Solution 4.2:1e
From Förberedande kurs i matematik 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. | ||
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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> | + | |
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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 | + |
Current revision
In the triangle, we seek the hypotenuse x, knowing the angle 35° 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 ({} \approx 19\textrm{.}2)\,\textrm{.} |