Respuesta :
Vectors
Given a vector of components A = (Ax, Ay), the magnitude of the vector is:
[tex]|A|=\sqrt[]{A^2_x+A^2_y}[/tex]And the angle in standard form (counterclockwise from the positive x-direction is given by:
[tex]\tan \theta=\frac{A_y}{A_x}[/tex]The vector A has components Ax = -5.50 m and Ay = 7.50 m. The magnitude is calculated as follows:
[tex]\begin{gathered} |A|=\sqrt[]{(-5,50)^2+7.50^2} \\ \text{Operating:} \\ |A|=\sqrt[]{30.25+56.25} \\ |A|=\sqrt[]{86.5} \\ |A|=9.3 \end{gathered}[/tex]The magnitude of A is 9.3 m.
Calculate the angle. It's important to notice the x-coordinate is negative and the y-coordinate is positive, thus the angle is in quadrant II (between 90° and 180°).
[tex]\tan \theta=\frac{7.50}{-5.50}=-1.3636[/tex]Calculating the angle with the inverse tangent function:
[tex]\theta=-53.746^o[/tex]This angle is not expressed correctly in the second quadrant, so we have to add 180°:
[tex]\theta=-53.746^o+180\~[/tex][tex]\theta=126.254^o[/tex]