[tex]\it c) Th.Pitagora\ \Rightarrow AB^2=BC^2-AC^2=(\sqrt3)^2-1^2=3-1=2\ \Rightarrow AB=\sqrt2\\ \\ sinB=\dfrac{AC}{BC}=\dfrac{^{\sqrt3)}1}{\ \sqrt3}=\dfrac{\sqrt3}{3}\\ \\ \\ cosB=\dfrac{AB}{BC}=\dfrac{^{\sqrt3)}\sqrt2}{\ \sqrt3}=\dfrac{\sqrt6}{3}\\ \\ \\ tgB=\dfrac{sinB}{cosB}=\dfrac{\dfrac{\sqrt3}{3}}{\dfrac{\sqrt6}{3}}=\dfrac{\sqrt3}{3}:\dfrac{\sqrt6}{3}=\dfrac{\sqrt3}{\not3}\cdot\dfrac{\not3}{\sqrt6}=\dfrac{\ \ \sqrt3^{(\sqrt3}}{\sqrt2\cdot\sqrt3}=\dfrac{^{\sqrt2}1}{\sqrt2}=\dfrac{\sqrt2}{2}[/tex]
[tex]\it d)\ \ Th. Pitagora \Rightarrow BC=a\sqrt2\\ \\ AB=AC=a \Rightarrow \Delta ABC-triunghi\ dreptunghic\ isoscel \Rightarrow \hat B=\hat C=45^o\\ \\ Func\c{\it t}iile\ \ trigonometrice\ se \ g\breve asesc\ la\ tabelul\ din\ carte\ (caiet)[/tex]
[tex]\it e)\ \ Th.\ Pitagora \Rightarrow AC^2=4a^2-a^2=3a^2 \Rightarrow AC=\sqrt{3a^2}=a\sqrt3\\ \\ BC=2AB \Rightarrow \hat C=30^o,\ \ \hat B=60^o[/tex]
