Lösung 4.2:6
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
			  			                                                      
		          
			| K  (Lösning 4.2:6 moved to Solution 4.2:6: Robot: moved page) | |||
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| - | { | + | We can work out the length we are looking for by taking the difference  | 
| - | < | + | <math>a-b\text{ }</math> | 
| - | + | of the sides  | |
| + | <math>a</math> | ||
| + | and  | ||
| + | <math>b</math> | ||
| + | in the triangles below: | ||
| + | |||
| [[Image:4_2_6_13.gif|center]] | [[Image:4_2_6_13.gif|center]] | ||
| [[Image:4_2_6_2.gif|center]] | [[Image:4_2_6_2.gif|center]] | ||
| + | |||
| + | If we take the tangent of the given angle in each triangle, we easily obtain  | ||
| + | <math>a</math> | ||
| + | and | ||
| + | <math>b</math>: | ||
| + | |||
| + | |||
| [[Image:4_2_6_13.gif|center]] | [[Image:4_2_6_13.gif|center]] | ||
| [[Image:4_2_6_4.gif|center]] | [[Image:4_2_6_4.gif|center]] | ||
| + | |||
| + | <math>a=1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}</math> | ||
Version vom 08:29, 29. Sep. 2008
We can work out the length we are looking for by taking the difference \displaystyle a-b\text{ } of the sides \displaystyle a and \displaystyle b in the triangles below:
If we take the tangent of the given angle in each triangle, we easily obtain \displaystyle a and \displaystyle b:
\displaystyle a=1\centerdot \tan 60^{\circ }=\frac{\sin 60^{\circ }}{\cos 60^{\circ }}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}
 
		  


