Three cities, A, B, and C, are located so that city A is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 miles from city B, how far is city A from city B ?
if i heard you right:
$$\small{\text{$
\begin{array}{rcl}
100^2 &=& 70^2+x^2-2\cdot 70 \cdot x \cdot \cos{(35\ensurement{^{\circ}} + 90\ensurement{^{\circ}} )} \qquad | \qquad\cos{(35\ensurement{^{\circ}} + 90\ensurement{^{\circ}} )} = -\sin{35\ensurement{^{\circ}}}\\
100^2 &=& 70^2+x^2 + 2\cdot 70 \cdot x \cdot \sin{(35\ensurement{^{\circ}} )} \\
100^2 &=& 70^2+x^2 + 140 \cdot x \cdot \sin{(35\ensurement{^{\circ}} )}\\
x^2 + 140 \cdot \sin{(35\ensurement{^{\circ}} )} \cdot x +70^2 - 100^2 &=& 0\\
x^2 + 80.3007010891 \cdot x -5100 &=& 0\\\
x_{1,2} &=& \dfrac{-80.3007010891 \pm \sqrt{ 80.3007010891^2-4(-5100)} } {2} \\\\
x_1 = \overline{AB}&=& 41.7767551583~\rm{miles} \qquad | \qquad x_2 < 0 \rm{~no~solution}
\end{array}
$}}$$
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