Aus Online Mathematik Brückenkurs 1

Version vom 11:54, 9. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:53, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 19:		Zeile 19:
	The second relation above gives that <math>y=x</math> and substituting this into the first relation gives		The second relation above gives that <math>y=x</math> and substituting this into the first relation gives
-	{{Displayed math||<math>x = (10+x)\frac{1}{\sqrt{3}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x = (10+x)\frac{1}{\sqrt{3}}\,\textrm{.}</math>}}
	Multiplying both sides by <math>\sqrt{3}</math> gives		Multiplying both sides by <math>\sqrt{3}</math> gives
-	{{Displayed math||<math>\sqrt{3}x=10+x</math>}}	+	{{Abgesetzte Formel||<math>\sqrt{3}x=10+x</math>}}
	moving all the ''x''-terms to the left-hand side gives		moving all the ''x''-terms to the left-hand side gives
-	{{Displayed math||<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}
	The answer is		The answer is
-	{{Displayed math||<math>x = \frac{10}{\sqrt{3}-1}\ \text{m}\approx 13\textrm{.}6\ \text{m.}</math>}}	+	{{Abgesetzte Formel||<math>x = \frac{10}{\sqrt{3}-1}\ \text{m}\approx 13\textrm{.}6\ \text{m.}</math>}}

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Version vom 08:53, 22. Okt. 2008

If we extend the line AB to a point D opposite C, we will get the right-angled triangle shown below, where the distance x between C and D is the desired distance.

The information in the exercise can be summarized by considering the two triangles ACD and BCD, and setting up relations for the tangents that the angles 30° and 45° gives rise to,

\displaystyle \begin{align} x &= (10+y)\tan 30^{\circ}\\[5pt] &= (10+y)\frac{1}{\sqrt{3}}\end{align}		\displaystyle \begin{align} x &= y\cdot\tan 45^{\circ}\\[5pt] &= y\cdot 1\end{align}

where y is the distance between B and D.

The second relation above gives that \displaystyle y=x and substituting this into the first relation gives

\displaystyle x = (10+x)\frac{1}{\sqrt{3}}\,\textrm{.}

Multiplying both sides by \displaystyle \sqrt{3} gives

\displaystyle \sqrt{3}x=10+x

moving all the x-terms to the left-hand side gives

\displaystyle (\sqrt{3}-1)x = 10\,\textrm{.}

The answer is

\displaystyle x = \frac{10}{\sqrt{3}-1}\ \text{m}\approx 13\textrm{.}6\ \text{m.}