From Förberedande kurs i matematik 1

Revision as of 09:41, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.2:1e moved to Solution 4.2:1e: Robot: moved page) ← Previous diff		Current revision (14:17, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	In the triangle, we seek the hypotenuse ''x'', knowing the angle 35° and that the adjacent has length 11.
-	<center> [[Image:4_2_1e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
	[[Image:4_2_1_e.gif|center]]		[[Image:4_2_1_e.gif|center]]
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		+	The definition of sine gives
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		+	{{Displayed math||<math>\sin 35^{\circ} = \frac{11}{x}</math>}}
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		+	and thus
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		+	{{Displayed math||<math>x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}</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{.}