Trigonometry

Lesson

A slightly lost hiker walks $300$300 m east before turning south walking another $800$800 m. What is its true bearing from its original position (to $1$1 decimal place)? And then what is this as a compass bearing?

**Think**: to do this questions a diagram is a great way to start. This is the diagram I drew.

The angle we want to find is in green. It is equal to 90° + the angle inside the triangle that I can find using trigonometry.

**Do**: Bearing is

$90+\tan^{-1}\frac{800}{300}$90+tan−1800300 |
$=$= | $90+\tan^{-1}\frac{8}{3}$90+tan−183 |

$=$= | $90+69.44$90+69.44 | |

$=$= | $159.44$159.44 |

and to 1 decimal place the final bearing is $159.4$159.4°N

As a compass bearing we need to know the acute angle with the North South line. This is $180-159.4=20.6$180−159.4=20.6

The compass points we will use will be South first, and then East, so the compass bearing is S$20.6$20.6°E

The position of a ship S is given to be $20$20 kilometres from P, on a true bearing of $0$0$49$49$^\circ$°T.

The position of the ship can also be given by its $\left(x,y\right)$(`x`,`y`) coordinates.

If the ship's $x$

`x`-coordinate is $x$`x`, find $x$`x`to one decimal place.If the ship's y-coordinate is $y$

`y`, find $y$`y`to one decimal place.