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{.} | 

