Distances from bearings
Draw north at the starting point and identify the reference line. A supplied right triangle lets you calculate distances; the bearing determines which acute angle belongs in it
- Choose an acute triangle angle Locate the journey in its quadrant. Use its acute angle to the reference line, not an unexamined full bearing
- Choose the matching sides If the angle is measured from the reference line, the perpendicular distance is opposite it. The along-line distance is adjacent \(\displaystyle \begin{aligned}d&=L\sin\theta\\a&=L\cos\theta\end{aligned}\)
- Interpret the sign Lengths are positive. In a signed component table, north and east are positive, while south and west are negative
- Use speed and duration when needed Match units before multiplying, and round only the requested result \(\displaystyle L=vt\)
- Northeast With positive east and north distances, calculate the acute angle from north \(\displaystyle b=\tan^{-1}\left(\frac{E}{N}\right)\)
- Other quadrants Use positive side lengths to find an acute angle from the nearby north or south ray. Then place that angle into the full clockwise turn
- Southwest example If the endpoint is west and south, add the acute angle west of south to 180° \(\displaystyle b=180^{\circ}+\tan^{-1}\left(\frac{W}{S}\right)\)
- Check the direction A southwest bearing is between 180° and 270°. Cardinal cases need no division: use 000°, 090°, 180° or 270°. Keep unrounded values until the final result
- Resolve each leg Use an acute reference angle and the correct signs. North/east are positive; south/west are negative
- Add matching components Sum eastward components and sum northward components separately. Opposite directions cancel \(\displaystyle \begin{aligned}E&=E_1+E_2\\N&=N_1+N_2\end{aligned}\)
- Find the direct length The net components form a new right triangle \(\displaystyle R=\sqrt{E^2+N^2}\)
- Find the bearing Locate the resultant’s quadrant, use an inverse ratio on its positive side lengths, and convert to a clockwise angle from north. A zero displacement has no direction
- State the model Assume straight legs on a local flat map with a common true-north reference. A heading is not necessarily the ground track if wind or current causes drift
Worked example
A straight east-west reference line passes through A. B is northeast of A. P is the perpendicular foot from B to the line. The bearing of B from A is 027 degrees. Calculate BP to one decimal place
- \(\displaystyle \theta=90^{\circ}-27^{\circ}\)Find the acute angle from the east-west line
- \(\displaystyle \theta=63^{\circ}\)This angle is at A between AP and AB
- \(\displaystyle \sin63^{\circ}=\frac{BP}{22.4}\)BP is opposite the angle and AB is the hypotenuse
- \(\displaystyle BP=22.4\sin63^{\circ}\)Multiply both sides by the journey length
- \(\displaystyle BP\approx19.95854614\ \mathrm{km}\)Evaluate in degree mode and keep the calculator value
- \(\displaystyle BP\approx20.0\ \mathrm{km}\)Round to one decimal place, including the final zero
Common mistakes
Difficulty levels
Questions about this topic
Must I always use sine?+
No. The ratio depends on your chosen acute angle. Sine with an angle from the reference line and cosine with its complement can calculate the same perpendicular distance
Why is the shortest distance perpendicular?+
Any sloping segment to another point on the line is the hypotenuse of a right triangle containing the perpendicular segment, so it is longer
Does a negative component mean a negative distance?+
No. It indicates a direction opposite to the chosen positive direction. Its magnitude is a positive length
How do I find a bearing from two components?+
Draw the quadrant, form an acute inverse-trigonometric angle using positive lengths, then convert to the clockwise angle from north. Check it against the quadrant
Can I add bearings to combine a route?+
No. Add signed north/east displacements using the leg lengths. Then calculate the resultant length and direction. A zero displacement has no bearing
Is travelled distance the same as displacement length?+
Usually not. Travelled distance adds every leg length. Displacement connects only the starting and finishing points
When is this navigation model an approximation?+
It uses a local flat map, straight legs and parallel true-north references. Long-distance Earth curvature, map distortion, wind and current can require a different model
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 9: Pythagoras and trigonometry