# Find area of triangle ABD

Let ABC be a right-angled triangle $B=90$. Let in a triangle pick a point $D$ inside the triangele such that $AD= 20, DC=15, DB=10$ and $AB=2BC$. What is the area of triangle $ABD$? Thanks!

i have got three equations $$5a^2=20^2+15^2-2\cdot 20\cdot 15\cos(2\pi-\alpha-\beta)$$ $$a^2=15^2+10^2-2\cdot 15\cdot 10\cos(\beta)$$ $$4a^2=10^2+20^2-2\cdot 10\cdot20\cos(\alpha)$$ solving this we get $$\alpha=\arctan\left(\frac{1}{2}\right)$$ and $$A_{\Delta ABD}=\frac{1}{2}20\cdot 10\sin\left(\arctan\left(\frac{1}{2}\right)\right)$$