Solution 16.7b

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(New page: From part a), <math>\mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j}</math> which gives, <math>\begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...)
Current revision (17:21, 10 March 2011) (edit) (undo)
 
Line 4: Line 4:
<math>\begin{align}
<math>\begin{align}
-
& \tan \alpha =\frac{4}{2\textrm{.}5} \\
+
& \tan \alpha =\frac{4}{2\textrm{.}5} \\
 +
& \\
& \alpha =58\textrm{.}0{}^\circ \\
& \alpha =58\textrm{.}0{}^\circ \\
\end{align}</math>
\end{align}</math>

Current revision

From part a),

\displaystyle \mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j} which gives,

\displaystyle \begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \\ & \alpha =58\textrm{.}0{}^\circ \\ \end{align}