# How does the angle of an isosceles triangle change as you increase the height

$$ABC$$ and $$ADB$$ are isosceles triangles. Given $$\beta,$$ $$R$$ and $$h$$, how can I find angle $$\alpha$$?

• $$\beta$$ is the top angle of the triangle $$ABC$$, so $$\angle{ACB}$$.

• $$h$$ is the change in height between $$ABC$$ and $$ADB$$.

• $$R$$ is the length of one of the legs of the isosceles triangle $$ADC$$, so $$\vec{|AD|}$$ and $$\vec{|DC|}$$.

• $$\alpha$$ is the top angle of the triangle $$ADB$$, so $$\angle{ADB}$$

My friend and I worked on it a bit, but we found some really complicated solution that was unusable. The problem itself seems pretty simple, so I feel like we did a wrong step at one point. Any help appreciated!

Here is an image of the Isosceles Triangles better describing the problem: • Could we see your "really complicated", "unusable" solution? Please edit it into the question for better visibility. Among other things, it would save people the trouble of writing up a solution that is no better than what you already have. Feb 16, 2022 at 3:08

Let $$M$$ be the middle of $$AB$$, so $$DM\perp AB$$. Then use trigonometric functions in $$\triangle CMB$$ and $$\triangle DMB$$: $$MB=R\sin\frac\alpha2\\h+CM=R\cos\frac\alpha2\\\tan\frac\beta2=\frac{MB}{CM}$$ Get $$CM$$ in terms of $$MB$$ and $$\beta$$ from the last equation and replace it in the second equation. The first and second equations now contain only two variables, $$MB$$ and $$\alpha$$. Calculate $$MB$$ first. To do that, square the equations and add them together. Use $$\sin^2\frac\alpha2+\cos^2\frac\alpha2=1$$. Then get $$\alpha$$ from the first equation.

• Thank you very much for your response. I’ve tried to solve the polynomial but although I get a solution it is not quite elegant. Is this the nature of the solutions or is there some simpler form that I could get to in the end? Feb 15, 2022 at 22:51
• Would you consider$\beta = 2\arctan (\tan (\alpha/2)(1/(1-(h/(R\cos (\alpha/2))))))$ more usable? Same equations but we solve for $MB$ and $CM$ in the first two then substitute into the third and do some factoring. Feb 16, 2022 at 0:57

One solution is let $$\space\gamma=\dfrac{\alpha}{2}$$

$$\space \sin\gamma = \dfrac{\dfrac{\overrightarrow{AB}}{2}}{R} \implies \gamma=\sin^{-1} \dfrac{\overrightarrow{AB}}{2R} \implies \alpha=2\sin^{-1} \dfrac{\overrightarrow{AB}}{2R}$$

An alternative is to let $$\space g \space$$ be the height from the base to point $$\space C.$$

Then $$\space\tan\dfrac{\beta}{2}=\dfrac{\overrightarrow{AB}}{2g} \implies g=\dfrac{\overrightarrow{AB}}{2\tan\dfrac{\beta}{2}}$$

Let $$\space H=g+h \space$$ which is the altitude of the larger triangle

Then $$\space\tan\dfrac{\alpha}{2}=\dfrac{\overrightarrow{AB}}{2H} \implies \alpha= 2\tan^{-1}\bigg( \dfrac{\overrightarrow{AB}}{2H} \bigg)$$

• We are not given $g$ or $H$. Feb 16, 2022 at 1:00
• @CyclotomicField I defined $\space g\space$ in the first "Then" statement and I defined $\space H\space$ in the following "Let" statement. Feb 16, 2022 at 1:14

Drop from point $$C$$ a line that's perpencular to $$AB$$ at $$H$$. As the triangle is isosceles, let $$HB = HA = a$$. Triangle $$CHB$$ and $$DHB$$ are right triangles.

With some knowledge of trigonometry, you have:

$$a = \sin(\alpha/2)R \tag1$$

$$(a\cot(\beta/2)+h)^2 +a^2 = R^2 \tag2$$

or $$a^2 (1+\cot(\beta/2)^2) +2h\cot(\beta/2)a + h^2-R^2 = 0$$

This is a quadratic equation and it should be easy to solve. Found $$a$$ and you can find $$\alpha = 2\arcsin(a/R).$$

• Thank you very much for your response. I’ve tried to solve the polynomial but although I get a solution it is not quite elegant. Is this the nature of the solutions or is there some simpler form that I could get to in the end? Feb 15, 2022 at 22:50
• I would say it's the nature of the solution. I would suggest you can draw some sample by hand and measure the degree to check. Hope this helps and tbh I think there would be a more optimized solution than mine! Feb 16, 2022 at 23:17

Let $$\theta = \angle BCD = \pi - \frac\beta2$$ and $$\phi = \angle CBD.$$ Then by the Law of Sines,

$$\frac{\sin\phi}{h} = \frac{\sin\theta}{R} = \frac{\sin(\beta/2)}{R}.$$ Therefore $$\phi = \arcsin\left(\frac hR \sin \frac\beta2\right).$$

But also $$\frac\alpha2 + \phi + \theta = \pi.$$

So \begin{align} \alpha &= 2(\pi - \theta - \phi) \\ &= 2 \left(\pi - \left(\pi - \frac\beta2\right) - \phi\right) \\ &= 2 \left(\frac\beta2 - \phi\right) \\ &= \beta - 2\phi \\ &= \beta - 2\arcsin\left(\frac hR \sin \frac\beta2\right) \\ \end{align}