I have little math problem that I wish to solve but somehow I am having a hard time working it out... Say we have an isoceles triangle with sides equal to r and a base equal to b. Solve for b in terms of r.
Is there a way to do it? Without needing info about the angles. I tried using a combination of The pythagorean theorem and Herons formula but it did not work out, even though it seems like it should. Hehe. XD
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As already noted, we need one more piece of information. But it doesn't have to be an angle. We could choose this to be the triangle's height. By Pythagoras, $b=2\sqrt{r^2-h^2}$. Or in terms of the area $\Delta=bh/2$,$$b^2=4(r^2-4\Delta^2/b^2)\implies b^4-4r^2b^2+16\Delta^2=0\implies b^2=2r^2\pm\sqrt{4r^4-16\Delta^2}.$$This actually gives two solutions, since replacing an apex angle $\theta$ with $\pi-\theta$ preserves the area.
Consider a circle of radius $r$. If you draw any two radii of this circle, you can draw an isosceles triangle with two sides equal to $r$. But you'll notice that you can make $b$ of any length from $0$ to $2r$! Since $b$ is not fixed, there is no expression of $b$ in terms of $r$.
If you add the angle $\theta$ between the two radii, then $b=2r\sin(\frac\theta2)$.