Lösung 4.2:8

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We start by drawing three auxiliary triangles, and calling the three vertical sides
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<center> [[Image:4_2_8-1(2).gif]] </center>
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<math>x,\ y</math>
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and
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<math>z</math>, as shown in the figure.
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<center> [[Image:4_2_8-2(2).gif]] </center>
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[[Image:4_2_8.gif|center]]
[[Image:4_2_8.gif|center]]
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Using the definition of cosine, we can work out
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<math>x\text{ }</math>
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and
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<math>y</math>
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from
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<math>x=a\cos \alpha </math>
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<math>y=b\cos \beta </math>
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and, for the same reason, we know that
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<math>z\text{ }</math>
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satisfies the relation
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<math>z=l\cos \gamma </math>
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In addition, we know that the lengths
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<math>x,\ y</math>
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and
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<math>z</math>
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satisfy the equality
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<math>z=x-y</math>
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If we substitute in the expressions for
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<math>x,\ y</math>
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and
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<math>z</math>, we obtain the trigonometric equation
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<math>l\cos \gamma =a\cos \alpha -b\cos \beta </math>
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where
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<math>\gamma </math>
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is the only unknown.

Version vom 09:40, 29. Sep. 2008

We start by drawing three auxiliary triangles, and calling the three vertical sides \displaystyle x,\ y and \displaystyle z, as shown in the figure.


Using the definition of cosine, we can work out \displaystyle x\text{ } and \displaystyle y from


\displaystyle x=a\cos \alpha


\displaystyle y=b\cos \beta

and, for the same reason, we know that \displaystyle z\text{ } satisfies the relation


\displaystyle z=l\cos \gamma


In addition, we know that the lengths \displaystyle x,\ y and \displaystyle z satisfy the equality


\displaystyle z=x-y


If we substitute in the expressions for \displaystyle x,\ y and \displaystyle z, we obtain the trigonometric equation


\displaystyle l\cos \gamma =a\cos \alpha -b\cos \beta


where \displaystyle \gamma is the only unknown.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:8“