This note extends trigonometry beyond right-angled triangles. It covers the sine rule, the cosine rule, the area formula \(\tfrac12 ab\sin C\), the graphs of sine, cosine and tangent, and how to find lengths and angles in three-dimensional shapes. These tools are used in surveying, navigation and engineering whenever a triangle has no right angle.
Key ideas
Label a triangle so that side \(a\) is opposite angle \(A\), side \(b\) is opposite angle \(B\) and side \(c\) is opposite angle \(C\). Capital letters are angles, small letters are sides.
Choosing the rule.
- Use the sine rule when you know a side and its opposite angle (a matching pair) and want another side or angle.
- Use the cosine rule to find a side when you know two sides and the angle between them, or to find an angle when you know all three sides.
- Use area = \(\tfrac12 ab\sin C\) when you know two sides and the angle between them.
The sine rule can be written with sides on top (to find a side) or with sines on top (to find an angle). Turn it upside down to suit the question.
Obtuse angles. The sine of an angle and the sine of \(180^\circ - \) that angle are equal, for example \(\sin 30^\circ = \sin 150^\circ = 0.5\). When you use the sine rule to find an angle, your calculator gives the acute answer. If the diagram shows an obtuse angle, the answer is \(180^\circ\) minus the calculator value. The cosine rule does not have this problem: a negative cosine gives an obtuse angle automatically.
Graphs. For \(0^\circ \le x \le 360^\circ\):
- \(y = \sin x\) starts at 0, rises to 1 at \(90^\circ\), returns to 0 at \(180^\circ\), falls to \(-1\) at \(270^\circ\), and is back at 0 at \(360^\circ\).
- \(y = \cos x\) starts at 1, is 0 at \(90^\circ\), is \(-1\) at \(180^\circ\), is 0 at \(270^\circ\), and is 1 at \(360^\circ\).
- \(y = \tan x\) starts at 0, rises without limit as \(x\) approaches \(90^\circ\), and is undefined at \(90^\circ\) and \(270^\circ\) (vertical asymptotes). It is 0 at \(0^\circ\), \(180^\circ\) and \(360^\circ\).
The sine and cosine curves have amplitude 1 (largest value 1, smallest \(-1\)) and a period of \(360^\circ\), which means the pattern repeats every \(360^\circ\). The tangent curve has period \(180^\circ\) and no amplitude. Because the curves repeat, an equation such as \(\sin x = 0.5\) has more than one solution in a range. Use the symmetry of the graph: if the calculator gives \(x\), the sine curve also gives \(180^\circ - x\), and the cosine curve also gives \(360^\circ - x\).
Three-dimensional problems. Find a right-angled triangle inside the solid, often by drawing a base diagonal first. Then use Pythagoras or SOH CAH TOA. The angle between a line and a plane is the angle between the line and its shadow on the plane, which is found by dropping a perpendicular from the top of the line to the plane. Redraw the triangle you are using as a flat sketch with its right angle marked.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) | Sine rule: a side and its opposite angle are known. |
| \(\dfrac{\sin A}{a} = \dfrac{\sin B}{b}\) | Sine rule rearranged for finding an angle. |
| \(a^2 = b^2 + c^2 - 2bc\cos A\) | Cosine rule for a side: two sides and the included angle. |
| \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\) | Cosine rule for an angle: all three sides known. |
| Area \(= \tfrac12 ab\sin C\) | Area of any triangle from two sides and the angle between them. |
| Amplitude and period | Half the distance between the highest and lowest points, and the length of one full cycle. |
| Space diagonal of a cuboid | \(\sqrt{l^2 + w^2 + h^2}\), from Pythagoras applied twice. |
Worked examples
Example 1 In triangle ABC, \(A = 48^\circ\), \(B = 65^\circ\) and \(a = 9.2\) cm. Find side \(b\).
Side \(a\) and angle \(A\) are a matching pair, and \(b\) is opposite the known angle \(B\), so use the sine rule.
\(\dfrac{b}{\sin 65^\circ} = \dfrac{9.2}{\sin 48^\circ}\)
\(b = \dfrac{9.2 \times \sin 65^\circ}{\sin 48^\circ} = \dfrac{9.2 \times 0.9063}{0.7431} = 11.22\ldots\)
Answer: \(b = 11.2\) cm.
