Lösung 4.3:6c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 4.3:6c moved to Solution 4.3:6c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
Because the angle
-
<center> [[Image:4_3_6c-1(2).gif]] </center>
+
<math>v</math>
-
{{NAVCONTENT_STOP}}
+
satisfies
-
{{NAVCONTENT_START}}
+
<math>\pi \le v\le \frac{3\pi }{2}</math>,
-
<center> [[Image:4_3_6c-2(2).gif]] </center>
+
<math>v</math>
-
{{NAVCONTENT_STOP}}
+
belongs to the third quadrant in the unit circle. Furthermore,
 +
<math>\text{tan }v=\text{3 }</math>
 +
gives that the line which corresponds to the angle
 +
<math>v</math>
 +
 
 +
<math>v</math>
 +
has a gradient of
 +
<math>\text{3}</math>.
 +
 
[[Image:4_3_6_c1.gif|center]]
[[Image:4_3_6_c1.gif|center]]
 +
 +
slope 3
 +
 +
 +
In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is
 +
<math>\text{1}</math>
 +
and the sides have a
 +
<math>\text{3}:\text{1 }</math>
 +
ratio.
 +
[[Image:4_3_6_c2.gif|center]]
[[Image:4_3_6_c2.gif|center]]
 +
 +
If we now use Pythagoras' theorem on the triangle, we see that the horizontal side
 +
<math>\text{a}</math>
 +
satisfies
 +
 +
 +
<math>a^{2}+\left( 3a \right)^{2}=1^{2}</math>
 +
 +
 +
which gives us that
 +
 +
 +
<math>10a^{2}=1</math>
 +
i.e.
 +
<math>a=\frac{1}{\sqrt{10}}</math>
 +
 +
 +
Thus, the angle
 +
<math>v</math>'s
 +
<math>x</math>
 +
-coordinate is
 +
<math>-\frac{1}{\sqrt{10}}</math>
 +
and
 +
<math>y</math>
 +
-coordinate is
 +
<math>-\frac{3}{\sqrt{10}}</math>, i.e.
 +
 +
<math>\cos v=--\frac{1}{\sqrt{10}}</math>
 +
 +
 +
<math>\sin v=-\frac{3}{\sqrt{10}}</math>

Version vom 09:38, 30. Sep. 2008

Because the angle \displaystyle v satisfies \displaystyle \pi \le v\le \frac{3\pi }{2}, \displaystyle v belongs to the third quadrant in the unit circle. Furthermore, \displaystyle \text{tan }v=\text{3 } gives that the line which corresponds to the angle \displaystyle v

\displaystyle v has a gradient of \displaystyle \text{3}.


slope 3


In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is \displaystyle \text{1} and the sides have a \displaystyle \text{3}:\text{1 } ratio.

If we now use Pythagoras' theorem on the triangle, we see that the horizontal side \displaystyle \text{a} satisfies


\displaystyle a^{2}+\left( 3a \right)^{2}=1^{2}


which gives us that


\displaystyle 10a^{2}=1 i.e. \displaystyle a=\frac{1}{\sqrt{10}}


Thus, the angle \displaystyle v's \displaystyle x -coordinate is \displaystyle -\frac{1}{\sqrt{10}} and \displaystyle y -coordinate is \displaystyle -\frac{3}{\sqrt{10}}, i.e.

\displaystyle \cos v=--\frac{1}{\sqrt{10}}


\displaystyle \sin v=-\frac{3}{\sqrt{10}}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:6c“