Lösung 4.3:6a

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{{NAVCONTENT_START}}
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If we think of the angle v as an angle in the unit circle, then
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<center> [[Image:4_3_6a-1(2).gif]] </center>
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<math>v</math>
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lies in the fourth quadrant and has
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<math>x</math>
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<center> [[Image:4_3_6a-2(2).gif]] </center>
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-coordinate
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<math>\frac{3}{4}</math>.
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[[Image:4_3_6_a1.gif|center]]
[[Image:4_3_6_a1.gif|center]]
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If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to
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<math>\text{1}</math>
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and an opposite side equal to
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<math>\frac{3}{4}</math>.
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[[Image:4_3_6_a2.gif|center]]
[[Image:4_3_6_a2.gif|center]]
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Using Pythagoras' theorem, it is possible to determine the remaining side from
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<math>b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}</math>
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which gives that
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<math>\begin{align}
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& b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\
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& b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\
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\end{align}</math>
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Because the angle
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<math>v</math>
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belongs to the fourth quadrant, its
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<math>y</math>
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-coordinate is negative and is therefore equal to
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<math>-b</math>, i.e.
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<math>\sin v=-\frac{\sqrt{7}}{4}</math>
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Thus, we have directly that
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<math>\tan v=\frac{\sin v}{\cos v}=\frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}}=-\frac{\sqrt{7}}{3}</math>

Version vom 09:12, 30. Sep. 2008

If we think of the angle v as an angle in the unit circle, then \displaystyle v lies in the fourth quadrant and has \displaystyle x -coordinate \displaystyle \frac{3}{4}.


If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to \displaystyle \text{1} and an opposite side equal to \displaystyle \frac{3}{4}.

Using Pythagoras' theorem, it is possible to determine the remaining side from


\displaystyle b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}


which gives that


\displaystyle \begin{align} & b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\ & b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\ \end{align}


Because the angle \displaystyle v belongs to the fourth quadrant, its \displaystyle y -coordinate is negative and is therefore equal to \displaystyle -b, i.e.


\displaystyle \sin v=-\frac{\sqrt{7}}{4}


Thus, we have directly that


\displaystyle \tan v=\frac{\sin v}{\cos v}=\frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}}=-\frac{\sqrt{7}}{3}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:6a“