Lösung 4.2:1e
Aus Online Mathematik Brückenkurs 1
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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 | ||
| - | <math>x</math>, knowing the angle 35o and that the adjacent has length 11. | ||
| - | |||
The definition of sine gives | The definition of sine gives | ||
| - | + | {{Displayed math||<math>\sin 35^{\circ} = \frac{11}{x}</math>}} | |
| - | <math>\sin 35^{\circ }=\frac{11}{x}</math> | + | |
| - | + | ||
and thus | and thus | ||
| - | + | {{Displayed math||<math>x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}</math>}} | |
| - | <math>x=\frac{11}{\sin 35^{\circ }}\quad | + | |
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
and thus
