Respuesta :
Step 1. The vector that we have is:
[tex]\vec{w}=-45i+28j[/tex]This vector is in the form
[tex]\vec{w}=ai+bj[/tex]where
[tex]\begin{gathered} a=-45 \\ b=28 \end{gathered}[/tex]And we need to find the trigonometric form of the vector.
Step 2. To find the trigonometric form we use this formula:
[tex]\vec{w}=|w|cos\theta i+|w|sin\theta j[/tex]Where |w| is the length of the vector and theta is the angle.
Step 3. Finding |w|, the length of the vector:
[tex]|w|=\sqrt{a^2+b^2}[/tex]in this case:
[tex]|w|=\sqrt{(-45)^2+28^2}[/tex]Solving the operations the result is:
[tex]\begin{gathered} \lvert w\rvert=\sqrt{2025+284} \\ |w|=53 \end{gathered}[/tex]Step 4. Now, we find the angle using:
[tex]\theta=\tan^{-1}(\frac{b}{a})[/tex]In this case:
[tex]\theta=\tan^{-1}(\frac{28}{-45})[/tex]Solving the operations and rounding to the nearest tenth of a degree:
[tex]\theta=-32[/tex]Step 5. Going back to the formula from step 2:
[tex]\vec{w}=\lvert w\rvert cos\theta\imaginaryI+\lvert w\rvert s\imaginaryI n\theta j[/tex]Substituting |w| and theta:
[tex]\vec{w}=53cos(-32)\imaginaryI+53s\imaginaryI n(-32)j[/tex]We can either leave the result as it is, or use the following properties of the cosine and sine to simplify:
[tex]\begin{gathered} cos(-A)=cosA \\ sin(-A)=-sinA \end{gathered}[/tex]And the result is simplified as follows:
[tex]\vec{w}=53cos(32)\imaginaryI-53s\imaginaryI n(32)j[/tex]Answer:
[tex]\vec{w}=53cos(32)\imaginaryI-53s\imaginaryI n(32)j[/tex]