Solution 4.2:6

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Current revision (11:28, 9 October 2008) (edit) (undo)
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We can work out the length we are looking for by taking the difference
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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.
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<math>a-b\text{ }</math>
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of the sides
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<math>a</math>
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and
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<math>b</math>
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in the triangles below:
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[[Image:4_2_6_13.gif|center]]
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[[Image:4_2_6_13.gif|center]][[Image:4_2_6_2.gif|center]]
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[[Image:4_2_6_2.gif|center]]
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If we take the tangent of the given angle in each triangle, we easily obtain
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If we take the tangent of the given angle in each triangle, we easily obtain <math>a</math> and <math>b</math>.
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<math>a</math>
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and
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<math>b</math>:
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{| width="100%"
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||[[Image:4_2_6_13.gif]]
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||<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>
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|-
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||[[Image:4_2_6_4.gif]]
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||<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>
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|}
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[[Image:4_2_6_13.gif|center]]
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Hence,
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[[Image:4_2_6_4.gif|center]]
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<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>
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{{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.

Image:4_2_6_13.gif \displaystyle a = 1\cdot\tan 60^{\circ} = \frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}
Image:4_2_6_4.gif \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{.}
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