Solution 4.2:6
From Förberedande kurs i matematik 1
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- | + | 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. | |
- | < | + | |
- | + | [[Image:4_2_6_13.gif|center]][[Image:4_2_6_2.gif|center]] | |
- | [[Image:4_2_6_13.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 | + | |
- | [[Image:4_2_6_4.gif | + | {| 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, | ||
+ | |||
+ | {{Displayed math||<math>x = a-b = \sqrt{3}-1\,\textrm{.}</math>}} |
Current revision
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,
\displaystyle x = a-b = \sqrt{3}-1\,\textrm{.} |