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Solution 4.2:7

From Förberedande kurs i matematik 1

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Current revision (11:54, 9 October 2008) (edit) (undo)
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If extend the line
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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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<math>\text{AB}</math>
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to a point
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<math>\text{D}</math>
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opposite
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<math>\text{C}</math>, we will get the right-angled triangle shown below, where the distance
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<math>x</math>
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between
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<math>\text{C}</math>
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and
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<math>\text{D}</math>
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is the desired distance.
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[[Image:4_2_7_1.gif|center]]
[[Image:4_2_7_1.gif|center]]
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The information in the exercise can be summarized by considering the two triangles
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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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<math>\text{ACD}</math>
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and
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<math>\text{BCD}</math>, and setting up relations for the tangents that the angles
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<math>\text{3}0^{\circ }</math>
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and
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<math>\text{45}^{\circ }</math>
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gives rise to,
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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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||
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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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[[Image:4_2_7_2.gif|center]]
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where ''y'' is the distance between B and D.
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<math>x=\left( 100+y \right)\tan 30^{\circ }=\left( 100+y \right)\frac{1}{\sqrt{3}}</math> <math>x=y\centerdot \tan 45^{\circ }=y\centerdot 1</math>
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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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where
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{{Displayed math||<math>\sqrt{3}x=10+x</math>}}
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<math>y</math>
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is the distance between B and D.
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The second relation above gives that
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moving all the ''x''-terms to the left-hand side gives
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<math>y=x</math>
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and substituting this into the first relation gives
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<math>x=\left( 100+x \right)\frac{1}{\sqrt{3}}</math>
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Multiplying both sides by
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<math>\sqrt{3}</math>
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gives
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<math>\sqrt{3}x=100+x</math>
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moving all the x-terms to the left-hand side gives
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<math>\left( \sqrt{3}-1 \right)x=100</math>
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{{Displayed math||<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}
The answer is
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>}}
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<math>x=\frac{100}{\sqrt{3}-1}\ \text{m}\quad \approx \quad \text{136}\text{.6}\ \text{m}</math>
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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
x=(10+y)tan30=(10+y)13 x=ytan45=y1

where y is the distance between B and D.

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

x=(10+x)13.

Multiplying both sides by 3  gives

3x=10+x 

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

(3−1)x=10. 

The answer is

x=103−1 m13.6 m.
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