Lösung 4.2:1e

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In the triangle, we seek the hypotenuse ''x'', knowing the angle 35° and that the adjacent has length 11.
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[[Image:4_2_1_e.gif|center]]
[[Image:4_2_1_e.gif|center]]
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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
The definition of sine gives
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{{Displayed math||<math>\sin 35^{\circ} = \frac{11}{x}</math>}}
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<math>\sin 35^{\circ }=\frac{11}{x}</math>
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and thus
and thus
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{{Displayed math||<math>x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}</math>}}
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<math>x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)</math>
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Version vom 14:17, 8. Okt. 2008

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

The definition of sine gives

Vorlage:Displayed math

and thus

Vorlage:Displayed math

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