Ian at 11:34, 27 March 2011

2011-03-27T11:34:15Z

←Older revision		Revision as of 11:34, 27 March 2011
Line 22:		Line 22:
	Also		Also

-	<math>\tan \alpha =\frac{-3}{4}~~=B1~~</math>	+	<math>\tan \alpha =\frac{-3}{4}</math>

	giving		giving

-	<math>\alpha =36\textrm{.}9{}^\circ</math>	+	<math>\alpha =-36\textrm{.}9{}^\circ</math>

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=-3\ \text{N}$ and $H=4\ \text{N}$ ....

2010-03-23T11:02:39Z

New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> and <math>\tan \alpha =\frac{V}{H} </math> Here <math>V=-3\ \text{N}</math> and <math>H=4\ \text{N}</math>....

New page

[[Image:teori4.gif]]

Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>

and

<math>\tan \alpha =\frac{V}{H} </math>

Here

<math>V=-3\ \text{N}</math> and
<math>H=4\ \text{N}</math>.

Substituting the given values

<math>{{F}^{2}}={{4}^{2}}+{{\left( -3 \right)}^{2}}=25</math>

giving

<math>F=5\ \text{N}</math>

Also

<math>\tan \alpha =\frac{-3}{4}=B1</math>

giving

<math>\alpha =36\textrm{.}9{}^\circ</math>

Solution 4.5b - Revision history

Ian at 11:34, 27 March 2011

Ian: New page: Image:teori4.gif Using {{F}^{2}}={{H}^{2}}+{{V}^{2}} and \tan \alpha =\frac{V}{H} Here V=-3\ \text{N} and H=4\ \text{N}....

Ian: New page: Image:teori4.gif Using ${{F}^{2}}={{H}^{2}}+{{V}^{2}}$ and $\tan \alpha =\frac{V}{H}$ Here $V=-3\ \text{N}$ and $H=4\ \text{N}$ ....