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		Revision as of 11:35, 28 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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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
		+	<math>\text{6}0^{\circ }</math>
		+	angle and the right-angle) and thus we can use the fact that the sum of all the angles in a triangle is
		+	<math>\text{18}0^{\circ }</math>,
		+
		+
		+	<math>v+60^{\circ }+90^{\circ }=180^{\circ }</math>
		+
		+
		+	which gives
		+
		+
		+	<math>v=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }</math>

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Revision as of 11:35, 28 September 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 }