Lösung 4.2:7
Aus Online Mathematik Brückenkurs 1
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Version vom 14:59, 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,
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| \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.} |
