Question

TG 39 admitere UPT2019

Answer 1

Vezi poza atasata!!!!!

Answer 2

$\it Fie\ BD\perp AC \Rightarrow \dfrac{AC\cdot BD}{2}=\mathcal{A}_{ABC} \Rightarrow \dfrac{8\cdot BD}{2}=16\sqrt2 \Rightarrow BD = 4\sqrt2\\ \\ Din\ \Delta ABD \Rightarrow sinA = \dfrac{BD}{AB} = \dfrac{4\sqrt2}{6} = \dfrac{2\sqrt2}{3}\\ \\ cosA=\sqrt{1-sin^2A} =\sqrt{1-\left(\dfrac{2\sqrt2}{3}\right)^2}=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\ \ \ \ (1)\\ \\ \\ Din\ \Delta ABD \Rightarrow cosA = \dfrac{AD}{AB} = \dfrac{AD}{6} \ \ \ \ (2)$

$\it (1),(2)\Rightarrow \dfrac{AD}{6}=\dfrac{1}{3} \Rightarrow AD=2\\ \\ CD=AC-AD=8-2=6\\ \\ Din\ \Delta DBC,\ cu\ teorema\ lui\ Pitagora,\ rezult\breve{a}\ \ BC=2\sqrt{17}\\ \\ Din\ \Delta DBC \Rightarrow sinC=\dfrac{4\sqrt2}{2\sqrt{17}}=\dfrac{2\sqrt{34}}{17}$