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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.

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  • $\begingroup$ 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

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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).

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  • $\begingroup$ 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

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