Lösung 4.2:1a

Aus Online Mathematik Brückenkurs 1

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<center> [[Image:4_2_1a.gif]] </center>
 
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[[Image:4_2_1_a.gif|center]]
[[Image:4_2_1_a.gif|center]]
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The definition of the tangent states that
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<math>\tan u=\frac{\text{opposite}}{\text{adjacent}}</math>
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In our case, this means that
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<math>\tan 27^{\circ }=\frac{x}{13}</math>
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which gives
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<math>x=\text{13}\centerdot \text{tan 27}^{\circ }</math>.
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NOTE: Using a calculator, we can work out what
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<math>x\text{ }</math>
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should be:
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<math>x=\text{13}\centerdot \text{tan 27}^{\circ }\approx 6.62</math>

Version vom 10:27, 28. Sep. 2008


The definition of the tangent states that


\displaystyle \tan u=\frac{\text{opposite}}{\text{adjacent}}


In our case, this means that


\displaystyle \tan 27^{\circ }=\frac{x}{13}


which gives \displaystyle x=\text{13}\centerdot \text{tan 27}^{\circ }.

NOTE: Using a calculator, we can work out what \displaystyle x\text{ } should be:


\displaystyle x=\text{13}\centerdot \text{tan 27}^{\circ }\approx 6.62

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