Solution 4.5b

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(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>....)
Current revision (11:34, 27 March 2011) (edit) (undo)
 
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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>

Current revision

Image:teori4.gif

Using \displaystyle {{F}^{2}}={{H}^{2}}+{{V}^{2}}

and

\displaystyle \tan \alpha =\frac{V}{H}

Here

\displaystyle V=-3\ \text{N} and \displaystyle H=4\ \text{N}.

Substituting the given values

\displaystyle {{F}^{2}}={{4}^{2}}+{{\left( -3 \right)}^{2}}=25

giving

\displaystyle F=5\ \text{N}

Also

\displaystyle \tan \alpha =\frac{-3}{4}

giving

\displaystyle \alpha =-36\textrm{.}9{}^\circ