From Förberedande kurs i matematik 1

Revision as of 09:43, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.2:2e moved to Solution 4.2:2e: Robot: moved page) ← Previous diff		Current revision (07:34, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	This exercise is somewhat of a trick question, because we don't need any trigonometry to solve it.
-	<center> [[Image:4_2_2e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	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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		+	which gives
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		+	{{Displayed math||<math>v = 180^{\circ} - 60^{\circ} - 90^{\circ} = 30^{\circ}\,\textrm{.}</math>}}

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Current revision

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{.}