Solution 4.2:6 - Revision history

Tek at 11:28, 9 October 2008

Tek — Thu, 09 Oct 2008 11:28:11 GMT

←Older revision		Revision as of 11:28, 9 October 2008
Line 1:		Line 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>}}

Ian — Mon, 29 Sep 2008 08:29:20 GMT

←Older revision		Revision as of 08:29, 29 September 2008
Line 1:		Line 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>

Tekbot — Tue, 09 Sep 2008 09:49:03 GMT

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

←Older revision

Revision as of 09:49, 9 September 2008

Tekbot — Thu, 21 Aug 2008 07:11:39 GMT

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

Madde — Mon, 18 Aug 2008 13:57:34 GMT

Oskar — Thu, 03 Apr 2008 08:06:59 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_6.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_2_6.gif]] </center>
{{NAVCONTENT_STOP}}