Lösung 4.2:2a

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The opposite and adjacent are given in the right-angled triangle and this means that the value of the tangent for the angle can be determined as the quotient between the opposite and the adjacent:
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<center> [[Image:4_2_2a.gif]] </center>
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<math>\text{tan }v\text{ }={2}/{5}\;~</math>
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[[Image:4_2_2_a.gif|center]]
[[Image:4_2_2_a.gif|center]]
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At the same time, this is a trigonometric equation for the angle
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<math>v</math>.
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NOTE: In the chapter on "Trigonometric equations", we will investigate more closely how to solve equations of this type.

Version vom 09:09, 10. Okt. 2008

The opposite and adjacent are given in the right-angled triangle and this means that the value of the tangent for the angle can be determined as the quotient between the opposite and the adjacent:


\displaystyle \text{tan }v\text{ }={2}/{5}\;~


At the same time, this is a trigonometric equation for the angle \displaystyle v.

NOTE: In the chapter on "Trigonometric equations", we will investigate more closely how to solve equations of this type.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:2a“