Is the opposite angle of all triangles with a known height and base always the same?

Question

It's been 15 years since my last trigonometry class, so bear with me here...

If I have variations of a triangle with the same base and the same height, is the opposite angle always the same no matter what the other two angles are? I guess another way of phrasing it is; is the sum of the two angles adjacent to the base always the same in a triangle with a given height?

I tried to illustrate in the image below. What I'm asking is if the angle a is the same in all three triangles, if x and y are constant?

Answer 1

That it cannot be the case is easily seen by making the height very small. Then when the tip is above the left base point, the angle is almost a right angle, while if it is in the middle, it's almost a straight angle.

Answer 2

No. An example:

For the first triangle, let $x=3, y=4$ so that when the hypotenuse is $5$ the triangle is right-angled. Then $$a=\tan^{-1}\frac34$$

For the third triangle, we still have the same $x$ and $y$. But now split the triangle in half vertically so that $$\frac a2=\tan^{-1}\frac{1.5}4$$ However, $$2\tan^{-1}\frac{1.5}4\neq\tan^{-1}\frac34$$ so it does not hold.

Answer 3

No. The extended version of the sine rule relates the length $x$ of the side to the sine of the opposite angle $a$ and the radius $R$ of the circumcircle of the triangle.

Here you would have $$\frac x{\sin a} =2R$$ so that $$R=\frac x{2\sin a}$$So if you are looking for the triangles with base $x$ and opposite angle $a$ you will find that the opposite vertex to the given side lies on an arc of a circle radius $R$.

In your diagram, you have suggested the angle remains the same for the far vertex on a straight line.

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You can prove the sine rule in the form I gave it quite easily, by adding the circumcentre to the diagram and dropping a perpendicular from the circumcentre to the base of the triangle. The circumcentre and base give you an isosceles triangle with two sides of length $R$. The angle at the centre is $2a$ because it is twice the angle at the circumference. The length of the base is $2R\sin a$, computed from the two right-angled triangles made by the perpendicular.