Question

DAU COROANA!!!
Urgent,va rog

Answer 1

$\tg(\angle B)=\tfrac{AM}{MB}\\ \tg(60°)=\tfrac{10}{MB}\\ \sqrt{3}=\tfrac{10}{MB}\\ \implies MB=\tfrac{10\sqrt{3}}{3}\\ \\ \\ AM²=MC×MB\\10²=MC×(\tfrac{10\sqrt{3}}{3})²\\ \tfrac{10\sqrt{3}×MC}{2}=100\\ MC=\tfrac{30}{\sqrt{3}}=10\sqrt{3}\\ \\ \implies BC=\tfrac{40\sqrt{3}}{3}\\ \\ AB²=10²+(\tfrac{10\sqrt{3}}{3})² \implies AB=\tfrac{20\sqrt{3}}{3}\\ \\ \iff BC²=AC²+AB²\\ \iff (\tfrac{40\sqrt{3}}{3})²=(\tfrac{20\sqrt{3}}{3})²+AC²\\ \implies AC=20\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ a)\: \: P_{ABC}=20+\tfrac{40\sqrt{3}}{3}+\tfrac{20\sqrt{3}}{3}=\color{red}20\sqrt{3}+400\: \: cm\\ \\ \\ b)\: \: \frac{A_{ABD}}{A_{ABC}}=\frac{\frac{c_{1}×c_{2}}{2}}{\tfrac{c_{1}×c_{2}}{2}}\\=\frac{50\sqrt{3}}{\tfrac{200\sqrt{3}}{2}}=\frac{50\sqrt{3}}{100\sqrt{3}}=\color{red}\frac{1}{2}$