Lösung 4.2:9 - Versionsgeschichte

Ombtut4 um 12:43, 7. Sep. 2009

Ombtut4 — Mon, 07 Sep 2009 12:43:27 GMT

← Nächstältere Version		Version vom 12:43, 7. Sep. 2009
Zeile 32:		Zeile 32:
	&= \sqrt{7^2+(12-2\sqrt{3})^2}\\[5pt]		&= \sqrt{7^2+(12-2\sqrt{3})^2}\\[5pt]
	&= \sqrt{49+(12^2-2\cdot 12\cdot 2\sqrt{3}+(2\sqrt{3})^2)}\\[5pt]		&= \sqrt{49+(12^2-2\cdot 12\cdot 2\sqrt{3}+(2\sqrt{3})^2)}\\[5pt]
-	&= \sqrt{205-38\sqrt{3}}\\[5pt]	+	&= \sqrt{205-48\sqrt{3}}\\[5pt]
	&\approx 11\textrm{.}0\ \text{km}\textrm{.}		&\approx 11\textrm{.}0\ \text{km}\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek: Replaced figures with metapost figures

Tek — Mon, 24 Aug 2009 14:56:31 GMT

Replaced figures with metapost figures

← Nächstältere Version		Version vom 14:56, 24. Aug. 2009
Zeile 1:		Zeile 1:
	Wir betrachten das Dreieck im unteren Bild. Der gesuchte Abstand ist die Hypotenuse ''c''.		Wir betrachten das Dreieck im unteren Bild. Der gesuchte Abstand ist die Hypotenuse ''c''.

-	~~[[Image~~:~~4_2_9_1~~.~~gif\|~~center]]	+	<center>{{:4.2.9 - Solution - An auxiliary triangle with vertices A to B}}</center>

	Die Hypotenuse können wir durch den Satz des Pythagoras bestimmen		Die Hypotenuse können wir durch den Satz des Pythagoras bestimmen
Zeile 9:		Zeile 9:
	Wir können die Katheten bestimmen, indem wir das Dreieck APR im nächsten Bild betrachten.		Wir können die Katheten bestimmen, indem wir das Dreieck APR im nächsten Bild betrachten.

-	~~[[Image~~:~~4_2_9_2~~.~~gif\|~~center]]	+	<center>{{:4.2.9 - Solution - An auxiliary triangle with vertices A to P}}</center>

	Nachdem <math>\text{AP}=4</math>, erhalten wir einfach ''x'' und ''y'':		Nachdem <math>\text{AP}=4</math>, erhalten wir einfach ''x'' und ''y'':
Zeile 21:		Zeile 21:

	{\| align="center"		{\| align="center"
-	\| align="center" \|~~[[Image:4_2_9_3~~-1.~~gif]]~~	+	\| align="center" valign="center"\|<math>\begin{align}a &= x+5 = 2+5 = 7,\\ b &= 12-y = 12-2\sqrt{3}.\end{align}</math>
	\| width="20px"\|		\| width="20px"\|
-	\| align="center" ~~\|[[Image:4_2_9_3-2.gif]]~~	+	\| align="center" valign="center"\|{{:4.2.9 - Solution - A figure with horizontal distances a, x and 5, and vertical distances 12, b and y}}
-	\|-	+
-	~~\| align~~="center~~" valign="top~~"\|~~<math>\begin~~{~~align}~~a &= x+5~~\\ &= 2+5 = 7\end{align}</math>~~	+
-	\|\|	+
-	~~\| align="center" valign="top"\|<math>\begin{align}~~b ~~&= 12-~~y~~\\ &= 12-2\sqrt{3~~}~~\end{align~~}~~</math>~~	+
	\|}		\|}
-		+
	Mit ''a'' und ''b'' erhalten wir c durch den Satz des Pythagoras		Mit ''a'' und ''b'' erhalten wir c durch den Satz des Pythagoras

Markus.bez: Sprache und Formulierung

Markus.bez — Thu, 18 Jun 2009 13:34:02 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 13:34, 18. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir betrachten das Dreieck im unteren Bild. Der ~~ersuchter~~ Abstand ist die Hypotenuse ''c''.	+	Wir betrachten das Dreieck im unteren Bild. Der gesuchte Abstand ist die Hypotenuse ''c''.

