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Solution 4.4b - Revision history
2026-05-28T11:29:32Z
Revision history for this page on the wiki
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Ian: New page: Image:teori4.gif <math>\tan \alpha =\frac{V}{H}</math> Here <math>V=-10\ \text{N}</math> and <math>H=9\ \text{N}</math>. Thus <math>\tan \alpha =\frac{-10}{9}=-1\textrm{.}1</mat...
2010-03-22T17:32:38Z
<p>New page: <a href="/wikis/2009/bridgecoursemechanics/index.php/Image:Teori4.gif" title="Image:Teori4.gif">Image:teori4.gif</a> <math>\tan \alpha =\frac{V}{H}</math> Here <math>V=-10\ \text{N}</math> and <math>H=9\ \text{N}</math>. Thus <math>\tan \alpha =\frac{-10}{9}=-1\textrm{.}1</mat...</p>
<p><b>New page</b></p><div>[[Image:teori4.gif]]<br />
<br />
<math>\tan \alpha =\frac{V}{H}</math><br />
<br />
Here <br />
<math>V=-10\ \text{N}</math><br />
and <br />
<math>H=9\ \text{N}</math>.<br />
<br />
Thus<br />
<br />
<math>\tan \alpha =\frac{-10}{9}=-1\textrm{.}1</math><br />
giving <br />
<math>\alpha =-48{}^\circ</math></div>
Ian