Solution 4.2:2b
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
|||
(3 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{.} |