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.
This exercise is somewhat of a trick question, because we don't need any trigonometry to solve it.
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Two angles are given in the triangle (the
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Two angles are given in the triangle (the 60° angle and the right-angle) and thus we can use the fact that the sum of all the angles in a triangle is 180°,
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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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{{Displayed math||<math>v + 60^{\circ} + 90^{\circ} = 180^{\circ}\,,</math>}}
which gives
which gives
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{{Displayed math||<math>v = 180^{\circ} - 60^{\circ} - 90^{\circ} = 30^{\circ}\,\textrm{.}</math>}}
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<math>v=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }</math>
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Version vom 07:34, 9. Okt. 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 60° angle and the right-angle) and thus we can use the fact that the sum of all the angles in a triangle is 180°,

Vorlage:Displayed math

which gives

Vorlage:Displayed math

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