Question

Os lados de um triangulo medem a=2√3, b=2√2,c=√6+√2 .Determine as medidas dos ângulos internos

Answer 1

Podemos usar o teorema do cosseno para achar um angulo e o teorema do seno para achar os outros:

$a^2=b^2+c^2-2bc\cos A\\\\12=8+8+4\sqrt3-2(2\sqrt2)(\sqrt6+\sqrt2)\cos A\\\\-4-4\sqrt3=(-8\sqrt3-8)\cos A\\\\\cos A=\dfrac{4+4\sqrt3}{8+8\sqrt3}\\\\\cos A=\dfrac12\\\\A=60^ \circ$

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}\\\\\dfrac{\frac{\pi}{3}}{2\sqrt3}=\dfrac{\sin B}{2\sqrt2}\\\\\sin B=\dfrac{\pi}{3}\dfrac{\sqrt2}{\sqrt3}=\dfrac{\pi}{9}\sqrt6\\\\B\approx58.76\\\\\\A+B+C=180\\\\60+58.76+C=180\\\\C=61.24$