Solve similar right triangles, given one's hypotenuse, the other's base, and the sum of their heights.

Question

I encountered this problem while trying to determine a generic equation for entasis, but this question is not about entasis.

$\theta$ is wanted—given this lovely figure

given that the two triangles are similar, and given $a$, $b$, and $h$.

I recognize that the sum of the heights of the triangles equals $h$, and that their ratio equals the scale factor, which seems like a likely avenue, but my trigonometry and geometry are weak and I can’t figure this one out.

Answer 1

It is pretty easy to see from the geometry of the figure that

$b \cot \theta + a \cos \theta = h, \tag 1$

whence,

$b \dfrac{\cos \theta}{\sin \theta} + a\cos \theta = h; \tag 2$

now using $\cos^2 \theta + \sin^2 \theta = 1$, i.e. $\cos \theta = \pm \sqrt{1 - \sin^2 \theta}$,

$b \dfrac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} + a \sqrt{1 - \sin^2 \theta} = h; \tag 3$

we choose the positive sign on $\pm \sqrt{1 - \sin^2 \theta}$ since the angle $\theta$ appears to be acute; next, we square:

$b^2 \dfrac{1 - \sin^2 \theta}{\sin^2 \theta} + 2ab \dfrac{1 - \sin^2 \theta}{\sin \theta} + a^2 (1 - \sin^2 \theta) = h^2; \tag 4$

we multiply by $\sin^2 \theta$:

$b^2 (1 - \sin^2 \theta)+ 2ab \sin \theta (1 - \sin^2 \theta) + a^2 \sin^2 \theta (1 - \sin^2 \theta) = h^2 \sin^2 \theta, \tag 5$

which may be written as a quartic equation in $\sin \theta$:

$-a^2 \sin^4 \theta -2ab \sin^3 \theta + (a^2 - b^2 - h^2)\sin^2 \theta + 2ab \sin \theta + b^2 = 0, \tag 6$

or

$a^2 \sin^4 \theta + 2ab \sin^3 \theta + (h^2 - a^2 - b^2)\sin^2 \theta - 2ab \sin \theta - b^2 = 0. \tag 7$

This is about as far as we can push things using elementary algebra and trigonometry. To find $\sin \theta$, we must solve this quartic, which may be done according to this wikipedia page.

Answer 2

Let $\mu$ be the ratio of the lower triangle’s altitude to the total height $h$. We then have $$\tan\theta = {b \over \mu h} \\ \cos\theta = {(1-\mu)h \over a}.$$ Eliminate $\mu$ to get the equation $$h = a\cos\theta + b\cot\theta.$$ Alternatively, you can apply the Pythagorean theorem to get the equation $$(\mu h)^2+b^2 = \left({\mu a\over 1-\mu}\right)^2,$$ a quartic in $\mu$.