Sine/Cosine Rules & Bearings
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Apply the sine and cosine rules and solve bearings problems in non-right-angled triangles.
Concrete scenario: A coastguard station at Mombasa port sees Ship A on a bearing of 030° (4 km away) and Ship B on a bearing of 110° (6 km away). How far apart are the two ships? You cannot drop a perpendicular neatly here — triangle OAB is not right-angled. You need general non-right triangle rules.
Geometric insight: Every non-right-angled triangle can be split into two right-angled triangles by dropping an altitude. When you apply Pythagoras to both halves, the altitude cancels out to yield the Cosine Rule: \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Equating the altitude in terms of sines yields the Sine Rule: \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).
Bearings & Triangles: A bearing is a clockwise angle measured from true North (000° to 360°). The interior angle formed between two routes from the same point is the absolute difference between their bearings (or \(360^{\circ}\) minus that difference if greater than \(180^{\circ}\)).
Interactive Bearings & Cosine Rule Explorer
Interior Angle \(\Delta\): \(110^{\circ} - 30^{\circ} = 80^{\circ}\)
Distance AB (Cosine Rule):
\(d = \sqrt{4^2 + 6^2 - 2(4)(6)\cos(80^{\circ})} \approx 6.45\,\text{km}\)
Key Formulas
Worked Examples
Problem: In triangle ABC, \(a=8\,\text{km}\), \(b=6\,\text{km}\), and \(\angle C=60^{\circ}\). Find side \(c\).
- Formula: \[c^{2}=a^{2}+b^{2}-2ab\cos C\]
- Substitute values: \[c^{2}=8^{2}+6^{2}-2(8)(6)\cos 60^{\circ}\]
- Calculate: \(\cos 60^{\circ}=0.5\), so \[c^{2}=64+36-48=52\]
- Solve for \(c\): \[c=\sqrt{52}\approx 7.21\,\text{km}\]
Problem: From Mombasa lighthouse O, Ship A is on bearing 045° at 120 nautical miles. Ship B is on bearing 125° at 80 nm. Find distance AB.
- Interior angle at O: \(\Delta = 125^{\circ} - 45^{\circ} = 80^{\circ}\).
- Apply Cosine Rule: \[D^{2}=120^{2}+80^{2}-2(120)(80)\cos 80^{\circ}\]
- Evaluate: \[D^{2}=14400+6400-19200(0.1736)=20800-3333.12=17466.88\]
- Square root: \[D=\sqrt{17466.88}\approx 132.2\,\text{nm}\]
Problem: A vessel sails 30 km from Port P on bearing 040° to point Q, then 40 km on bearing 130° to point R. Find the total direct bearing from Port P to point R.
- Recognize the right angle at Q: The interior turn angle at Q is \(180^{\circ} - (130^{\circ} - 40^{\circ}) = 90^{\circ}\).
- Calculate triangle angle \(\angle QPR\): \[\tan(\angle QPR) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{40}{30} = 1.3333 \implies \angle QPR = \tan^{-1}(1.3333) \approx 53.13^{\circ}\]
- Add initial bearing to angle: \[\text{Bearing of R from P} = 40^{\circ} + 53.13^{\circ} = 93.13^{\circ} \approx 93^{\circ}\]
Common Mistakes
Real World
Practice