Răspuns:
[tex]b-a=2\\B-A=30^*\\c=2+2\sqrt{3}[/tex]
Teorema sinusului:
[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=>\frac{b-a}{sinB-sinA}=\frac{c}{sinC}[/tex]
[tex]\frac{2}{2sin\frac{B-A}{2}*cos\frac{B+A}{2} }=\frac{2(1+\sqrt{3}) }{sin(2*\frac{C}{2}) }[/tex]
[tex]\frac{1}{sin15*cos\frac{B+A}{2} }=\frac{2(1+\sqrt{3} )}{2sin\frac{C}{2}*cos\frac{C}{2} }[/tex]
[tex]A+B+C=180=>C=180-(A+B)[/tex]
[tex]sin\frac{C}{2}=sin(\frac{180-(A+B)}{2} )=sin(90-\frac{A+B}{2} )=cos\frac{A+B}{2}[/tex]
Inlocuim:
[tex]2cos\frac{B+A}{2} *cos\frac{C}{2}=2(1+\sqrt{3} )*\frac{\sqrt{6}-\sqrt{2} }{4} *cos\frac{B+A}{2} | :(cos\frac{B+A}{2})[/tex] (Am impartit la cos[B+2/2])
[tex]2cos\frac{C}{2} =\frac{(1+\sqrt{3} )(\sqrt{6} -\sqrt{2} )}{2}[/tex]
[tex]4cos\frac{C}{2}=\sqrt{6}-\sqrt{2}+\sqrt{18}-\sqrt{6}[/tex]
[tex]4cos\frac{C}{2}=-\sqrt{2}+3\sqrt{2}[/tex]
[tex]4cos\frac{C}{2}=2\sqrt{2}[/tex]
[tex]cos\frac{C}{2}=\frac{\sqrt{2} }{2}=> \frac{C}{2}=45^*=>C=90^*[/tex]
[tex]A+B+C=180^*=>C=180^*-B-A[/tex] (Si cum B=A+30) =>
[tex]C=180^*-A-30^*-A=> C = 150-2A=> 2A=150-C <=> 2A=150-90[/tex]
[tex]=60^*=>A=30^*[/tex]
[tex]A=30^*; C=90^* => B =60^*[/tex]
Etapa 2) Acum vom afla laturile:
[tex]sinA=\frac{a}{2+2\sqrt{3} }=\frac{1}{2}=>a=\frac{2(1+\sqrt{3} )}{2}=1+\sqrt{3}[/tex]
[tex]sinB=\frac{b}{2(1+\sqrt{3} )}=\frac{\sqrt{3} }{2}=>b=\frac{2(1+\sqrt{3} )*\sqrt{3} }{2}=\sqrt{3}+3[/tex]
And we are done!! :)