	[[Image:4_2_9_1.gif\|center]]		[[Image:4_2_9_1.gif\|center]]

-	Die Hypotenuse können wir durch ~~das Gesetz~~ des Pythagoras bestimmen	+	Die Hypotenuse können wir durch den Satz des Pythagoras bestimmen

	{{Abgesetzte Formel\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}
Zeile 11:		Zeile 11:
	[[Image:4_2_9_2.gif\|center]]		[[Image:4_2_9_2.gif\|center]]

-	Nachdem <math>\text{AP}=4</math> erhalten wir einfach ''x'' und ''y'',	+	Nachdem <math>\text{AP}=4</math>, erhalten wir einfach ''x'' und ''y'':

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	Mit ''x'' und ''y'' erhalten wir die Katheten ''a'' und ''b'' indem wir die horizontalen und vertikalen Abstände in der Figur betrachten.	+	Mit ''x'' und ''y'' erhalten wir die Katheten ''a'' und ''b'', indem wir die horizontalen und vertikalen Abstände in der Figur betrachten.

	{\| align="center"		{\| align="center"
Zeile 30:		Zeile 30:
	\|}		\|}

-	Mit ''a'' und ''b'' erhalten wir c durch ~~das Gesetz~~ des Pythagoras	+	Mit ''a'' und ''b'' erhalten wir c durch den Satz des Pythagoras

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Martin um 16:55, 4. Apr. 2009

Martin — Sat, 04 Apr 2009 16:55:37 GMT

← Nächstältere Version		Version vom 16:55, 4. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If we introduce the dashed triangle below, the distance as the crow flies between A and B is equal to the triangle's hypotenuse,~~ ''c''.	+	Wir betrachten das Dreieck im unteren Bild. Der ersuchter Abstand ist die Hypotenuse ''c''.

	[[Image:4_2_9_1.gif\|center]]		[[Image:4_2_9_1.gif\|center]]

-	~~One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since the Pythagorean theorem then gives~~	+	Die Hypotenuse können wir durch das Gesetz des Pythagoras bestimmen

	{{Abgesetzte Formel\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}

-	~~In turn~~, ~~we can determine the opposite and adjacent by introducing another triangle~~ APR~~, where R is the point on the line PQ which the dashed triangle's side of length ''a'' cuts the line~~.	+	Wir können die Katheten bestimmen, indem wir das Dreieck APR im nächsten Bild betrachten.

	[[Image:4_2_9_2.gif\|center]]		[[Image:4_2_9_2.gif\|center]]

-	~~Because we know that~~ <math>\text{AP}=4</math> ~~and the angle at P, simple trigonometry shows that~~ ''x'' ~~and~~ ''y'' ~~are given by~~	+	Nachdem <math>\text{AP}=4</math> erhalten wir einfach ''x'' und ''y'',

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	~~We can now start to look for the solution. Since~~ ''x'' ~~and~~ ''y'' ~~have been calculated, we can determine~~ ''a'' ~~and~~ ''b'' ~~by considering the horizontal and vertical distances~~ in ~~the figure~~.	+	Mit ''x'' und ''y'' erhalten wir die Katheten ''a'' und ''b'' indem wir die horizontalen und vertikalen Abstände in der Figur betrachten.

	{\| align="center"		{\| align="center"
Zeile 30:		Zeile 30:
	\|}		\|}

-	~~With~~ ''a'' ~~and~~ ''b'' ~~given, the Pythagorean theorem leads to~~	+	Mit ''a'' und ''b'' erhalten wir c durch das Gesetz des Pythagoras

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: hat „Solution 4.2:9“ nach „Lösung 4.2:9“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:00:24 GMT

hat „Solution 4.2:9“ nach „Lösung 4.2:9“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:00, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:53:58 GMT

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

← Nächstältere Version		Version vom 08:53, 22. Okt. 2008
Zeile 5:		Zeile 5:
	One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since the Pythagorean theorem then gives		One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since the Pythagorean theorem then gives

-	{{~~Displayed math~~\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}

	In turn, we can determine the opposite and adjacent by introducing another triangle APR, where R is the point on the line PQ which the dashed triangle's side of length ''a'' cuts the line.		In turn, we can determine the opposite and adjacent by introducing another triangle APR, where R is the point on the line PQ which the dashed triangle's side of length ''a'' cuts the line.
Zeile 13:		Zeile 13:
	Because we know that <math>\text{AP}=4</math> and the angle at P, simple trigonometry shows that ''x'' and ''y'' are given by		Because we know that <math>\text{AP}=4</math> and the angle at P, simple trigonometry shows that ''x'' and ''y'' are given by

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x &= 4\sin 30^{\circ } = 4\cdot \frac{1}{2} = 2,\\[5pt]		x &= 4\sin 30^{\circ } = 4\cdot \frac{1}{2} = 2,\\[5pt]
	y &= 4\cos 30^{\circ } = 4\cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\,\textrm{.}		y &= 4\cos 30^{\circ } = 4\cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\,\textrm{.}
Zeile 32:		Zeile 32:
	With ''a'' and ''b'' given, the Pythagorean theorem leads to		With ''a'' and ''b'' given, the Pythagorean theorem leads to

