Lösung 4.2:9

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K (Lösning 4.2:9 moved to Solution 4.2:9: Robot: moved page)
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If we introduce the dashed triangle below, the distance as the crow flies between
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<center> [[Image:4_2_9-1(3).gif]] </center>
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<math>\text{A}</math>
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and
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<math>\text{B}</math>
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<center> [[Image:4_2_9-2(3).gif]] </center>
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is equal to the triangle's hypotenuse,
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<math>c</math>.
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<center> [[Image:4_2_9-3(3).gif]] </center>
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[[Image:4_2_9_1.gif|center]]
[[Image:4_2_9_1.gif|center]]
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One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives
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<math>c^{2}=a^{2}+b^{2}</math>
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In turn, we can determine the opposite and adjacent by introducing another triangle
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<math>\text{APR}</math>, where
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<math>\text{R}</math>
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is the point on the line
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<math>\text{PQ}</math>
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which the dashed triangle's side of length
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<math>a</math>
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cuts the line.
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[[Image:4_2_9_2.gif|center]]
[[Image:4_2_9_2.gif|center]]
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Because we know that
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<math>\text{AP}=\text{4}</math>
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and the angle at P, simple trigonometry shows that
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<math>x</math>
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and
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<math>y</math>
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are given by
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<math>\begin{align}
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& x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\
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& y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\
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\end{align}</math>
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We can now start to look for the solution. Since
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<math>x</math>
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and
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<math>y</math>
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have been calculated, we can determine
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<math>a</math>
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and b by considering the horizontal and vertical distances in the figure.
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[[Image:4_2_9_3.gif|center]]
[[Image:4_2_9_3.gif|center]]
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<math>a=x+5=2+5=7</math>
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<math>b=12-y=12-2\sqrt{3}</math>
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With a and
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<math>b</math>
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given, Pythagoras' theorem leads to
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<math>\begin{align}
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& c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+\left( 12-2\sqrt{3} \right)^{2}} \\
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& =\sqrt{49+\left( 12^{2}-2\centerdot 12\centerdot 2\sqrt{3}+\left( 2\sqrt{3} \right)^{2} \right)} \\
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& =\sqrt{205-38\sqrt{3}}\quad \approx \quad 11.0\quad \text{km}\text{.} \\
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\end{align}</math>

Version vom 09:58, 29. Sep. 2008

If we introduce the dashed triangle below, the distance as the crow flies between \displaystyle \text{A} and \displaystyle \text{B} is equal to the triangle's hypotenuse, \displaystyle c.


One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives


\displaystyle c^{2}=a^{2}+b^{2}


In turn, we can determine the opposite and adjacent by introducing another triangle \displaystyle \text{APR}, where \displaystyle \text{R} is the point on the line \displaystyle \text{PQ} which the dashed triangle's side of length \displaystyle a cuts the line.

Because we know that \displaystyle \text{AP}=\text{4} and the angle at P, simple trigonometry shows that \displaystyle x and \displaystyle y are given by


\displaystyle \begin{align} & x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\ & y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\ \end{align}


We can now start to look for the solution. Since \displaystyle x and \displaystyle y have been calculated, we can determine \displaystyle a and b by considering the horizontal and vertical distances in the figure.


\displaystyle a=x+5=2+5=7

\displaystyle b=12-y=12-2\sqrt{3}


With a and \displaystyle b given, Pythagoras' theorem leads to


\displaystyle \begin{align} & c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+\left( 12-2\sqrt{3} \right)^{2}} \\ & =\sqrt{49+\left( 12^{2}-2\centerdot 12\centerdot 2\sqrt{3}+\left( 2\sqrt{3} \right)^{2} \right)} \\ & =\sqrt{205-38\sqrt{3}}\quad \approx \quad 11.0\quad \text{km}\text{.} \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:9“