Example 2 A triangular plot of land has sides 6 m, 9 m and 11 m. (a) Find the largest angle. (b) Find the area of the plot using this angle.
(a) The largest angle is opposite the longest side. Call it \(A\), so \(a = 11\), \(b = 6\), \(c = 9\).
\(\cos A = \dfrac{6^2 + 9^2 - 11^2}{2 \times 6 \times 9} = \dfrac{36 + 81 - 121}{108} = \dfrac{-4}{108}\)
\(A = \cos^{-1}(-0.037037) = 92.12^\circ\ldots\) (obtuse, as the cosine is negative).
(b) Area \(= \tfrac12 \times 6 \times 9 \times \sin 92.12^\circ = 27 \times 0.99931 = 26.98\ldots\)
Answer: (a) \(92.1^\circ\). (b) \(27.0\ \text{m}^2\).
Example 3 A storage box is a cuboid 8 cm long, 6 cm wide and 5 cm high. Find (a) the length of the diagonal of the base, (b) the length of the longest straight rod that fits inside, and (c) the angle between that longest rod and the base.
(a) The base is a rectangle, so the diagonal \(d\) satisfies \(d^2 = 8^2 + 6^2 = 100\) and \(d = 10\) cm.
(b) The longest rod runs from one bottom corner to the opposite top corner. It is the hypotenuse of a right-angled triangle with sides 10 cm (base diagonal) and 5 cm (height). \(\ell^2 = 10^2 + 5^2 = 125\), so \(\ell = 11.18\ldots\) cm.
(c) The angle \(\theta\) between the rod and the base is at the bottom corner, between the rod and the base diagonal. The height 5 cm is opposite \(\theta\) and the base diagonal 10 cm is adjacent, so \(\tan\theta = \dfrac{5}{10}\) and \(\theta = 26.57^\circ\ldots\)
Answer: (a) 10 cm. (b) 11.2 cm. (c) \(26.6^\circ\).
Common errors
- Common error: using the sine rule when there is no matching side and angle pair. Check for a pair first, and use the cosine rule if you do not have one.
- Common error: in the cosine rule, evaluating \(b^2 + c^2 - 2bc\cos A\) in the wrong order. Do the multiplication \(2bc\cos A\) first and subtract it, rather than subtracting \(2bc\) from \(b^2 + c^2\) first.
- Common error: forgetting that an obtuse angle may be needed with the sine rule, and giving the acute value from the calculator.
- Common error: having the calculator in radian mode. Check it is in degrees.
- Common error: stating only one solution of an equation like \(\cos x = -0.5\) when the range \(0^\circ \le x \le 360^\circ\) has two.
Exam tip: Before using any formula, label the triangle with \(a\), \(b\), \(c\) and \(A\), \(B\), \(C\) so each side is opposite its angle. In 3D questions, draw the flat right-angled triangle you are using separately and write its lengths on it.
Check yourself
- In triangle ABC, \(a = 14\) cm, \(b = 11\) cm and \(A = 70^\circ\), with \(B\) acute. Find \(B\).
Answer
\(\sin B = \dfrac{11\sin 70^\circ}{14} = \dfrac{11 \times 0.93969}{14} = 0.7383\), so \(B = 47.6^\circ\). - In a triangle, \(b = 7.4\) cm, \(c = 10.3\) cm and \(A = 112^\circ\). Find \(a\).
Answer
\(a^2 = 7.4^2 + 10.3^2 - 2(7.4)(10.3)\cos 112^\circ = 54.76 + 106.09 + 57.11 = 217.96\) (the cosine is negative, so subtracting adds), so \(a = 14.8\) cm. - Find the area of a triangle with sides 12 cm and 15 cm and an angle of \(38^\circ\) between them.
Answer
Area \(= \tfrac12 \times 12 \times 15 \times \sin 38^\circ = 90 \times 0.61566 = 55.4\ \text{cm}^2\). - State the amplitude and period of \(y = \cos x\), and find both solutions of \(\sin x = 0.5\) for \(0^\circ \le x \le 360^\circ\).
Answer
Amplitude 1, period \(360^\circ\). The calculator gives \(x = 30^\circ\), and the second solution is \(180^\circ - 30^\circ = 150^\circ\).