Ian at 17:21, 10 March 2011

2011-03-10T17:21:20Z

←Older revision		Revision as of 17:21, 10 March 2011
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>

Ian: New page: From part a), $\mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j}$ which gives, $\begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...$

2010-09-23T18:57:21Z

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 \\ \...

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 \\
\end{align}</math>

Solution 16.7b - Revision history

Ian at 17:21, 10 March 2011

Ian: New page: From part a), \mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j} which gives, \begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...

Ian: New page: From part a), $\mathbf{v}=2\textrm{.}5\mathbf{i}+4\mathbf{j}$ which gives, $\begin{align} & \tan \alpha =\frac{4}{2\textrm{.}5} \\ & \alpha =58\textrm{.}0{}^\circ \\ \...$