From Förberedande kurs i matematik 1

Revision as of 11:12, 28 September 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (14:17, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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		+	In the triangle, we seek the hypotenuse ''x'', knowing the angle 35° and that the adjacent has length 11.
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	[[Image:4_2_1_e.gif|center]]		[[Image:4_2_1_e.gif|center]]
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-	In the triangle, we seek the hypotenuse
-	<math>x</math>, knowing the angle 35o and that the adjacent has length 11.
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	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>	+

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Current revision

In the triangle, we seek the hypotenuse x, knowing the angle 35° and that the adjacent has length 11.

The definition of sine gives

\displaystyle \sin 35^{\circ} = \frac{11}{x}

and thus

\displaystyle x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}