Lösung 4.3:6b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 9: Zeile 9:
We can determine the triangle's remaining side by using the Pythagorean theorem,
We can determine the triangle's remaining side by using the Pythagorean theorem,
-
{{Displayed math||<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}
+
{{Abgesetzte Formel||<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}
which gives that
which gives that
-
{{Displayed math||<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}
This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have
This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have
-
{{Displayed math||<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
+
{{Abgesetzte Formel||<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
and thus
and thus
-
{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}

Version vom 08:56, 22. Okt. 2008

We draw an angle \displaystyle v in the unit circle, and the fact that \displaystyle \sin v = 3/10 means that its y-coordinate equals \displaystyle 3/10.

With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.

We can determine the triangle's remaining side by using the Pythagorean theorem,

\displaystyle a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2

which gives that

\displaystyle a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}

This means that the angle's x-coordinate is \displaystyle -a, i.e. we have

\displaystyle \cos v=-\frac{\sqrt{91}}{10}

and thus

\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:6b“