Solution 4.2:2e
From Förberedande kurs i matematik 1
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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. | ||
- | 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°, |
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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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+ | {{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{.} |