Lösung 4.2:2e

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This exercise is somewhat of a trick question, because we don't need any trigonometry to solve it.
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<center> [[Image:4_2_2e.gif]] </center>
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Two angles are given in the triangle (the
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<math>\text{6}0^{\circ }</math>
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angle and the right-angle) and thus we can use the fact that the sum of all the angles in a triangle is
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<math>\text{18}0^{\circ }</math>,
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<math>v+60^{\circ }+90^{\circ }=180^{\circ }</math>
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which gives
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<math>v=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }</math>

Version vom 11:35, 28. Sep. 2008

This exercise is somewhat of a trick question, because we don't need any trigonometry to solve it.

Two angles are given in the triangle (the \displaystyle \text{6}0^{\circ } angle and the right-angle) and thus we can use the fact that the sum of all the angles in a triangle is \displaystyle \text{18}0^{\circ },


\displaystyle v+60^{\circ }+90^{\circ }=180^{\circ }


which gives


\displaystyle v=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }

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