Solution 4.2:2e

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (07:34, 9 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
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.
-
Two angles are given in the triangle (the
+
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°,
-
<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>
+
 +
{{Displayed math||<math>v + 60^{\circ} + 90^{\circ} = 180^{\circ}\,,</math>}}
which gives
which gives
-
 
+
{{Displayed math||<math>v = 180^{\circ} - 60^{\circ} - 90^{\circ} = 30^{\circ}\,\textrm{.}</math>}}
-
<math>v=180^{\circ }-60^{\circ }-90^{\circ }=30^{\circ }</math>
+

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{.}