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	c &= \sqrt{a^2+b^2}\\[5pt]		c &= \sqrt{a^2+b^2}\\[5pt]
	&= \sqrt{7^2+(12-2\sqrt{3})^2}\\[5pt]		&= \sqrt{7^2+(12-2\sqrt{3})^2}\\[5pt]

Tek um 12:46, 9. Okt. 2008

Tek — Thu, 09 Oct 2008 12:46:00 GMT

← Nächstältere Version		Version vom 12:46, 9. Okt. 2008
Zeile 1:		Zeile 1:
-	If we introduce the dashed triangle below, the distance as the crow flies between	+	If we introduce the dashed triangle below, the distance as the crow flies between A and B is equal to the triangle's hypotenuse, ''c''.
-	~~<math>\text{~~A~~}</math>~~	+
-	and	+
-	~~<math>\text{~~B~~}</math>~~	+
-	is equal to the triangle's hypotenuse,	+
-	~~<math>~~c~~</math>~~.	+
-		+

	[[Image:4_2_9_1.gif\|center]]		[[Image:4_2_9_1.gif\|center]]

-	One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since ~~Pythagoras'~~ theorem then gives	+	One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since the Pythagorean theorem then gives

		+	{{Displayed math\|\|<math>c^2 = a^2 + b^2\,\textrm{.}</math>}}

-	~~<math>c^{2}=a^{2}+b^{2}</math>~~	+	In turn, we can determine the opposite and adjacent by introducing another triangle APR, where R is the point on the line PQ which the dashed triangle's side of length ''a'' cuts the line.
-		+
-		+
-	In turn, we can determine the opposite and adjacent by introducing another triangle	+
-	~~<math>\text{~~APR~~}</math>~~, where	+
-	~~<math>\text{~~R~~}</math>~~	+
-	is the point on the line	+
-	~~<math>\text{~~PQ~~}</math>~~	+
-	which the dashed triangle's side of length	+
-	~~<math>~~a~~</math>~~	+
-	cuts the line.	+

	[[Image:4_2_9_2.gif\|center]]		[[Image:4_2_9_2.gif\|center]]

-	Because we know that	+	Because we know that <math>\text{AP}=4</math> and the angle at P, simple trigonometry shows that ''x'' and ''y'' are given by
-	<math>\text{AP}=~~\text{~~4}</math>	+
-	and the angle at P, simple trigonometry shows that	+
-	~~<math>~~x~~</math>~~	+
-	and	+
-	~~<math>~~y~~</math>~~	+
-	are given by	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\~~	+
-	~~& y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~We can now start to look for the solution. Since~~	+
-	~~<math>x</math>~~	+
-	~~and~~	+
-	~~<math>y</math>~~	+
-	~~have been calculated, we can determine~~	+
-	~~<math>a</math>~~	+
-	~~and b by considering the horizontal and vertical distances in the figure.~~	+
-		+
-		+
-	~~[[Image:4_2_9_3.gif\|center]]~~	+

-	<math>a=x+5=2+5=~~7</math>~~	+	{{Displayed math\|\|<math>\begin{align}
-		+	x &= 4\sin 30^{\circ } = 4\cdot \frac{1}{2} = 2,\\[5pt]
-	~~<math>b=12-~~y=~~12-~~2\sqrt{3}</math>	+	y &= 4\cos 30^{\circ } = 4\cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

		+	We can now start to look for the solution. Since ''x'' and ''y'' have been calculated, we can determine ''a'' and ''b'' by considering the horizontal and vertical distances in the figure.

-	~~With~~ a ~~and~~	+	{\| align="center"
-	<math>b</math>	+	\| align="center" \|[[Image:4_2_9_3-1.gif]]
-	~~given, Pythagoras' theorem leads to~~	+	\| width="20px"\|
		+	\| align="center" \|[[Image:4_2_9_3-2.gif]]
		+	\|-
		+	\| align="center" valign="top"\|<math>\begin{align}a &= x+5\\ &= 2+5 = 7\end{align}</math>
		+	\|\|
		+	\| align="center" valign="top"\|<math>\begin{align}b &= 12-y\\ &= 12-2\sqrt{3}\end{align}</math>
		+	\|}

