[tex]\it \Delta ABD-dr,\ \ \widehat{BDA}=90^o,\ \ \hat B=45^o \Rightarrow \Delta ABD-dr.\ is. \Rightarrow AD=AD=4,5\ cm\\ \\ \\ \hat C=180^o-(105^o+45^o)=180^o-150^o=30^o[/tex]
(acum urmează spectacolul, cu ascunzișul învârtit după cireș)
[tex]\bf Din\ \Delta ADC\ \Rightarrow\ tgC=\dfrac{cateta\ \ opus\breve a}{cateta\ \ al\breve a turat\breve a}\ \Rightarrow\ tg30^o= \dfrac{AD}{CD}\ \Rightarrow\ \\ \\ \\ \ \Rightarrow\ \dfrac{\sqrt3}{3}=\dfrac{4,5}{CD}\ \Rightarrow\ CD=\dfrac{^{\sqrt3)}3\cdot4,5}{\ \sqrt3}=\dfrac{\not3\cdot4.5\sqrt3}{\not3}=4,5\sqrt3\ cm[/tex]
[tex]\bf \Delta ADC-dr,\ \ \hat D=90^o,\ \ \stackrel{T.P.}{\Longrightarrow}\ \ AC^2=AD^2+CD^2\ \Rightarrow\ \\ \\ \ \Rightarrow\ AC^2=4,5^2+(4,5\sqrt3)^2=20,25+20,25\cdot3=20,25(1+3)=20,25\cdot4=\\ \\ =81\ \Rightarrow\ AC=\sqrt{81}=9\ cm[/tex]
