Solution 4.2:2a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
(Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_2a.gif </center> {{NAVCONTENT_STOP}})
Current revision (12:16, 10 October 2008) (edit) (undo)
m
 
(4 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The opposite and adjacent are given in the right-angled triangle and this means that the value of the tangent for the angle can be determined as the quotient between the opposite and the adjacent:
-
<center> [[Bild:4_2_2a.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{| width="100%"
 +
| width="50%" align="center"|<math>\tan v = 2/5</math>
 +
| width="50%" align="left"|[[Image:4_2_2_a.gif]]
 +
|}
 +
 
 +
At the same time, this is a trigonometric equation for the angle ''v''.
 +
 
 +
 
 +
Note: In the chapter on "Trigonometric equations", we will investigate more closely how to solve equations of this type.

Current revision

The opposite and adjacent are given in the right-angled triangle and this means that the value of the tangent for the angle can be determined as the quotient between the opposite and the adjacent:

\displaystyle \tan v = 2/5 Image:4_2_2_a.gif

At the same time, this is a trigonometric equation for the angle v.


Note: In the chapter on "Trigonometric equations", we will investigate more closely how to solve equations of this type.