		+	With ''a'' and ''b'' given, the Pythagorean theorem leads to

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+~~\left~~( 12-2\sqrt{3} ~~\right~~)^{2}} \\	+	c &= \sqrt{a^2+b^2}\\[5pt]
-	& =\sqrt{49+~~\left~~( 12^{2}-2\~~centerdot~~ 12\~~centerdot~~ 2\sqrt{3}+~~\left~~( 2\sqrt{3} ~~\right~~)^{2~~} \right~~)} \\	+	&= \sqrt{7^2+(12-2\sqrt{3})^2}\\[5pt]
-	& =\sqrt{205-38\sqrt{3}}\~~quad~~ \~~approx~~ \~~quad~~ 11.0\~~quad~~ \text{km}\~~text~~{.} \\	+	&= \sqrt{49+(12^2-2\cdot 12\cdot 2\sqrt{3}+(2\sqrt{3})^2)}\\[5pt]
-	\end{align}</math>	+	&= \sqrt{205-38\sqrt{3}}\\[5pt]
		+	&\approx 11\textrm{.}0\ \text{km}\textrm{.}
		+	\end{align}</math>}}

Ian um 09:58, 29. Sep. 2008

Ian — Mon, 29 Sep 2008 09:58:23 GMT

← Nächstältere Version		Version vom 09:58, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	If we introduce the dashed triangle below, the distance as the crow flies between
-	<~~center~~> ~~[[Image:4_2_9-1(3).gif]]~~ </~~center~~>	+	<math>\text{A}</math>
-	~~{{NAVCONTENT_STOP}}~~	+	and
-	{~~{NAVCONTENT_START~~}}	+	<math>\text{B}</math>
-	~~<center> [[Image:4_2_9-2(3).gif]]~~ </~~center~~>	+	is equal to the triangle's hypotenuse,
-	~~{{NAVCONTENT_STOP}}~~	+	<math>c</math>.
-	~~{{NAVCONTENT_START}}~~	+
-	<~~center~~> ~~[[Image:4_2_9-3(3).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+


	[[Image:4_2_9_1.gif\|center]]		[[Image:4_2_9_1.gif\|center]]
		+
		+	One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives
		+
		+
		+	<math>c^{2}=a^{2}+b^{2}</math>
		+
		+
		+	In turn, we can determine the opposite and adjacent by introducing another triangle
		+	<math>\text{APR}</math>, where
		+	<math>\text{R}</math>
		+	is the point on the line
		+	<math>\text{PQ}</math>
		+	which the dashed triangle's side of length
		+	<math>a</math>
		+	cuts the line.
		+
	[[Image:4_2_9_2.gif\|center]]		[[Image:4_2_9_2.gif\|center]]
		+
		+	Because we know that
		+	<math>\text{AP}=\text{4}</math>
		+	and the angle at P, simple trigonometry shows that
		+	<math>x</math>
		+	and
		+	<math>y</math>
		+	are given by
		+
		+
		+	<math>\begin{align}
		+	& x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\
		+	& y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\
		+	\end{align}</math>
		+
		+
		+	We can now start to look for the solution. Since
		+	<math>x</math>
		+	and
		+	<math>y</math>
		+	have been calculated, we can determine
		+	<math>a</math>
		+	and b by considering the horizontal and vertical distances in the figure.
		+
		+
	[[Image:4_2_9_3.gif\|center]]		[[Image:4_2_9_3.gif\|center]]
		+
		+	<math>a=x+5=2+5=7</math>
		+
		+	<math>b=12-y=12-2\sqrt{3}</math>
		+
		+
		+	With a and
		+	<math>b</math>
		+	given, Pythagoras' theorem leads to
		+
		+
		+	<math>\begin{align}
		+	& c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+\left( 12-2\sqrt{3} \right)^{2}} \\
		+	& =\sqrt{49+\left( 12^{2}-2\centerdot 12\centerdot 2\sqrt{3}+\left( 2\sqrt{3} \right)^{2} \right)} \\
		+	& =\sqrt{205-38\sqrt{3}}\quad \approx \quad 11.0\quad \text{km}\text{.} \\
		+	\end{align}</math>

Tekbot: Lösning 4.2:9 moved to Solution 4.2:9: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:50:03 GMT

Lösning 4.2:9 moved to Solution 4.2:9: Robot: moved page

← Nächstältere Version

Version vom 09:50, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:12:09 GMT

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

← Nächstältere Version		Version vom 07:12, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_9-1(3).gif]] </center>	+	<center> [[Image:4_2_9-1(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_9-2(3).gif]] </center>	+	<center> [[Image:4_2_9-2(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_9-3(3).gif]] </center>	+	<center> [[Image:4_2_9-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}


-	[[~~Bild~~:4_2_9_1.gif\|center]]	+	[[Image:4_2_9_1.gif\|center]]
-	[[~~Bild~~:4_2_9_2.gif\|center]]	+	[[Image:4_2_9_2.gif\|center]]
-	[[~~Bild~~:4_2_9_3.gif\|center]]	+	[[Image:4_2_9_3.gif\|center]]