Lösung 4.2:1d

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[[Image:4_2_1_d.gif|center]]
[[Image:4_2_1_d.gif|center]]
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The side marked
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<math>x</math>
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is the hypotenuse in the right-angled triangle and the side of length
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<math>\text{16}</math>
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is the adjacent to the angle of
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<math>\text{2}0^{\circ }</math>.
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By writing the quotient for
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<math>\text{cos20}^{\circ }</math>, we obtain the relation
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<math>\text{cos20}^{\circ }=\frac{16}{x}</math>
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and this gives
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<math>x=\frac{16}{\text{cos20}^{\circ }}\quad \left( \approx 17.0 \right).</math>

Version vom 11:08, 28. Sep. 2008

The side marked \displaystyle x is the hypotenuse in the right-angled triangle and the side of length \displaystyle \text{16} is the adjacent to the angle of \displaystyle \text{2}0^{\circ }.


By writing the quotient for \displaystyle \text{cos20}^{\circ }, we obtain the relation


\displaystyle \text{cos20}^{\circ }=\frac{16}{x}


and this gives


\displaystyle x=\frac{16}{\text{cos20}^{\circ }}\quad \left( \approx 17.0 \right).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:1d“