Lösung 4.2:6
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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- | We can work out the length we are looking for by taking the difference | + | We can work out the length we are looking for by taking the difference <math>a-b</math> of the sides <math>a</math> and <math>b</math> in the triangles below. |
- | <math>a-b | + | |
- | 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 | + | If we take the tangent of the given angle in each triangle, we easily obtain <math>a</math> and <math>b</math>. |
- | <math>a</math> | + | |
- | and | + | |
- | <math>b</math> | + | |
+ | {| width="100%" | ||
+ | ||[[Image:4_2_6_13.gif]] | ||
+ | ||<math>a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}</math> | ||
+ | |- | ||
+ | ||[[Image:4_2_6_4.gif]] | ||
+ | ||<math>b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1</math> | ||
+ | |} | ||
- | + | Hence, | |
- | + | ||
- | <math>a= | + | {{Displayed math||<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}} |
Version vom 11:28, 9. Okt. 2008
We can work out the length we are looking for by taking the difference \displaystyle a-b 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.
Hence,