# I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y. I kn

I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.
I know the length of the long sides, X.
I also know the acute, vertex angle opposite the base Y, lets
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euromillionsna
Step 1
Cut the iscoles triangle in half to get a two right triangles with opposite side $\frac{1}{2}y$ and hypotenuse x.
$\frac{1}{2}Y=\mathrm{sin}\left(\frac{1}{2}a\right)x$
So apparently this is claiming
$2\mathrm{sin}\left(\frac{1}{2}a\right)=\mathrm{tan}a$
which isnt
###### Not exactly what you’re looking for?
ataill0k

Step 1
Bisect angle a
$\mathrm{sin}\frac{a}{2}=\frac{y}{2x}$
$\mathrm{cos}\frac{a}{2}=\frac{\sqrt{4{x}^{2}-{y}^{2}}}{2x}$
$\mathrm{tan}\frac{a}{2}=\frac{y}{\sqrt{4{x}^{2}-{y}^{2}}}$
$\mathrm{tan}a=\frac{2\mathrm{tan}\frac{a}{2}}{1-{\mathrm{tan}}^{2}\frac{a}{2}}$
$\mathrm{tan}a=\frac{2y}{\sqrt{4{x}^{2}-{y}^{2}}\left(1-\frac{{y}^{2}}{4{x}^{2}-{y}^{2}}\right)}$
$\mathrm{tan}a=\frac{y\sqrt{4{x}^{2}-{y}^{2}}}{2{x}^{2}-{y}^{2}}$
$\frac{y}{x}$ is small, $\frac{\sqrt{4{x}^{2}-{y}^{2}}}{2{x}^{2}-{y}^{2}}\approx \frac{1}{x}$
and $\mathrm{tan}a\approx \frac{y}{x}$