Răspuns:
folosește triunghiul dreptunghic care are unghiurile ascuțite de B=30 și C=60 grade.
La acesta exista o teorema care spune ca la unghiul de 30° se opune cateta AC=ipotenuza/2=BC/2
notam AC=l
BC=2l
aplic Pitagora
AB=l^2-l^2/4=3l^2/4
AB=lrad3 /2
Cum sinB=sin30=AC/BC=1/2
sinC=sin60=AB/BC=rad3 /2
[tex]\displaystyle\bf\\Construim~\Delta ABC~echilateral~cu~inaltimea~AD~D\in BC.\\Latura~\Delta ABC~o~notam~cu~L.\\AB=LVezi~desenul~atasat.\\In~triunghiul~echilateral~inaltimea~este~si~bisectoare\\si~mediana~si~mediatoare.\\\\BD=CD~deoarece~inaltimea~este~si~mediana.\\\\\implies~BD=\frac{BC}{2}=\frac{L}{2}~unde~L=latura~triunghiului~echilateral\\\\\angle ADB=\angle ADC=90^o~deoarece~AD\perp BC~fiind~inaltime.\\\\<BAD=<CAD=\frac{<BAC}{2}=\frac{60^o}{2}=30^o \\deoarece~inaltimea~AD~este~si~bisectoare[/tex]
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[tex]\displaystyle\bf\\In~triunghuil~dreptunghic~ABD~avem"\\\\AB=L\\\\BD=\frac{L}{2}~calculata~mai~sus.\\\\AD=\sqrt{AB^2-BC^2}=\sqrt{L^2-\bigg(\frac{L}{2} \bigg)^2}=\sqrt{L^2-\frac{L^2}{4}}=\\\\\\=\sqrt{\frac{4L^2-L^2}{4}}=\sqrt{\frac{3L^2}{4}}=\frac{L\sqrt{3}}{2}\\\\m(\angle ABD)=60^o~fiind~un~unghi~al~triunghiului~lateral.\\m(\angle BAD)=30^o~~(Calculat~mai~sus.)AB=ipotenuza\\AD~si~BD~~sunt~catete.[/tex]
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[tex]\displaystyle\bf\\Calculam~sin30^o~~si~~sin60^o\\\\sin30^o=sin(<BAD)=\frac{cateta~opusa}{ipotenuza}=\frac{BD}{AB}=\frac{~~\dfrac{L}{2}~~}{L}=\frac{1}{2}\\\\sin60^o=\frac{cateta~alaturata}{ipotenuza}=\frac{AD}{AB}=\frac{~~\dfrac{L\sqrt{3} }{2}~~}{L}=\frac{\sqrt{3}}{2}\\\\Raspuns~final:\\\\\boxed{\bf sin30^o=\frac{1}{2}}\\\\\boxed{\bf sin60^o=\frac{\sqrt{3}}{2}}[/tex]
