Solution 4.2:9 - Revision history

Tek at 12:46, 9 October 2008

Tek — Thu, 09 Oct 2008 12:46:00 GMT

←Older revision		Revision as of 12:46, 9 October 2008
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-	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 at 09:58, 29 September 2008

Ian — Mon, 29 Sep 2008 09:58:23 GMT

←Older revision		Revision as of 09:58, 29 September 2008
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-	~~{{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

←Older revision

Revision as of 09:50, 9 September 2008

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

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

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

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-	<center> [[~~Bild~~:4_2_9-1(3).gif]] </center>	+	<center> [[Image:4_2_9-1(3).gif]] </center>
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-	[[~~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]]

Madde at 14:06, 18 August 2008

Madde — Mon, 18 Aug 2008 14:06:14 GMT

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Oskar — Thu, 03 Apr 2008 08:11:50 GMT

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