Question

Să se rezolve un triunghi ABC în care
b-a = 2, B - A = 30°, c = 2 + 12.​

Answer 1

Răspuns:

$b-a=2\\B-A=30^*\\c=2+2\sqrt{3}$

Teorema sinusului:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=>\frac{b-a}{sinB-sinA}=\frac{c}{sinC}$

$\frac{2}{2sin\frac{B-A}{2}*cos\frac{B+A}{2} }=\frac{2(1+\sqrt{3}) }{sin(2*\frac{C}{2}) }$

$\frac{1}{sin15*cos\frac{B+A}{2} }=\frac{2(1+\sqrt{3} )}{2sin\frac{C}{2}*cos\frac{C}{2} }$

$A+B+C=180=>C=180-(A+B)$

$sin\frac{C}{2}=sin(\frac{180-(A+B)}{2} )=sin(90-\frac{A+B}{2} )=cos\frac{A+B}{2}$

Inlocuim:

$2cos\frac{B+A}{2} *cos\frac{C}{2}=2(1+\sqrt{3} )*\frac{\sqrt{6}-\sqrt{2} }{4} *cos\frac{B+A}{2} | :(cos\frac{B+A}{2})$ (Am impartit la cos[B+2/2])

$2cos\frac{C}{2} =\frac{(1+\sqrt{3} )(\sqrt{6} -\sqrt{2} )}{2}$

$4cos\frac{C}{2}=\sqrt{6}-\sqrt{2}+\sqrt{18}-\sqrt{6}$

$4cos\frac{C}{2}=-\sqrt{2}+3\sqrt{2}$

$4cos\frac{C}{2}=2\sqrt{2}$

$cos\frac{C}{2}=\frac{\sqrt{2} }{2}=> \frac{C}{2}=45^*=>C=90^*$

$A+B+C=180^*=>C=180^*-B-A$ (Si cum B=A+30) =>

$C=180^*-A-30^*-A=> C = 150-2A=> 2A=150-C <=> 2A=150-90$

$=60^*=>A=30^*$

$A=30^*; C=90^* => B =60^*$

Etapa 2) Acum vom afla laturile:

$sinA=\frac{a}{2+2\sqrt{3} }=\frac{1}{2}=>a=\frac{2(1+\sqrt{3} )}{2}=1+\sqrt{3}$

$sinB=\frac{b}{2(1+\sqrt{3} )}=\frac{\sqrt{3} }{2}=>b=\frac{2(1+\sqrt{3} )*\sqrt{3} }{2}=\sqrt{3}+3$

And we are done!! :)