# Evolution of angle between two triangles as they tilted towards each other

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?

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

• 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. Mar 9, 2021 at 6:24
• 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.
– ap21
Mar 9, 2021 at 6:29
• Just edited my comment above. The axis is the line that bisects $\theta_i$
– ap21
Mar 9, 2021 at 6:30
• @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)?
– ap21
Mar 9, 2021 at 6:35

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 $$\cos^{-1}(1-2\sin^2\frac12\theta_i\cos^2\alpha)$$.