Geometry and Trigonometry - Bearings

A ship sails $12$ km on a bearing of $090^\circ$ and then $5$ km on a bearing of $180^\circ$. Find the bearing of the ship from its starting point.

Let the starting point be $A$. The ship moves $12$ km East to point $B$ and $5$ km South to point $C$. In triangle $ABC$, $AB = 12$ km and $BC = 5$ km. We need the angle $\theta$ clockwise from North at point $A$. The internal angle at $A$ (let's call it $\alpha$) is found using: $\tan(\alpha) = \frac{BC}{AB} = \frac{5}{12}$ $\alpha = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.6^\circ$ Since the first movement was $090^\circ$ (East), the total bearing from North is: $90^\circ + 22.6^\circ = 112.6^\circ$

We model the journey as a right-angled triangle. Since the ship first travels East ($090^\circ$) and then South ($180^\circ$), the paths are perpendicular. We calculate the internal angle using trigonometry and add it to the initial $90^\circ$ displacement from North.

The bearing of point $Q$ from point $P$ is $065^\circ$. Calculate the bearing of $P$ from $Q$.

The given bearing $\theta = 065^\circ$. Since $065^\circ < 180^\circ$, we add $180^\circ$: $Bearing = 065^\circ + 180^\circ = 245^\circ$

To find a back bearing (the direction looking back at the start), we add or subtract $180^\circ$. This is because North lines are parallel, making the interior angles between the two points supplementary ($180^\circ$).

Calculate the distance between two points $X$ and $Y$ if $Y$ is on a bearing of $040^\circ$ from $X$ and the horizontal distance (East) between them is $15$ km.

The bearing $040^\circ$ forms an angle of $40^\circ$ with the North line. This means the angle with the East line is $90^\circ - 40^\circ = 50^\circ$. However, it is simpler to use the $40^\circ$ angle from North. The East distance ($15$ km) is the side opposite to the $40^\circ$ angle. Let $d$ be the direct distance (hypotenuse): $\sin(40^\circ) = \frac{15}{d}$ $d = \frac{15}{\sin(40^\circ)} \approx 23.34\text{ km}$

By drawing a right-angled triangle where the hypotenuse is the direct path and the 'Opposite' side is the Eastward displacement, we use the Sine ratio to find the total distance.

$BC = 150$ $AB = 200$ Angle at $B$ can be found using interior angles. The angle between the North line at $B$ and the line $BA$ is $180^\circ - 120^\circ = 60^\circ$. The angle between North and $BC$ is $30^\circ$. Thus, the total angle $ABC$ is $60^\circ + 30^\circ = 90^\circ$. Using Pythagoras: $AC = \sqrt{200^2 + 150^2}$ $AC = \sqrt{40000 + 22500}$ $AC = \sqrt{62500}$ $AC = 250\text{ km}$

By drawing North lines at $A$ and $B$, we find that the path forms a right-angled triangle because the interior angle relative to the south-bound line and the next bearing add up to $90^\circ$.

The angle measured clockwise from North is $210^\circ$. The angle from the South line $(180^\circ)$ is $210^\circ - 180^\circ = 30^\circ$. In the right-angled triangle formed with the vertical South line: $\sin(30^\circ) = \frac{\text{Opposite (West distance)}}{\text{Hypotenuse}}$ $\sin(30^\circ) = \frac{W}{40}$ $W = 40 \times \sin(30^\circ)$ $W = 40 \times 0.5 = 20\text{ km}$

To find the 'West' component, we determine the angle between the path and the North-South axis and use the sine function.