 9, 2021 at 6:29
  • $\begingroup$ Just edited my comment above. The axis is the line that bisects $\theta_i$ $\endgroup$
    – ap21
    Mar 9, 2021 at 6:30
  • $\begingroup$ @user10354138 Also, would the result change if the axis is instead moved to the point at the middle of the triangle base (i.e. at distance w/2 from the red point)? $\endgroup$
    – ap21
    Mar 9, 2021 at 6:35

1 Answer 1


Pick coordinates centred at the red point with the $z$-axis being the internal angle bisector of $\theta_i$, $x$-axis being the orthogonal on the plane and $y$-axis the vertical out-of-plane direction. So the two initial apex are $(\pm L\sin\frac12\theta_i,0,L\cos\frac12\theta_i)$. If the bases of the two triangles are on the same line, then we have $\sin\frac12\theta_i=w/(2L)$. It isn't clear whether this is the case from your description.

Then rotate the triangles by angle $\pm\alpha$ about $z$-axis so $y$ becomes positive, we have the two points $(\pm L\sin\frac12\theta_i\cos\alpha,L\sin\frac12\theta_i\sin\alpha,L\cos\frac12\theta_i)$. The angle they made is $\cos^{-1}(1-2\sin^2\frac12\theta_i\cos^2\alpha)$.

If you rotate about other axes parallel to this, you will get parallel line, so the angle is unchanged (except you get skewed lines instead of intersecting lines in general).

  • $\begingroup$ Sorry for the confusion, but the bases of the 2 triangles are NOT on the same line. I added a picture above to make this clear. I don't know if this will make a difference to your answer though. $\endgroup$
    – ap21
    Mar 9, 2021 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.