Lösung 4.2:1e

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In the triangle, we seek the hypotenuse
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<math>x</math>, knowing the angle 35o and that the adjacent has length 11.
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The definition of sine gives
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<math>\sin 35^{\circ }=\frac{11}{x}</math>
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and thus
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<math>x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)</math>

Version vom 11:12, 28. Sep. 2008

In the triangle, we seek the hypotenuse \displaystyle x, knowing the angle 35o and that the adjacent has length 11.


The definition of sine gives


\displaystyle \sin 35^{\circ }=\frac{11}{x}


and thus


\displaystyle x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)

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