Worked example
A boat leaves the harbour and travels 9 km north west, then 12 km north. Calculate:
a) its distance from the harbour
b) the bearing it now has to sail to get back to the harbour
Answer
We need to sketch the problem first.
Find the size of angle R.
The boat travels from the harbour in a NW direction. This is halfway between west and north so it is at an angle of 45掳 to the axis for east-west and the axis for north-south. Angles of 45掳 are formed as shown in the diagram below.
When the boat's direction changes to N, the internal angle formed can be found by adding 45掳 and 90掳.
The 90掳 comes from the angle between the north-south and east-west axes
This angle is 135掳, as shown below.
The distance of the boat from the harbour is the missing side.
We know two sides and included angle, so we use the cosine rule.
\({a^2} = {b^2} + {c^2} - 2bcCosA\)
\(= {9^2} + {12^2} - (2 \times 9 \times 12 \times \cos 135^\circ )\)
\(= 81 + 144 鈥 (-152.74)\)
\(= 81 + 144 + 152.74\)
\(= 377.74\)
\(a=\sqrt{377.74}\)
\(a=19.4\,\,km\)
We now have to find the new bearing to sail back to the harbour.
Let's look again at the sketch.
If we find the angle \(x^{\circ}\) then we can find our required angle by subtracting this answer from 180掳. The angles in a straight line add up to 180掳, in this case the north line.
We can use the sine rule or the cosine rule as we now have so much information. The sine rule is usually quicker.
\(\frac{{19.4}}{{\sin 135^\circ }} = \frac{9}{{\sin x^\circ }}\)
\(19.4\sin x^\circ = 9\sin 135^\circ\)
\(\sin x^\circ = \frac{{9\sin 135^\circ }}{{19.4}}\)
\(sinx^{\circ}=0.328\)
\(x^{\circ}=sin^{-1}\,0.328\)
\(x^{\circ}=19.1^{\circ}\)
Therefore the angle we require is: \(180^{\circ}-19.1^{\circ}=159.9^{\circ}\).