A man travels:

$12$ kilometres northeast

then:

$20$ kilometres $30 \degrees$ west of north

then:

$18$ kilometres $60 \degrees$ south of west.

Assuming the curvature of the Earth to be negligible at this scale, at the end of this travel, he is $14.7$ kilometres in a direction $45 \degrees 49'$ west of north from his starting point.

## Proof 1

Let the route of the traveller be embedded in the complex plane.

Let $P$ be the final location of the traveller.

The given directions can be translated into absolute arguments thus:

Northeast is $45 \degrees$ north of east, and therefore an argument of $45 \degrees$
$30 \degrees$ west of north is $30 + 90 \degrees$ north of east, and therefore an argument of $120 \degrees$
$60 \degrees$ south of west is $60 + 180 \degrees$ north of east, and therefore an argument of $240 \degrees$.

Thus the problem reduces to finding the sum:

 $\ds P$ $=$ $\ds 12 \cis 45 \degrees + 20 \cis 120 \degrees + 18 \cis 240 \degrees$ $\ds$ $=$ $\ds \paren {12 \cos 45 \degrees + 20 \cos 120 \degrees + 18 \cos 240 \degrees} + i \paren {12 \sin 45 \degrees + 20 \sin 120 \degrees + 18 \sin 240 \degrees}$ $\ds$ $=$ $\ds \paren {12 \frac {\sqrt 2} 2 + 20 \paren {-\dfrac 1 2} + 18 \paren {-\dfrac 1 2} } + i \paren {12 \sin 45 \degrees + 20 \sin 120 \degrees + 18 \sin 240 \degrees}$ Cosine of $45 \degrees$, Cosine of $120 \degrees$, Cosine of $240 \degrees$ $\ds$ $=$ $\ds \paren {12 \frac {\sqrt 2} 2 + 20 \paren {-\dfrac 1 2} + 18 \paren {-\dfrac 1 2} } + i \paren {12 \frac {\sqrt 2} 2 + 20 \frac {\sqrt 3} 2 + 18 \paren {-\frac {\sqrt 3} 2} }$ Sine of $45 \degrees$, Sine of $120 \degrees$, Sine of $240 \degrees$ $\ds$ $=$ $\ds \paren {6 \sqrt 2 - 19} + i \paren {6 \sqrt 2 + \sqrt 3}$ after algebra

Then:

 $\ds R \cis \theta$ $=$ $\ds \paren {6 \sqrt 2 - 19} + i \paren {6 \sqrt 2 + \sqrt 3}$ $\ds \leadsto \ \$ $\ds R$ $=$ $\ds \sqrt {\paren {6 \sqrt 2 - 19}^2 + \paren {6 \sqrt 2 + \sqrt 3}^2}$ $\ds$ $\approx$ $\ds 14.7$ $\ds \leadsto \ \$ $\ds \theta$ $=$ $\ds \cos^{-1} \dfrac {6 \sqrt 2 - 19} {14.7}$ $\ds$ $\approx$ $\ds \map {\cos^{-1} } {- 0 \cdotp 717}$ $\ds$ $\approx$ $\ds 135 \degrees 49'$ $\ds$ $\approx$ $\ds 45 \degrees 49' \text { west of north}$

$\blacksquare$

## Proof 2

By plotting the points in a graphics package, or on paper with a ruler and protractor:

$\blacksquare$