Solution 4.2:2b
From Förberedande kurs i matematik 1
(Difference between revisions)
m (Robot: Automated text replacement (-[[Bild: +[[Image:)) |
m |
||
| (2 intermediate revisions not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{ | + | 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]] | ||
| + | |} | ||
| + | |||
| + | The right-hand side in this equation can be simplified, so that we get | ||
| + | |||
| + | {{Displayed math||<math>\sin v = \frac{7}{11}\,\textrm{.}</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{.} |

