Solution 4.2:7

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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.
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<center> [[Bild:4_2_7-1(2).gif]] </center>
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<center> [[Bild:4_2_7-2(2).gif]] </center>
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[[Image:4_2_7_1.gif|center]]
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[[Bild:4_2_7_1.gif|center]]
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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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[[Bild:4_2_7_2.gif|center]]
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{| align="center"
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|align="center"|[[Image:4_2_7_2-1.gif]]
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|width="20px"|&nbsp;
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|align="center"|[[Image:4_2_7_2-2.gif]]
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|-
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|align="center" valign="top"|<math>\begin{align} x &= (10+y)\tan 30^{\circ}\\[5pt] &= (10+y)\frac{1}{\sqrt{3}}\end{align}</math>
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|align="center" valign="top"|<math>\begin{align} x &= y\cdot\tan 45^{\circ}\\[5pt] &= y\cdot 1\end{align}</math>
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|}
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where ''y'' is the distance between B and D.
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The second relation above gives that <math>y=x</math> and substituting this into the first relation gives
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{{Displayed math||<math>x = (10+x)\frac{1}{\sqrt{3}}\,\textrm{.}</math>}}
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Multiplying both sides by <math>\sqrt{3}</math> gives
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{{Displayed math||<math>\sqrt{3}x=10+x</math>}}
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moving all the ''x''-terms to the left-hand side gives
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{{Displayed math||<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}
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The answer is
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{{Displayed math||<math>x = \frac{10}{\sqrt{3}-1}\ \text{m}\approx 13\textrm{.}6\ \text{m.}</math>}}

Current revision

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,

Image:4_2_7_2-1.gif   Image:4_2_7_2-2.gif
\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.}
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