Solution 4.2:7

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.2:7 moved to Solution 4.2:7: Robot: moved page)
Current revision (11:54, 9 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
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.
-
<center> [[Image:4_2_7-1(2).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
{{NAVCONTENT_START}}
+
-
<center> [[Image:4_2_7-2(2).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
-
 
+
[[Image:4_2_7_1.gif|center]]
[[Image:4_2_7_1.gif|center]]
-
[[Image:4_2_7_2.gif|center]]
+
 
 +
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,
 +
 
 +
{| align="center"
 +
|align="center"|[[Image:4_2_7_2-1.gif]]
 +
|width="20px"|&nbsp;
 +
|align="center"|[[Image:4_2_7_2-2.gif]]
 +
|-
 +
|align="center" valign="top"|<math>\begin{align} x &= (10+y)\tan 30^{\circ}\\[5pt] &= (10+y)\frac{1}{\sqrt{3}}\end{align}</math>
 +
||
 +
|align="center" valign="top"|<math>\begin{align} x &= y\cdot\tan 45^{\circ}\\[5pt] &= y\cdot 1\end{align}</math>
 +
|}
 +
 
 +
where ''y'' is the distance between B and D.
 +
 
 +
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>}}
 +
 
 +
Multiplying both sides by <math>\sqrt{3}</math> gives
 +
 
 +
{{Displayed math||<math>\sqrt{3}x=10+x</math>}}
 +
 
 +
moving all the ''x''-terms to the left-hand side gives
 +
 
 +
{{Displayed math||<math>(\sqrt{3}-1)x = 10\,\textrm{.}</math>}}
 +
 
 +
The answer is
 +
 
 +
{{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.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.2:7"