Respuesta :
[tex]\bf \begin{cases}
r=8\\
\theta =30^o
\end{cases}\qquad \qquad \qquad \qquad
\begin{cases}
tan(\theta )=\frac{b}{a}\\
r^2=a^2+b^2
\end{cases}\\\\
-------------------------------\\\\
tan(30^o)=\cfrac{b}{a}\implies \cfrac{\quad\frac{1}{2} \quad }{\frac{\sqrt{3}}{2}}=\cfrac{b}{a}\implies \cfrac{1}{\sqrt{3}}=\cfrac{b}{a}\implies \boxed{\cfrac{a}{\sqrt{3}}=b}\\\\
-------------------------------[/tex]
[tex]\bf 8^2=a^2+b^2\implies 64=a^2+b^2\implies 64=a^2+\left( \boxed{\cfrac{a}{\sqrt{3}}} \right)^2 \\\\\\ 64=a^2+\cfrac{a^2}{3}\implies 64=\cfrac{4a^2}{3}\implies 16=\cfrac{a^2}{3}\implies 48=a^2 \\\\\\ \sqrt{48}=a\implies \boxed{4\sqrt{3}=a}\\\\ -------------------------------\\\\ \cfrac{a}{\sqrt{3}}=b\implies \cfrac{4\sqrt{3}}{\sqrt{3}}=b\implies \boxed{4=b}\\\\ -------------------------------\\\\ v~=~4\sqrt{3}i~+~4j[/tex]
[tex]\bf 8^2=a^2+b^2\implies 64=a^2+b^2\implies 64=a^2+\left( \boxed{\cfrac{a}{\sqrt{3}}} \right)^2 \\\\\\ 64=a^2+\cfrac{a^2}{3}\implies 64=\cfrac{4a^2}{3}\implies 16=\cfrac{a^2}{3}\implies 48=a^2 \\\\\\ \sqrt{48}=a\implies \boxed{4\sqrt{3}=a}\\\\ -------------------------------\\\\ \cfrac{a}{\sqrt{3}}=b\implies \cfrac{4\sqrt{3}}{\sqrt{3}}=b\implies \boxed{4=b}\\\\ -------------------------------\\\\ v~=~4\sqrt{3}i~+~4j[/tex]
If the magnitude is 8 and the angle (θ) is 30°, then the vector v is given as v = (4√3)i + 4j. Then the correct option is C.
What is a vector?
The quantity which has magnitude, direction, and follows the law of vector addition is called a vector.
Write the vector v in terms of i and j whose magnitude and direction angle θ are given.
Magnitude (A) = 8, θ = 30°
The vector equation is given as
v = (A cos θ)i + (A sin θ)j
v = (8 × cos 30°)i + (8 × sin 30°)j
v = (4√3)i + 4j
More about the vector link is given below.
https://brainly.com/question/13188123
