Solution 4.2:2b
From Förberedande kurs i matematik 1
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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 ''v'', | |
- | + | {| width="100%" | |
- | + | |width="50%" align="center"|<math>\sin v = \frac{70}{110}\,\textrm{.}</math> | |
- | + | |width="50%" align="center"|[[Image:4_2_2_b.gif]] | |
- | + | |} | |
- | <math>\ | + | |
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The right-hand side in this equation can be simplified, so that we get | The right-hand side in this equation can be simplified, so that we get | ||
- | + | {{Displayed math||<math>\sin v = \frac{7}{11}\,\textrm{.}</math>}} | |
- | <math>\ | + |
Current revision
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 v,
\displaystyle \sin v = \frac{70}{110}\,\textrm{.} | ![]() |
The right-hand side in this equation can be simplified, so that we get
\displaystyle \sin v = \frac{7}{11}\,\textrm{.} |