Applications of the Sine Rule and Cosine Rule

Once you know the sine rule and cosine rule, the next step is knowing which one to use and how to turn a word problem into a triangle you can actually solve. This lesson walks through both formulas, shows how to decide between them, and applies each one to realistic problems.

Both rules use the same labeling convention: each lowercase side sits opposite the uppercase angle with the same letter. Side \(a\) is opposite angle \(A\), side \(b\) is opposite angle \(B\), and side \(c\) is opposite angle \(C\).

The sine rule states:

\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

Use it whenever you know a pair made up of an angle and its opposite side, plus one more piece of information, that is, the AAS (angle-angle-side) or ASA (angle-side-angle) case.

Worked example: AAS with the sine rule

In triangle \(ABC\), \(A = 42^\circ\), \(B = 65^\circ\), and \(a = 10\) cm\(\). Find side \(b\).

\(\frac{a}{\sin A} = \frac{b}{\sin B} \quad \Rightarrow \quad b = \frac{a \sin B}{\sin A} = \frac{10 \sin 65^\circ}{\sin 42^\circ} \approx \frac{10(0.9063)}{0.6691} \approx 13.5\) cm\(\)

The cosine rule states:

\(a^2 = b^2 + c^2 - 2bc\cos A\)

and its rearranged forms for the other two sides. It is the right tool for the SAS (side-angle-side) case, when you know two sides and the angle between them, and for the SSS (side-side-side) case, when you know all three sides but no angles. For the full derivation and proof of this formula, see the dedicated law of cosines lesson.

Worked example: SAS with the cosine rule

In triangle \(ABC\), \(b = 8\) cm\(\), \(c = 6\) cm\(\), and the included angle \(A = 75^\circ\). Find side \(a\).

\(a^2 = 8^2 + 6^2 - 2(8)(6)\cos 75^\circ \approx 64 + 36 - 96(0.2588) \approx 75.2\)

\(a \approx \sqrt{75.2} \approx 8.67\) cm\(\)

Once one side is found this way, the sine rule can be used to find the remaining angles.

When the cosine rule is rearranged to solve for an angle, for example \(\cos A = \frac{b^2+c^2-a^2}{2bc}\), a negative result means the angle is obtuse, while a positive result means it is acute. This matters because cosine is positive for angles between \(0^\circ\) and \(90^\circ\) but negative between \(90^\circ\) and \(180^\circ\). If this sign behavior is unfamiliar, the ASTC rule lesson explains exactly which ratios are positive in which range.

Most application problems boil down to sketching a triangle from the description, labeling the known sides and angles, and picking the matching rule.

Worked example: surveying a lake with the cosine rule

A surveyor stands at a point where the two lines of sight to opposite ends of a lake measure \(250\) m\(\) and \(310\) m\(\), with an angle of \(78^\circ\) between them. Find the distance across the lake.

Here two sides and the included angle are known, so this is an SAS case for the cosine rule. Let \(d\) be the distance across the lake:

\(d^2 = 250^2 + 310^2 - 2(250)(310)\cos 78^\circ \approx 62500 + 96100 - 155000(0.2079) \approx 126375\)

\(d \approx \sqrt{126375} \approx 355.5\) m\(\)

Worked example: a navigation problem with the sine rule

Two lighthouses \(A\) and \(B\) are \(20\) km\(\) apart. From \(A\), a ship is sighted at an angle of \(50^\circ\) to the line \(AB\), and from \(B\) the same ship is sighted at an angle of \(65^\circ\) to the line \(AB\). How far is the ship from lighthouse \(A\)?

The triangle formed has two known angles and the included side, an ASA case. The third angle is \(180^\circ - 50^\circ - 65^\circ = 65^\circ\). Using the sine rule with the ship's position as vertex \(C\):

\(\frac{AC}{\sin B} = \frac{AB}{\sin C} \quad \Rightarrow \quad AC = \frac{20 \sin 65^\circ}{\sin 65^\circ} = 20\) km\(\)

(In this particular case the two base angles happen to be equal, making the triangle isosceles; in general the same setup works for any pair of sighting angles.)

When a problem gives two sides and an angle that is not between them (SSA), the sine rule can produce two different valid triangles, one with an acute angle and one with its obtuse supplement. After solving \(\sin B = k\) for angle \(B\), always check both \(B\) and \(180^\circ - B\) to see which (or both) give a triangle whose angles sum to \(180^\circ\). Reviewing how a reference angle is used to find related angles makes this check much faster.

As a rule of thumb: reach for the sine rule when an angle and its opposite side are both known (AAS or ASA), and reach for the cosine rule when the known information is two sides with the included angle (SAS) or all three sides (SSS). Many multi-step problems use one rule to find a missing piece and then switch to the other to finish the triangle.

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