From Förberedande kurs i matematik 1

Revision as of 09:49, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.2:6 moved to Solution 4.2:6: Robot: moved page) ← Previous diff		Current revision (11:28, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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.
-	<center> [[Image:4_2_6.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	[[Image:4_2_6_13.gif|center]][[Image:4_2_6_2.gif|center]]
-	[[Image:4_2_6_13.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_4.gif|center]]	+	{| 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>
		+	|}
		+
		+	Hence,
		+
		+	{{Displayed math||<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}}

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Current revision

We can work out the length we are looking for by taking the difference \displaystyle a-b of the sides \displaystyle a and \displaystyle b in the triangles below.

If we take the tangent of the given angle in each triangle, we easily obtain \displaystyle a and \displaystyle b.

	\displaystyle a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}
	\displaystyle b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1

Hence,

\displaystyle x = a-b = \sqrt{3}-1\,\textrm{.}