MHB Eb 4.d determine the magnitude and direction of V

[karush]

$\tiny{\\ eb 4.d}$
$\textsf{If $v_x=-9.80$ units and $v_y=-6.40$ units,}\\$
$\textit{determine the magnitude and direction of $V$.}$
\begin{align*}\displaystyle
magnitude&=\sqrt{(-9.8)^2+(-6.4)^2}
=\color{red}{11.7047}\\
direction&=\arctan{\left[\frac{6.4}{-9.8}\right]}
=\color{red}{33.15^o}
\end{align*}

ok the direction is actually in the Q4 but $\arctan{\left[\frac{-6.4}{-9.8}\right]}$
is in Q1! Do we just add $180^o$ for that or is it $-33.15^o$

Also is there symbols for magnitude and direction other than an arrow on the graph

Also where is the latex for degree instead of ^o

 

[MarkFL]

$\tiny{\\ eb 4.d}$
$\textsf{If $v_x=-9.80$ units and $v_y=-6.40$ units,}\\$
$\textit{determine the magnitude and direction of $V$.}$
\begin{align*}\displaystyle
magnitude&=\sqrt{(-9.8)^2+(-6.4)^2}
=\color{red}{11.7047}\\
direction&=\arctan{\left[\frac{6.4}{-9.8}\right]}
=\color{red}{33.15^o}
\end{align*}

ok the direction is actually in the Q4 but $\arctan{\left[\frac{-6.4}{-9.8}\right]}$
is in Q1! Do we just add $180^o$ for that or is it $-33.15^o$

The direction would be in quadrant III, and so recognizing that both components are negative, you would in fact add $\pi$ to the angle:

$$\theta=\pi+\arctan\left(\frac{v_y}{v_x}\right)$$

Also is there symbols for magnitude and direction other than an arrow on the graph

Also where is the latex for degree instead of ^o

I use ^{\circ} for the degree symbol.

 

[karush]

The direction would be in quadrant III, and so recognizing that both components are negative, you would in fact add $\pi$ to the angle:

$$\theta=\pi+\arctan\left(\frac{v_y}{v_x}\right)$$

$\textsf{ok so it}$ $$33.15^\circ + 180^\circ = 213.15^\circ $$
that would place it Q3

I use ^{\circ} for the degree symbol.

actually why don't we have \degree on the Symbol/Command Set:

 

[MarkFL]

It was an oversight I suppose, but adding it now would require a lot of effort. :)

 

[karush]

I completely understand