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

Question

$ABC$ 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:

Answer 1

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.

Answer 2

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

Answer 3

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 $\alpha = 2\arcsin(a/R).$

Answer 4

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}