3.2A

Sine & Cosine Rules

  • The Sine Rule: Use when given two sides and an angle opposite one of them, or two angles and any side: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
  • The Cosine Rule: Use for SAS (two sides and included angle) or SSS (three sides): $$c^2 = a^2 + b^2 - 2ab\cos C \iff \cos C = \frac{a^2 + b^2 - c^2}{2ab}$$
  • Area of Any Triangle: $$\text{Area} = \frac{1}{2}ab\sin C$$
  • 3-Figure Navigation Bearings: Measured clockwise from true North ($000^\circ$ to $360^\circ$). Always draw a North arrow at each station.

Choose the Rule Before Calculating

  • Lower-case $a,b,c$ are opposite the matching angles $A,B,C$. Mark each opposite pair on the diagram.
  • Use the sine rule when an opposite side–angle pair is known. Use the cosine rule for two sides and their included angle (SAS), or three sides (SSS).
  • For area $\frac12ab\sin C$, the angle must lie between the two selected sides. Triangle angles add to $180^\circ$.
  • Worked example: $a=5$, $A=30^\circ$, $B=45^\circ$ give $b=5\sin45^\circ/\sin30^\circ=5\sqrt2\approx7.07$.
  • SSS example: With sides $5,7,8$, the angle opposite $8$ satisfies $\cos C=(25+49-64)/(2\cdot5\cdot7)=1/7$, giving $C\approx81.8^\circ$.

Bearings and Navigation Diagrams

  • A bearing is measured clockwise from north and written with three digits: $045^\circ$, $090^\circ$, $270^\circ$.
  • Reverse bearing: add $180^\circ$ and reduce modulo $360^\circ$. If B is on bearing $070^\circ$ from A, A is on bearing $250^\circ$ from B.
  • Draw a north line at every station. Convert bearings to interior triangle angles before applying a rule.
  • Worked example: Walk $10$ km east, then $10$ km north. The displacement is $\sqrt{10^2+10^2}=14.1$ km on bearing $045^\circ$. The return bearing is $225^\circ$.
  • For displacement length $d$ on bearing $\beta$, the east component is $d\sin\beta$ and north component $d\cos\beta$. Add signed components before finding a net distance; choose the correct bearing quadrant.
ABCN10 km east10 km
Distance travelled is 20 km; displacement is the straight segment AC.

Elevation, Depression and Valid Triangles

  • Elevation is measured upward from the observer’s horizontal line. Depression is measured downward; parallel horizontal lines give equal alternate angles.
  • Add the observer’s eye height when the triangle gives height above their eyes. Use horizontal distance in $h=d\tan\theta$.
  • Worked example: At $40$ m horizontal distance, an elevation of $35^\circ$ from eyes $1.6$ m above the ground gives tower height $40\tan35^\circ+1.6\approx29.6$ m.
  • Check triangle inequality: the sum of any two sides must exceed the third. If sine-rule calculation gives $\sin B>1$, the proposed triangle does not exist.
  • Two possible angles: If $\sin B=s$ with $0<s<1$, test $B=\sin^{-1}s$ and $180^\circ-B$. Keep only those making all triangle angles positive. An SSA problem can have zero, one or two triangles.
  • Use degrees for bearings. Keep intermediate values unrounded and state lengths, areas and angles with appropriate units.

10 Progressive Practice Exercises

Choose the method, draw a diagram, and explain each step in surveying and navigation problems.

Exercise 1 (Paper 1 — 3 Marks): In triangle $ABC$, $a = 7.0\text{ cm}$, $b = 9.0\text{ cm}$, and angle $C = 42^\circ$. Calculate the length of side $c$, correct to 3 significant figures.
Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C = 7^2 + 9^2 - 2(7)(9)\cos(42^\circ)$ [M1]
$c^2 = 49 + 81 - 126(0.743145) = 130 - 93.636 = 36.364$ [A1]
$c = \sqrt{36.364} \approx 6.030 \approx 6.03\text{ cm}$ [A1]

Answer: $6.03\text{ cm}$

Exercise 2: In triangle ABC, $a=5$ cm, $A=30^\circ$ and $B=45^\circ$. Find $b$ and $C$.

