Lösung 4.2:2b

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[[Image:4_2_2_b.gif|center]]
[[Image:4_2_2_b.gif|center]]
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In this right-angled triangle, the opposite and the hypotenuse are given. This means that we can directly set up a relation for the sine of the angle
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<math>v</math>,
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<math>\text{sin }v\text{ }=\text{ }{70}/{110}\;</math>.
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The right-hand side in this equation can be simplified, so that we get
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<math>\text{sin }v\text{ }=\text{ }{7}/{11}\;</math>.

Version vom 11:26, 28. Sep. 2008

In this right-angled triangle, the opposite and the hypotenuse are given. This means that we can directly set up a relation for the sine of the angle \displaystyle v,


\displaystyle \text{sin }v\text{ }=\text{ }{70}/{110}\;.


The right-hand side in this equation can be simplified, so that we get


\displaystyle \text{sin }v\text{ }=\text{ }{7}/{11}\;.

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