Lösung 4.3:5

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An often-used technique to calculate
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<center> [[Image:4_3_5-1(2).gif]] </center>
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<math>\text{cos }v</math>
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and
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<math>\text{tan }v</math>, given the sine value of an acute angle, is to draw the angle
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<center> [[Image:4_3_5-2(2).gif]] </center>
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<math>v</math>
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in a right-angled triangle which has two sides arranged so that
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<math>\text{sin }v={5}/{7}\;</math>.
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[[Image:4_3_5_1.gif|center]]
[[Image:4_3_5_1.gif|center]]
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Using Pythagoras' theorem, we can determine the length of the third side in the triangle.
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[[Image:4_3_5_2.gif|center]]
[[Image:4_3_5_2.gif|center]]
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<math>x^{2}+5^{2}=7^{2}</math>
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which gives that
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<math>x=\sqrt{7^{2}-5^{2}}=\sqrt{24}=2\sqrt{6}</math>
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Then, using the definition of cosine and tangent,
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<math>\begin{align}
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& \cos v=\frac{x}{7}=\frac{2\sqrt{6}}{7}, \\
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& \tan v=\frac{5}{x}=\frac{5}{2\sqrt{6}} \\
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\end{align}</math>
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NOTE: Note that the right-angled triangle that we use is just a tool and has nothing to do with the triangle that is referred to in the question.

Version vom 12:11, 29. Sep. 2008

An often-used technique to calculate \displaystyle \text{cos }v and \displaystyle \text{tan }v, given the sine value of an acute angle, is to draw the angle \displaystyle v in a right-angled triangle which has two sides arranged so that \displaystyle \text{sin }v={5}/{7}\;.


Using Pythagoras' theorem, we can determine the length of the third side in the triangle.



\displaystyle x^{2}+5^{2}=7^{2} which gives that \displaystyle x=\sqrt{7^{2}-5^{2}}=\sqrt{24}=2\sqrt{6}

Then, using the definition of cosine and tangent,


\displaystyle \begin{align} & \cos v=\frac{x}{7}=\frac{2\sqrt{6}}{7}, \\ & \tan v=\frac{5}{x}=\frac{5}{2\sqrt{6}} \\ \end{align}


NOTE: Note that the right-angled triangle that we use is just a tool and has nothing to do with the triangle that is referred to in the question.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:5“