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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!

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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 $$A_{\Delta ABD}=\frac{1}{2}20\cdot 10\sin\left(\arctan\left(\frac{1}{2}\right)\right)$$

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  • $\begingroup$ Excuse me, can you give me tips for solving equations? $\endgroup$ – Erdös Oct 11 '17 at 18:55

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