Solution 4.2:7 - Revision history

Tek at 11:54, 9 October 2008

Tek — Thu, 09 Oct 2008 11:54:47 GMT

←Older revision		Revision as of 11:54, 9 October 2008
Line 1:		Line 1:
-	If extend the line	+	If we extend the line AB to a point D opposite C, we will get the right-angled triangle shown below, where the distance ''x'' between C and D is the desired distance.
-	~~<math>\text{~~AB~~}</math>~~	+
-	to a point	+
-	~~<math>\text{~~D~~}</math>~~	+
-	opposite	+
-	~~<math>\text{~~C~~}</math>~~, we will get the right-angled triangle shown below, where the distance	+
-	~~<math>~~x~~</math>~~	+
-	between	+
-	~~<math>\text{~~C~~}</math>~~	+
-	and	+
-	~~<math>\text{~~D~~}</math>~~	+
-	is the desired distance.	+
-		+

	[[Image:4_2_7_1.gif\|center]]		[[Image:4_2_7_1.gif\|center]]

-	The information in the exercise can be summarized by considering the two triangles	+	The information in the exercise can be summarized by considering the two triangles ACD and BCD, and setting up relations for the tangents that the angles 30° and 45° gives rise to,
-	~~<math>\text{~~ACD~~}</math>~~	+
-	and	+
-	~~<math>\text{~~BCD~~}</math>~~, and setting up relations for the tangents that the angles	+
-	~~<math>\text{3}0^{\circ }</math>~~	+
-	and	+
-	~~<math>\text{45}^{\circ }</math>~~	+
-	gives rise to,	+

		+	{\| align="center"
		+	\|align="center"\|[[Image:4_2_7_2-1.gif]]
		+	\|width="20px"\|
		+	\|align="center"\|[[Image:4_2_7_2-2.gif]]
		+	\|-
		+	\|align="center" valign="top"\|<math>\begin{align} x &= (10+y)\tan 30^{\circ}\\[5pt] &= (10+y)\frac{1}{\sqrt{3}}\end{align}</math>
		+	\|\|
		+	\|align="center" valign="top"\|<math>\begin{align} x &= y\cdot\tan 45^{\circ}\\[5pt] &= y\cdot 1\end{align}</math>
		+	\|}

-	~~[[Image:4_2_7_2~~.~~gif\|center]]~~	+	where ''y'' is the distance between B and D.

-	<math>~~x=\left( 100+~~y ~~\right)\tan 30^{\circ }~~=~~\left( 100+y \right)\frac{1}{\sqrt{3}}</math> <math>~~x~~=y\centerdot \tan 45^{\circ }=y\centerdot 1~~</math>	+	The second relation above gives that <math>y=x</math> and substituting this into the first relation gives

		+	{{Displayed math\|\|<math>x = (10+x)\frac{1}{\sqrt{3}}\,\textrm{.}</math>}}

		+	Multiplying both sides by <math>\sqrt{3}</math> gives

-	~~where~~	+	{{Displayed math\|\|<math>\sqrt{3}x=10+x</math>}}
-	<math>y</math>	+
-	~~is the distance between B and D.~~	+

-	~~The second relation above gives that~~	+	moving all the ''x''-terms to the left-hand side gives
-	~~<math>y=x</math>~~	+
-	~~and substituting this into the first relation gives~~	+
-		+
-		+
-	~~<math>x=\left( 100+x \right)\frac{1}{\sqrt{3}}</math>~~	+
-		+
-		+
-	~~Multiplying both sides by~~	+
-	~~<math>\sqrt{3}</math>~~	+
-	~~gives~~	+
-		+
-		+
-	~~<math>\sqrt{3}x=100+x</math>~~	+
-		+
-		+
-	moving all the x-terms to the left-hand side gives	+
-		+
-		+
-	~~<math>\left( \sqrt{3}-1 \right)x=100</math>~~	+

		+	{{Displayed math\|\|<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}

	The answer is		The answer is

-		+	{{Displayed math\|\|<math>x = \frac{10}{\sqrt{3}-1}\ \text{m}\approx 13\textrm{.}6\ \text{m.}</math>}}
-	<math>x=\frac{~~100~~}{\sqrt{3}-1}\ \text{m}~~\quad~~ \approx \~~quad \text~~{~~136~~}~~\text{.~~6}\ \text{m}</math>	+

