Question

(TRIGONOMETRIA) (LEI DOS COSSENOS) (Unesp) : Os lados de um triângulo medem 2√3,√6 e 3 + √3. Determine o ângulo oposto ao lado que mede √6 Me enrolei na parte algébrica... Alguém pode responder? PLEASE!  Gabarito: 30°

Answer 1

$a^2=b^2+c^2+2.b.c.cos \alpha\\ \\ (\sqrt6)^2=(2\sqrt3)^2+(3+\sqrt3)^2-2.2\sqrt3.(3+\sqrt3).cos \alpha\\ \\ 6=12+9+6\sqrt3+3-4\sqrt3(3+\sqrt3).cos \alpha\\ \\ 6=12+9+6\sqrt3+3-(12\sqrt3+12).cos \alpha\\ \\ (12\sqrt3+12).cos \alpha=12+9+3-6+6\sqrt3\\ \\ (12\sqrt3+12).cos \alpha=18+6\sqrt3\\ \\ cos \alpha=\frac{18+6\sqrt3}{12+12\sqrt3}=\frac{3+\sqrt3}{2+2\sqrt3}\\ \\ cos \alpha=\frac{3+\sqrt3}{2+2\sqrt3}.\frac{2-2\sqrt3}{2-2\sqrt3}=\frac{-4\sqrt3}{4-12}=\frac{-4\sqrt3}{-8}=\frac{\sqrt3}{2}$

Logo α = 30°