Solution 4.2:1e - Revision history

Tek at 14:17, 8 October 2008

Tek — Wed, 08 Oct 2008 14:17:12 GMT

←Older revision		Revision as of 14:17, 8 October 2008
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		+	In the triangle, we seek the hypotenuse ''x'', knowing the angle 35° and that the adjacent has length 11.
		+
	[[Image:4_2_1_e.gif\|center]]		[[Image:4_2_1_e.gif\|center]]
-
-	In the triangle, we seek the hypotenuse
-	<math>x</math>, knowing the angle 35o and that the adjacent has length 11.
-

	The definition of sine gives		The definition of sine gives

-		+	{{Displayed math\|\|<math>\sin 35^{\circ} = \frac{11}{x}</math>}}
-	<math>\sin 35^{\circ }=\frac{11}{x}</math>	+
-		+

	and thus		and thus

-		+	{{Displayed math\|\|<math>x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}</math>}}
-	<math>x=\frac{11}{\sin 35^{\circ }}\quad ~~\left~~( \approx 19.2 ~~\right~~)</math>	+

Ian — Sun, 28 Sep 2008 11:12:47 GMT

←Older revision		Revision as of 11:12, 28 September 2008
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-	{{NAVCONTENT_START}}
-	<center> [[Image:4_2_1e.gif]] </center>
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	[[Image:4_2_1_e.gif\|center]]		[[Image:4_2_1_e.gif\|center]]
		+
		+	In the triangle, we seek the hypotenuse
		+	<math>x</math>, knowing the angle 35o and that the adjacent has length 11.
		+
		+
		+	The definition of sine gives
		+
		+
		+	<math>\sin 35^{\circ }=\frac{11}{x}</math>
		+
		+
		+	and thus
		+
		+
		+	<math>x=\frac{11}{\sin 35^{\circ }}\quad \left( \approx 19.2 \right)</math>

Tekbot — Tue, 09 Sep 2008 09:41:03 GMT

←Older revision

Revision as of 09:41, 9 September 2008

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

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

Madde — Fri, 15 Aug 2008 12:51:04 GMT

Oskar — Wed, 02 Apr 2008 15:43:57 GMT

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