Solution 4.2:2b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-[[Bild: +[[Image:))
Current revision (14:29, 8 October 2008) (edit) (undo)
m
 
(2 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
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'',
-
<center> [[Image:4_2_2b.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{| width="100%"
-
[[Image:4_2_2_b.gif|center]]
+
|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{.} Image:4_2_2_b.gif

The right-hand side in this equation can be simplified, so that we get

\displaystyle \sin v = \frac{7}{11}\,\textrm{.}