Aus Online Mathematik Brückenkurs 1

If we introduce the dashed triangle below, the distance as the crow flies between \displaystyle \text{A} and \displaystyle \text{B} is equal to the triangle's hypotenuse, \displaystyle c.

One way to determine the hypotenuse is to know the triangle's opposite and adjacent sides, since Pythagoras' theorem then gives

\displaystyle c^{2}=a^{2}+b^{2}

In turn, we can determine the opposite and adjacent by introducing another triangle \displaystyle \text{APR}, where \displaystyle \text{R} is the point on the line \displaystyle \text{PQ} which the dashed triangle's side of length \displaystyle a cuts the line.

Because we know that \displaystyle \text{AP}=\text{4} and the angle at P, simple trigonometry shows that \displaystyle x and \displaystyle y are given by

\displaystyle \begin{align} & x=4\sin 30^{\circ }=4\centerdot \frac{1}{2}=2, \\ & y=4\cos 30^{\circ }=4\centerdot \frac{\sqrt{3}}{2}=2\sqrt{3} \\ \end{align}

We can now start to look for the solution. Since \displaystyle x and \displaystyle y have been calculated, we can determine \displaystyle a and b by considering the horizontal and vertical distances in the figure.

\displaystyle a=x+5=2+5=7

\displaystyle b=12-y=12-2\sqrt{3}

With a and \displaystyle b given, Pythagoras' theorem leads to

\displaystyle \begin{align} & c=\sqrt{a^{2}+b^{2}}=\sqrt{7^{2}+\left( 12-2\sqrt{3} \right)^{2}} \\ & =\sqrt{49+\left( 12^{2}-2\centerdot 12\centerdot 2\sqrt{3}+\left( 2\sqrt{3} \right)^{2} \right)} \\ & =\sqrt{205-38\sqrt{3}}\quad \approx \quad 11.0\quad \text{km}\text{.} \\ \end{align}