This note covers trigonometry in right-angled triangles, then the sine rule, the cosine rule and the area formula for triangles with no right angle. Together these let you find any missing side or angle in any triangle, which is the basis of surveying, navigation and engineering design.
Key ideas
In a right-angled triangle, label the sides relative to the angle \(\theta\) you are using. The hypotenuse is opposite the right angle. The opposite side faces \(\theta\). The adjacent side touches \(\theta\) and is not the hypotenuse.
- \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\).
- To find a missing angle, use the inverse functions \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) on your GDC.
- For triangles with no right angle, label the vertices \(A, B, C\) and the opposite sides \(a, b, c\). Side \(a\) is opposite angle \(A\).
- Use the sine rule when you know a side and the angle opposite it (a matching pair) plus one more piece of information.
- Use the cosine rule when you know two sides and the angle between them, or all three sides.
- The angles in a triangle add up to \(180^\circ\), which gives the third angle once two are known.
Choosing the rule: ask "do I have a matching side and angle pair?" If yes, try the sine rule. If not, use the cosine rule. The case of two sides and a non-included angle can give two triangles, and is covered in the note on the unit circle and the ambiguous case.
Useful checks for any triangle answer:
- The longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. If your answer breaks this pattern, look for a slip.
- Two sides of a triangle must together be longer than the third side.
- An angle in a triangle is always less than \(180^\circ\), and the cosine rule gives a negative \(\cos C\) exactly when \(C\) is obtuse.
- The sine rule can also be written upside down, \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\), which is easier when you are finding an angle.
Multi-step problems. Many questions need two rules in a row. A typical pattern is to use the cosine rule to find a missing side, then the sine rule to find an angle, then the area formula. When you do this, store each value on your GDC at full accuracy rather than retyping a rounded number. Also draw each triangle separately if a diagram contains more than one.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sin\theta = \dfrac{o}{h},\ \cos\theta = \dfrac{a}{h},\ \tan\theta = \dfrac{o}{a}\) | Right-angled triangles only. |
| \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) | Sine rule. Use with a matching side and angle pair. |
| \(c^2 = a^2 + b^2 - 2ab\cos C\) | Cosine rule for a side. \(C\) is the angle between sides \(a\) and \(b\). |
| \(\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\) | Cosine rule rearranged to find an angle from three sides. |
| \(\text{Area} = \dfrac{1}{2}ab\sin C\) | Area from two sides and the angle between them. |
Worked examples
Example 1 A ladder of length 5.5 m leans against a wall and makes an angle of \(68^\circ\) with the horizontal ground. Find how high up the wall it reaches and how far its foot is from the wall.
The ladder is the hypotenuse. The height is opposite the \(68^\circ\) angle, and the distance from the wall is adjacent to it.
Height: \(\sin 68^\circ = \dfrac{h}{5.5}\), so \(h = 5.5 \sin 68^\circ \approx 5.10\) m.
Distance: \(\cos 68^\circ = \dfrac{d}{5.5}\), so \(d = 5.5 \cos 68^\circ \approx 2.06\) m.
Answer: height \(\approx 5.10\) m; foot \(\approx 2.06\) m from the wall.
Example 2 In triangle \(ABC\), \(A = 48^\circ\), \(B = 71^\circ\) and \(a = 12.5\) cm. Find side \(b\) and side \(c\).
Side \(a\) and angle \(A\) form a matching pair, so use the sine rule.
\(\dfrac{b}{\sin 71^\circ} = \dfrac{12.5}{\sin 48^\circ}\), so \(b = \dfrac{12.5 \sin 71^\circ}{\sin 48^\circ} \approx 15.9\) cm.
Third angle: \(C = 180^\circ - 48^\circ - 71^\circ = 61^\circ\).
\(c = \dfrac{12.5 \sin 61^\circ}{\sin 48^\circ} \approx 14.7\) cm.
Answer: \(b \approx 15.9\) cm, \(c \approx 14.7\) cm.
Example 3 A triangular plot of land has two sides of length 7.2 m and 9.5 m with an angle of \(112^\circ\) between them. Find the third side and the area of the plot.
Two sides and the included angle are known, so use the cosine rule.
\(c^2 = 7.2^2 + 9.5^2 - 2(7.2)(9.5)\cos 112^\circ\)
\(\cos 112^\circ \approx -0.37461\), so \(-2(7.2)(9.5)\cos 112^\circ = -136.8 \times (-0.37461) \approx 51.25\).
\(c^2 \approx 51.84 + 90.25 + 51.25 = 193.34\).
\(c \approx 13.9\) m.
Area \(= \dfrac{1}{2}(7.2)(9.5)\sin 112^\circ \approx 31.7\) m².
Answer: third side \(\approx 13.9\) m; area \(\approx 31.7\) m².
Common errors
- Having the calculator in radian mode when the angles are in degrees.
- Using the cosine rule with an angle that is not between the two sides in the formula. The angle in the cosine term must be opposite the side on the left of the equation.
- In the cosine rule, the minus sign applies to the whole term \(2ab\cos C\). When \(C\) is obtuse, \(\cos C\) is negative, so the term adds to the total.
- Rounding the intermediate values too early. Keep full calculator values and round only the final answer.
- Using \(\frac{1}{2}ab\sin C\) with an angle that is not between sides \(a\) and \(b\).
Exam tip: Write the triangle with all known values on it before choosing a rule. If a question has several parts, check whether an earlier answer gives you the matching pair you need, and use the unrounded value for later steps.
Check yourself
- A ramp rises 1.2 m over a horizontal distance of 9.0 m. Find the angle the ramp makes with the horizontal.
Answer
\(\tan\theta = \frac{1.2}{9.0}\), so \(\theta = \tan^{-1}(0.1333) \approx 7.59^\circ\). - In triangle \(PQR\), \(p = 8\) cm, \(q = 11\) cm and \(r = 15\) cm. Find the largest angle.
Answer
The largest angle is opposite the longest side, \(R\). \(\cos R = \frac{8^2 + 11^2 - 15^2}{2(8)(11)} = \frac{-40}{176}\), so \(R \approx 103^\circ\). - In triangle \(ABC\), \(B = 40^\circ\), \(C = 65^\circ\) and \(b = 9\) cm. Find \(c\).
Answer
\(c = \frac{9 \sin 65^\circ}{\sin 40^\circ} = \frac{9 \times 0.90631}{0.64279} \approx 12.7\) cm. - Find the area of a triangle with sides 10 cm and 13 cm and an included angle of \(35^\circ\).
Answer
\(\frac{1}{2}(10)(13)\sin 35^\circ = 65 \times 0.57358 \approx 37.3\) cm².