Question

DETERMINE O VALOR DE X NO TRIÂNGULO ABAIXO, SABENDO QUE COS120⁰= COS60⁰

A)10

B)12

C)16

D)-10

E)-16​

Answer 1

Resposta:

Explicação passo a passo:

X²+6² -2.x.6.cos120 = 14²

X²+36 -12x.(-1/2) = 196

X² + 6x - 160 = 0

∆=36+640= 676. ==> √∆=26

x'= -b+√∆/2a = -6+26/2 = 20/2= 10 ✓

Answer 2

$\large\boxed{\begin{array}{l}\rm 14^2=6^2+x^2-2\cdot 6\cdot x\cdot cos(120^\circ)\\\rm 196=36+x^2-\backslash\!\!\!2\cdot6\cdot x\cdot\dfrac{1}{\backslash\!\!\!2}\\\\\rm 196=36+x^2-6x\\\rm x^2-6x+36-196=0\end{array}}$

$\large\boxed{\begin{array}{l}\rm x^2-6x-160=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-6)^2-4\cdot1\cdot(-160)\\\rm\Delta=36+640\\\rm\Delta=676\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-6)\pm\sqrt{676}}{2\cdot1}\\\\\rm x=\dfrac{6\pm26}{2}\\\\\\\rm x=\dfrac{6+26}{2}=\dfrac{32}{2}=16\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~C} }}}}\end{array}}$