Lösung 4.2:1f

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<center> [[Image:4_2_1f.gif]] </center>
 
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[[Image:4_2_1_f.gif|center]]
[[Image:4_2_1_f.gif|center]]
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The side adjacent to the angle of
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<math>\text{5}0^{\circ }</math>
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is marked
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<math>x</math>
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and the opposite is the side of length
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<math>\text{19}</math>.
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If we write the tangent for the angle, this gives a relation which contains
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<math>x</math>
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as the only unknown,
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<math>\tan 50^{\circ }=\frac{19}{x}</math>
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This gives
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<math>x=\frac{19}{\tan 50^{\circ }}\quad \left( \approx 15.9 \right)</math>

Version vom 11:17, 28. Sep. 2008

The side adjacent to the angle of \displaystyle \text{5}0^{\circ } is marked \displaystyle x and the opposite is the side of length \displaystyle \text{19}.


If we write the tangent for the angle, this gives a relation which contains \displaystyle x as the only unknown,


\displaystyle \tan 50^{\circ }=\frac{19}{x}


This gives


\displaystyle x=\frac{19}{\tan 50^{\circ }}\quad \left( \approx 15.9 \right)

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:1f“