Ian at 08:55, 29 September 2008

Ian — Mon, 29 Sep 2008 08:55:59 GMT

←Older revision		Revision as of 08:55, 29 September 2008
Line 1:		Line 1:
-	~~{{NAVCONTENT_START}}~~	+	If extend the line
-	<~~center~~> ~~[[Image:4_2_7-1(2).gif]]~~ </~~center~~>	+	<math>\text{AB}</math>
-	{~~{NAVCONTENT_STOP}~~}	+	to a point
-	{~~{NAVCONTENT_START}~~}	+	<math>\text{D}</math>
-	<~~center~~> ~~[[Image:4_2_7~~-~~2(2).gif]]~~ </~~center~~>	+	opposite
-	{{~~NAVCONTENT_STOP}~~}	+	<math>\text{C}</math>, we will get the right-angled triangle shown below, where the distance
		+	<math>x</math>
		+	between
		+	<math>\text{C}</math>
		+	and
		+	<math>\text{D}</math>
		+	is the desired distance.


	[[Image:4_2_7_1.gif\|center]]		[[Image:4_2_7_1.gif\|center]]
		+
		+	The information in the exercise can be summarized by considering the two triangles
		+	<math>\text{ACD}</math>
		+	and
		+	<math>\text{BCD}</math>, and setting up relations for the tangents that the angles
		+	<math>\text{3}0^{\circ }</math>
		+	and
		+	<math>\text{45}^{\circ }</math>
		+	gives rise to,
		+
		+
	[[Image:4_2_7_2.gif\|center]]		[[Image:4_2_7_2.gif\|center]]
		+
		+	<math>x=\left( 100+y \right)\tan 30^{\circ }=\left( 100+y \right)\frac{1}{\sqrt{3}}</math> <math>x=y\centerdot \tan 45^{\circ }=y\centerdot 1</math>
		+
		+
		+
		+	where
		+	<math>y</math>
		+	is the distance between B and D.
		+
		+	The second relation above gives that
		+	<math>y=x</math>
		+	and substituting this into the first relation gives
		+
		+
		+	<math>x=\left( 100+x \right)\frac{1}{\sqrt{3}}</math>
		+
		+
		+	Multiplying both sides by
		+	<math>\sqrt{3}</math>
		+	gives
		+
		+
		+	<math>\sqrt{3}x=100+x</math>
		+
		+
		+	moving all the x-terms to the left-hand side gives
		+
		+
		+	<math>\left( \sqrt{3}-1 \right)x=100</math>
		+
		+
		+	The answer is
		+
		+
		+	<math>x=\frac{100}{\sqrt{3}-1}\ \text{m}\quad \approx \quad \text{136}\text{.6}\ \text{m}</math>

Tekbot: Lösning 4.2:7 moved to Solution 4.2:7: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:49:23 GMT

Lösning 4.2:7 moved to Solution 4.2:7: Robot: moved page

←Older revision

Revision as of 09:49, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:11:49 GMT

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

←Older revision		Revision as of 07:11, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_7-1(2).gif]] </center>	+	<center> [[Image:4_2_7-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_7-2(2).gif]] </center>	+	<center> [[Image:4_2_7-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}


-	[[~~Bild~~:4_2_7_1.gif\|center]]	+	[[Image:4_2_7_1.gif\|center]]
-	[[~~Bild~~:4_2_7_2.gif\|center]]	+	[[Image:4_2_7_2.gif\|center]]

Madde at 14:02, 18 August 2008

Madde — Mon, 18 Aug 2008 14:02:28 GMT

←Older revision		Revision as of 14:02, 18 August 2008
Line 5:		Line 5:
	<center> [[Bild:4_2_7-2(2).gif]] </center>		<center> [[Bild:4_2_7-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+
		+	[[Bild:4_2_7_1.gif\|center]]
		+	[[Bild:4_2_7_2.gif\|center]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_2_7-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_2_7-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 08:07:59 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_7-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_2_7-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_2_7-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:4_2_7-2(2).gif]] </center>
{{NAVCONTENT_STOP}}