$b=5\sin45^\circ/\sin30^\circ=5\sqrt2\approx7.07$ cm. $C=180^\circ-30^\circ-45^\circ=105^\circ$.

Exercise 3: A triangle has sides $5$, $7$ and $8$ cm. Find its largest angle.

The largest angle is opposite the longest side $8$. $\cos C=(5^2+7^2-8^2)/(2\cdot5\cdot7)=1/7$, hence $C\approx81.8^\circ$.

Exercise 4: Two sides of a triangle are $8$ cm and $10$ cm with included angle $30^\circ$. Find the area.

$A=\frac12(8)(10)\sin30^\circ=20\text{ cm}^2$. The included angle is between the selected sides.

Exercise 5: B is on a bearing of $070^\circ$ from A. Find the bearing of A from B and explain why.

Reverse direction differs by $180^\circ$. $070^\circ+180^\circ=250^\circ$; both north lines are parallel.

Exercise 6: A walker travels $10$ km east from A to B, then $10$ km north to C. Find AC, the bearing of C from A and the return bearing.

$AC=\sqrt{10^2+10^2}=10\sqrt2\approx14.1$ km. Equal east/north displacements give bearing $045^\circ$. The return bearing is $225^\circ$, not $315^\circ$.

Exercise 7: Observation stations A and B are $10$ km apart, with B due east of A. A lighthouse P is on bearing $060^\circ$ from A and $330^\circ$ from B. Find AP.

At A the interior angle is $90^\circ-60^\circ=30^\circ$. At B it is $330^\circ-270^\circ=60^\circ$, so the angle at P is $90^\circ$. $AP/\sin60^\circ=10/\sin90^\circ$, hence $AP=5\sqrt3\approx8.66$ km.

Exercise 8: A person stands $40$ m horizontally from a tower. The top’s elevation is $35^\circ$ from eyes $1.6$ m above level ground. Find the tower height.

Height above the eyes is $40\tan35^\circ\approx28.0083$ m. Add eye height: $H=40\tan35^\circ+1.6\approx29.6$ m.

Exercise 9: A proposed triangle has $A=30^\circ$, opposite side $a=4$ cm and side $b=10$ cm. Can it exist? Justify your answer.

Sine rule gives $\sin B=b\sin A/a=10(0.5)/4=1.25$. Since sine cannot exceed 1, no such triangle exists. Do not continue with an inverse-sine error.

Exercise 10: A triangular plot has two boundary lengths $70$ m and $90$ m meeting at $60^\circ$. Find its area and the fencing length needed if a $4$ m gate is left unfenced. Give final results to 3 significant figures.

Area $=\frac12(70)(90)\sin60^\circ\approx2727.98\text{ m}^2$, so $2730\text{ m}^2$. Third side $=\sqrt{70^2+90^2-2(70)(90)\cos60^\circ}=\sqrt{6700}\approx81.8535$ m. Fencing $=70+90+\sqrt{6700}-4\approx237.854$ m, hence $238$ m. This assumes straight boundaries and a gate measured along a boundary.

Formulae

Non-Right-Angled Trig Formulas

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ $$c^2 = a^2 + b^2 - 2ab\cos C$$ $$\text{Area} = \frac{1}{2}ab\sin C$$

Method and Diagram Checklist

  • Label opposite side–angle pairs; mark the included angle for cosine rule or area.
  • Draw north at both stations; write reverse bearings with three digits.
  • Keep all valid triangle angles, reject impossible values, and distinguish displacement from distance travelled.
  • Draw the horizontal eye line for elevation and depression, and include observer height when needed.
Topic 3.2 Check

Exam Mini Test (5 Questions)

Score: 0 / 5

1. In $\triangle ABC$, $a=6, b=8,$ and $C=30^\circ$. What is the area?

2. B is on bearing 315° from A. What is A’s bearing from B?

3. Which rule directly finds an angle when all three sides are known?

4. A triangle has a = 4, b = 10 and A = 30°. How many triangles fit?

5. From eye height 1.5 m, elevation is 45° at horizontal distance 20 m. What is the tower height?