Lösung 4.2:6
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
K (hat „Solution 4.2:6“ nach „Lösung 4.2:6“ verschoben: Robot: moved page) |
Version vom 14:59, 22. 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.
|
| \displaystyle a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3} |
| \displaystyle b = 1\cdot\tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1 |
Hence,
| \displaystyle x = a-b = \sqrt{3}-1\,\textrm{.} |
