Lösung 4.2:6 - Versionsgeschichte

Tek: Replaced figures with metapost figures

2009-08-24T14:14:42Z

Replaced figures with metapost figures

← Nächstältere Version		Version vom 14:14, 24. Aug. 2009
Zeile 1:		Zeile 1:
	Wir können die Länge ''x'' berechnen, indem wir die Differenz <math>a-b</math> der Seiten <math>a</math> und <math>b</math> berechnen.		Wir können die Länge ''x'' berechnen, indem wir die Differenz <math>a-b</math> der Seiten <math>a</math> und <math>b</math> berechnen.

-	~~[[Image~~:~~4_2_6_13~~.~~gif\|center]][[Image:4_2_6_2~~.~~gif\|~~center]]	+	<center>{{:4.2.6 - Solution - Two triangles with angles 60° and 45°, respectively}}</center>

	Berechnen wir den Tangens der beiden Winkel, erhalten wir <math>a</math> und <math>b</math>.		Berechnen wir den Tangens der beiden Winkel, erhalten wir <math>a</math> und <math>b</math>.

	{\| width="100%"		{\| width="100%"
-	\|\|~~[[Image~~:~~4_2_6_13~~.~~gif]]~~	+	\|\|{{:4.2.6 - Solution - A triangle with angle 60°}}
	\|\|<math>a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}</math>		\|\|<math>a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}</math>
	\|-		\|-
-	\|\|~~[[Image~~:~~4_2_6_4~~.~~gif]]~~	+	\|\|{{:4.2.6 - Solution - A triangle with angle 45°}}
	\|\|<math>b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1</math>		\|\|<math>b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1</math>
	\|}		\|}

2009-07-23T17:24:24Z

← Nächstältere Version		Version vom 17:24, 23. Jul. 2009
Zeile 3:		Zeile 3:
	[[Image:4_2_6_13.gif\|center]][[Image:4_2_6_2.gif\|center]]		[[Image:4_2_6_13.gif\|center]][[Image:4_2_6_2.gif\|center]]

-	Berechnen wir den Tangens der beiden ~~Winkeln~~, erhalten wir <math>a</math> und <math>b</math>.	+	Berechnen wir den Tangens der beiden Winkel, erhalten wir <math>a</math> und <math>b</math>.

	{\| width="100%"		{\| width="100%"

2009-06-18T13:27:41Z

← Nächstältere Version		Version vom 13:27, 18. Jun. 2009
Zeile 13:		Zeile 13:
	\|}		\|}

-	Also ist,	+	Also ist

	{{Abgesetzte Formel\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}

2009-04-04T16:08:08Z

← Nächstältere Version		Version vom 16:08, 4. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We can work out the length we are looking for by taking the difference~~ <math>a-b</math> ~~of the sides~~ <math>a</math> ~~and~~ <math>b</math> ~~in the triangles below~~.	+	Wir können die Länge ''x'' berechnen, indem wir die Differenz <math>a-b</math> der Seiten <math>a</math> und <math>b</math> berechnen.

	[[Image:4_2_6_13.gif\|center]][[Image:4_2_6_2.gif\|center]]		[[Image:4_2_6_13.gif\|center]][[Image:4_2_6_2.gif\|center]]

-	~~If we take the tangent of the given angle in each triangle~~, ~~we easily obtain~~ <math>a</math> ~~and~~ <math>b</math>.	+	Berechnen wir den Tangens der beiden Winkeln, erhalten wir <math>a</math> und <math>b</math>.

	{\| width="100%"		{\| width="100%"
Zeile 13:		Zeile 13:
	\|}		\|}

-	~~Hence~~,	+	Also ist,

	{{Abgesetzte Formel\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}

2008-10-22T14:59:24Z

hat „Solution 4.2:6“ nach „Lösung 4.2:6“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:59, 22. Okt. 2008

2008-10-22T08:53:28Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:53, 22. Okt. 2008
Zeile 15:		Zeile 15:
	Hence,		Hence,

-	{{~~Displayed math~~\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}

2008-10-09T11:28:11Z

← Nächstältere Version		Version vom 11:28, 9. Okt. 2008
Zeile 1:		Zeile 1:
-	We can work out the length we are looking for by taking the difference	+	We can work out the length we are looking for by taking the difference <math>a-b</math> of the sides <math>a</math> and <math>b</math> in the triangles below.
-	<math>a-b~~\text{ }~~</math>	+
-	of the sides	+
-	<math>a</math>	+
-	and	+
-	<math>b</math>	+
-	in the triangles below:	+

-	[[Image:4_2_6_13.gif\|center]]	+	[[Image:4_2_6_13.gif\|center]][[Image:4_2_6_2.gif\|center]]
-	[[Image:4_2_6_2.gif\|center]]	+

-	If we take the tangent of the given angle in each triangle, we easily obtain	+	If we take the tangent of the given angle in each triangle, we easily obtain <math>a</math> and <math>b</math>.
-	<math>a</math>	+
-	and	+
-	<math>b</math>:	+

		+	{\| width="100%"
		+	\|\|[[Image:4_2_6_13.gif]]
		+	\|\|<math>a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}</math>
		+	\|-
		+	\|\|[[Image:4_2_6_4.gif]]
		+	\|\|<math>b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1</math>
		+	\|}

-	~~[[Image:4_2_6_13.gif\|center]]~~	+	Hence,
-	~~[[Image:4_2_6_4.gif\|center]]~~	+

-	<math>a=~~1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{~~\sqrt{3}~~}{2}}{\frac{~~1~~}{2}}=~~\~~sqrt~~{3}</math>	+	{{Displayed math\|\|<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}

2008-09-29T08:29:20Z

← Nächstältere Version		Version vom 08:29, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	{~~{NAVCONTENT_START}~~}	+	We can work out the length we are looking for by taking the difference
-	<~~center~~> ~~[[Image:4_2_6.gif]]~~ </~~center~~>	+	<math>a-b\text{ }</math>
-	~~{{NAVCONTENT_STOP}}~~	+	of the sides
		+	<math>a</math>
		+	and
		+	<math>b</math>
		+	in the triangles below:
		+
	[[Image:4_2_6_13.gif\|center]]		[[Image:4_2_6_13.gif\|center]]
	[[Image:4_2_6_2.gif\|center]]		[[Image:4_2_6_2.gif\|center]]
		+
		+	If we take the tangent of the given angle in each triangle, we easily obtain
		+	<math>a</math>
		+	and
		+	<math>b</math>:
		+
		+
	[[Image:4_2_6_13.gif\|center]]		[[Image:4_2_6_13.gif\|center]]
	[[Image:4_2_6_4.gif\|center]]		[[Image:4_2_6_4.gif\|center]]
		+
		+	<math>a=1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}</math>

2008-09-09T09:49:03Z

Lösning 4.2:6 moved to Solution 4.2:6: Robot: moved page

← Nächstältere Version

Version vom 09:49, 9. Sep. 2008

2008-08-21T07:11:39Z

Robot: Automated text replacement (-[[Bild: +[[Image:)