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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

enter image description here

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.

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  • $\begingroup$ Seems doable in principle. If you let $\mu$ be the ratio of the lower altitude to $h$, you can set up a quartic equation in $\mu$. $\endgroup$ – amd Dec 21 '18 at 23:50
  • $\begingroup$ @amd So $\mu = \frac{\text{lower altitude}}{h}$? How from there to a quartic? or, what gets raised to the fourth? $\endgroup$ – holomenicus Dec 21 '18 at 23:55
  • $\begingroup$ It’s a quartic in $\mu$. Alternatively, you can set up a pair of equations involving trig function of $\theta$ and then eliminate $\mu$ from them. $\endgroup$ – amd Dec 22 '18 at 0:08
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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.

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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 $\mu$.

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  • $\begingroup$ For a $\mu$-free solution, isolate $b\cot\theta$ in your $h$ equation, and multiply-through by $\sin\theta$ ... $$\begin{align} \sin\theta (h-a\cos\theta)= b \cos\theta &\quad\to\quad \sin^2\theta (h-a\cos\theta)^2 = b^2 \cos^2\theta \\ &\quad\to\quad (1-\cos^2\theta) (h-a\cos\theta)^2 = b^2 \cos^2\theta \end{align}$$ Abbreviating $\cos\theta$ as $k$, this gives $$a^2 k^4- 2 a h k^3+ ( h^2 - a^2 + b^2 )k^2+ 2 a h k - h^2 = 0$$ This quartic, in general, has ugly roots. $\endgroup$ – Blue Dec 22 '18 at 1:29
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    $\begingroup$ @Blue You've got that right. $ echo "solve(a^2 * k^4 - 2*a*h*k^3 + (h^2 - a^2 + b^2)*k^2 + 2*a*h*k - h^2 = 0, k);" | maxima | wc -l prints 295 lines of equation to express the roots. $\endgroup$ – holomenicus Dec 22 '18 at 1:39
  • $\begingroup$ @Blue I suspect that a quartic is unavoidable. $\endgroup$ – amd Dec 22 '18 at 2:32
  • $\begingroup$ @amd: Agreed. By the way, under the substitution $\sin\theta = 2t/(1+t^2)$, $\cos\theta = (1-t^2)/(1+t^2)$ in the $h$ equation, we get this quartic $$b t^4 + 2 (a+h) t^3 - 2 (a-h) t - b = 0$$ The roots are still ugly, but at least the coefficients are nicer. :) $\endgroup$ – Blue Dec 22 '18 at 2:38

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