Solution 4.2:9
From Förberedande kurs i matematik 1
m (Lösning 4.2:9 moved to Solution 4.2:9: Robot: moved page) |
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- | + | If we introduce the dashed triangle below, the distance as the crow flies between | |
- | < | + | <math>\text{A}</math> |
- | + | and | |
- | { | + | <math>\text{B}</math> |
- | + | is equal to the triangle's hypotenuse, | |
- | + | <math>c</math>. | |
- | + | ||
- | < | + | |
- | + | ||
[[Image:4_2_9_1.gif|center]] | [[Image:4_2_9_1.gif|center]] | ||
+ | |||
+ | One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives | ||
+ | |||
+ | |||
+ | <math>c^{2}=a^{2}+b^{2}</math> | ||
+ | |||
+ | |||
+ | In turn, we can determine the opposite and adjacent by introducing another triangle | ||
+ | <math>\text{APR}</math>, where | ||
+ | <math>\text{R}</math> | ||
+ | is the point on the line | ||
+ | <math>\text{PQ}</math> | ||
+ | which the dashed triangle's side of length | ||
+ | <math>a</math> | ||
+ | cuts the line. | ||
+ | |||
[[Image:4_2_9_2.gif|center]] | [[Image:4_2_9_2.gif|center]] | ||
+ | |||
+ | Because we know that | ||
+ | <math>\text{AP}=\text{4}</math> | ||
+ | and the angle at P, simple trigonometry shows that | ||
+ | <math>x</math> | ||
+ | and | ||
+ | <math>y</math> | ||
+ | are given by | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\ | ||
+ | & y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | |||
+ | We can now start to look for the solution. Since | ||
+ | <math>x</math> | ||
+ | and | ||
+ | <math>y</math> | ||
+ | have been calculated, we can determine | ||
+ | <math>a</math> | ||
+ | and b by considering the horizontal and vertical distances in the figure. | ||
+ | |||
+ | |||
[[Image:4_2_9_3.gif|center]] | [[Image:4_2_9_3.gif|center]] | ||
+ | |||
+ | <math>a=x+5=2+5=7</math> | ||
+ | |||
+ | <math>b=12-y=12-2\sqrt{3}</math> | ||
+ | |||
+ | |||
+ | With a and | ||
+ | <math>b</math> | ||
+ | given, Pythagoras' theorem leads to | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+\left( 12-2\sqrt{3} \right)^{2}} \\ | ||
+ | & =\sqrt{49+\left( 12^{2}-2\centerdot 12\centerdot 2\sqrt{3}+\left( 2\sqrt{3} \right)^{2} \right)} \\ | ||
+ | & =\sqrt{205-38\sqrt{3}}\quad \approx \quad 11.0\quad \text{km}\text{.} \\ | ||
+ | \end{align}</math> |
Revision as of 09:58, 29 September 2008
If we introduce the dashed triangle below, the distance as the crow flies between
One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives
In turn, we can determine the opposite and adjacent by introducing another triangle
Because we know that
=4
21=2
y=4cos30
=4
2
3=2
3
We can now start to look for the solution. Since
3
With a and
a2+b2=
72+
12−2
3
2=
49+
122−2
12
2
3+
2
3
2
=
205−38
3
11
0km.