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Solution 4.2:7

From Förberedande kurs i matematik 1

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m (Lösning 4.2:7 moved to Solution 4.2:7: Robot: moved page)
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{{NAVCONTENT_START}}
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If extend the line
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<center> [[Image:4_2_7-1(2).gif]] </center>
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<math>\text{AB}</math>
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{{NAVCONTENT_STOP}}
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to a point
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{{NAVCONTENT_START}}
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<math>\text{D}</math>
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<center> [[Image:4_2_7-2(2).gif]] </center>
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opposite
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{{NAVCONTENT_STOP}}
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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.
[[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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<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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[[Image:4_2_7_2.gif|center]]
[[Image:4_2_7_2.gif|center]]
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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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 +
 +
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where
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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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<math>y=x</math>
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and substituting this into the first relation gives
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 +
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<math>x=\left( 100+x \right)\frac{1}{\sqrt{3}}</math>
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 +
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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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 +
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moving all the x-terms to the left-hand side gives
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 +
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<math>\left( \sqrt{3}-1 \right)x=100</math>
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 +
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The answer is
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 +
 +
<math>x=\frac{100}{\sqrt{3}-1}\ \text{m}\quad \approx \quad \text{136}\text{.6}\ \text{m}</math>

Revision as of 08:55, 29 September 2008

If 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,


x=100+ytan30=100+y13  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=100+x13 


Multiplying both sides by 3  gives


3x=100+x 


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


3−1x=100 


The answer is


x=1003−1 m136.6 m

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