Aus Online Mathematik Brückenkurs 1

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{.}