How to find the direct distance and bearing from a journey diagram

Key takeaway: Journey diagrams often hide a triangle in plain sight. Once you identify the sides and the included angle, the distance part is routine; the bearing part needs a careful angle chase.

The route forms a triangle with sides $7\text{ km}$ and $5\text{ km}$, and the included angle shown in the diagram is $30^\circ$.

1) Find the direct distance from camp to waterfall

Use the cosine rule:

$C^2=a^2+b^2-2ab\cos C.$

Substitute the known values:

$d^2=7^2+5^2-2(7)(5)\cos 30^\circ.$

So

$d^2=49+25-70\left(\frac{\sqrt{3}}{2}\right),$

$d^2\approx 13.36,$

$d\approx 3.65\text{ km}.$

2) Find the bearing of the camp from the waterfall

First determine the angle at the waterfall using the sine rule if needed, or by using the triangle angles after the direct side is found.

A common approach is:

• find the angle between the line to the lookout and the direct line to camp,
• then convert that angle to a bearing measured clockwise from north.

If the diagram gives a north line and a bearing of $300^\circ$ at camp, then the reverse bearing from waterfall to camp is $300^\circ-180^\circ=120^\circ$.

So the bearing of the camp from the waterfall is

$120^\circ.$

Final answers

• Direct distance from camp to waterfall: $3.65\text{ km}$
• Bearing of the camp from the waterfall: $120^\circ$

If your class uses a slightly different angle interpretation from the diagram labels, re-check which angle is measured at each vertex before applying the cosine rule.

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Pitfalls the pros know 👇 The biggest mistake here is reading the $30^\circ$ or $20^\circ$ label at the wrong vertex. In travel problems, one angle usually belongs to the triangle, while another is a compass bearing. Mixing those up changes everything.

Also, bearing is always written as a three-figure angle measured clockwise from north. People often answer with a plain interior angle instead, which is not the same thing.

What if the problem changes? If the direct path were asked from lookout to waterfall instead of camp to waterfall, you would build a different triangle and use the same cosine rule idea.

If the question asked for the bearing of the waterfall from the camp, you would use the forward bearing rather than the reverse bearing. The reverse and forward bearings differ by $180^\circ$.

Tags: bearing, cosine rule, triangle