the bottom or base edge b, the top left edge a, the

the bottom or base edge b,
the top left edge a,
the top right edge c.
Let a=c+2 and b=10.
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Isla Klein
Taking the origin as the intersection of a and b, with the equation in terms of the angle $\theta$ at that vertex:
Through the cosine law we have:
$\left(a-2{\right)}^{2}={a}^{2}+100-20a\mathrm{cos}\theta$
$-4a+4=100-20a\mathrm{cos}\theta$
$a=\frac{96}{20\mathrm{cos}\theta -4}$
With our particular setup, a is the radius. So this is a polar equation, where
$r\left(\theta \right)=\frac{96}{20\mathrm{cos}\theta -4}=\frac{24}{5\mathrm{cos}\theta -1}$
we can convert to cartesian coordinates if you want:
$5r\mathrm{cos}\theta -r=24$
$\left(5r\mathrm{cos}\theta -24{\right)}^{2}={r}^{2}$
$\left(5x-24{\right)}^{2}={x}^{2}+{y}^{2}$
${y}^{2}=24{x}^{2}-240x+576$
${y}^{2}=24\left(x-5{\right)}^{2}-24$
$\left(x-5{\right)}^{2}-\frac{{y}^{2}}{24}=1$
which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\mathrm{cos}\theta <\frac{1}{5}$