Lösung 4.2:2e

Aus Online Mathematik Brückenkurs 1

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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°,
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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{{Displayed math||<math>v + 60^{\circ} + 90^{\circ} = 180^{\circ}\,,</math>}}
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{{Abgesetzte Formel||<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>}}
+
{{Abgesetzte Formel||<math>v = 180^{\circ} - 60^{\circ} - 90^{\circ} = 30^{\circ}\,\textrm{.}</math>}}

Version vom 08:51, 22. 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°,

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

which gives

\displaystyle v = 180^{\circ} - 60^{\circ} - 90^{\circ} = 30^{\circ}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:2e“