The Sine Rule: Unlocking the Secrets of Any Triangle
The Core Statement and Its Derivation
At its heart, the Sine Rule expresses a beautifully simple relationship for any triangle. For a triangle labeled $ABC$, where side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$, the rule is:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$
This can also be written in its reciprocal form, which is often more useful when finding an angle: $$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} $$
But where does this come from? Let's derive it intuitively. Consider any triangle $ABC$. The most straightforward derivation involves dropping an altitude from one vertex to the opposite side, creating two right-angled triangles. For example, dropping an altitude $h$ from vertex $C$ to side $c$ gives us two right triangles.
In triangle $ACD$ (where $D$ is the foot of the altitude), we have $\sin A = \frac{h}{b}$, so $h = b \sin A$.
In triangle $BCD$, we have $\sin B = \frac{h}{a}$, so $h = a \sin B$.
Since both expressions equal $h$, we can set them equal: $b \sin A = a \sin B$. Rearranging gives us the first part of the rule: $$ \frac{a}{\sin A} = \frac{b}{\sin B} $$. Repeating this process by dropping an altitude from a different vertex will include side $c$ and angle $C$, completing the full rule.
When and How to Apply the Sine Rule
The Sine Rule is your go-to tool when you are dealing with triangles that are not right-angled. Specifically, it is most useful when you know:
- Two angles and one side (AAS or ASA): You can find the missing side directly.
- Two sides and a non-included angle (SSA): You can find a missing angle, but this leads to the "ambiguous case" which we will discuss separately.
It is not the best choice when you know all three sides (SSS) or two sides and the included angle (SAS). For those situations, the Cosine Rule is more efficient.
1. Label your triangle with $A$, $B$, $C$ for angles and $a$, $b$, $c$ for the opposite sides.
2. Identify what you know and what you need to find.
3. Write down the part of the Sine Rule that contains three known quantities and one unknown.
4. Solve for the unknown using cross-multiplication.
Example 1 (Finding a Side): In triangle $ABC$, angle $A = 55^\circ$, angle $B = 72^\circ$, and side $b = 10$ cm. Find side $a$.
We know: $A=55^\circ$, $B=72^\circ$, $b=10$ cm. We need $a$.
Use the ratio: $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Plug in values: $\frac{a}{\sin 55^\circ} = \frac{10}{\sin 72^\circ}$.
Solve: $a = \frac{10 \times \sin 55^\circ}{\sin 72^\circ}$.
Using a calculator: $\sin 55^\circ \approx 0.8192$, $\sin 72^\circ \approx 0.9511$.
So, $a \approx \frac{10 \times 0.8192}{0.9511} \approx 8.61$ cm.
Navigating the Ambiguous Case (SSA)
The "ambiguous case" occurs when we are given two sides and a non-included angle (SSA). This situation can yield zero, one, or two possible triangles. Why? Because the sine function has a special property: $\sin \theta = \sin (180^\circ - \theta)$. For example, $\sin 30^\circ = \sin 150^\circ = 0.5$.
When we use the Sine Rule to find an angle, we get a sine value (e.g., 0.5). The calculator gives us the acute angle (e.g., $30^\circ$), but the obtuse angle $150^\circ$ also has the same sine. Both could potentially be valid angles in a triangle, depending on the side lengths.
The table below helps determine the number of possible triangles given sides $a$, $b$, and angle $A$ (where $a$ is opposite $A$):
| Condition | Relative to $h = b \sin A$ | Number of Triangles |
|---|---|---|
| $a < h$ | Side $a$ is too short to reach the base. | 0 |
| $a = h$ | Side $a$ just reaches the base, forming a right angle. | 1 (Right triangle) |
| $h < a < b$ | Side $a$ can swing to two different positions. | 2 |
| $a \ge b$ | Side $a$ is too long to swing; it locks into one position. | 1 |
Example 2 (Ambiguous Case): Triangle $ABC$ has $A=30^\circ$, $a=7$ cm, $b=10$ cm. Find possible values for angle $B$.
First, find $h = b \sin A = 10 \times \sin 30^\circ = 10 \times 0.5 = 5$ cm.
We have $a=7$, and $h < a < b$ (since $5 < 7 < 10$). According to the table, this is the ambiguous case with two possible triangles.
Use the Sine Rule: $\frac{\sin B}{b} = \frac{\sin A}{a}$ -> $\frac{\sin B}{10} = \frac{\sin 30^\circ}{7}$ -> $\sin B = \frac{10 \times 0.5}{7} = \frac{5}{7} \approx 0.7143$.
The acute angle from the calculator is $B_1 = \arcsin(0.7143) \approx 45.6^\circ$.
The obtuse possibility is $B_2 = 180^\circ - 45.6^\circ = 134.4^\circ$.
We must check if $B_2$ is valid: $A + B_2 = 30^\circ + 134.4^\circ = 164.4^\circ$, which is less than $180^\circ$, so a triangle can be formed. Therefore, two possible sets of angles exist: $(30^\circ, 45.6^\circ, 104.4^\circ)$ and $(30^\circ, 134.4^\circ, 15.6^\circ)$.
Real-World Applications and Problem Solving
The Sine Rule is not just an abstract mathematical concept; it is a practical tool used in various professions to solve real problems without direct measurement.
Application 1: Surveying and Mapping
A surveyor wants to measure the width of a river. She stands at point $A$ on one bank and identifies a distinctive tree at point $C$ on the opposite bank. She walks 50 meters along her bank to point $B$, and measures angle $ABC$ to be $80^\circ$ and angle $BAC$ to be $60^\circ$. What is the width of the river (the perpendicular distance from $C$ to line $AB$)?
First, find the third angle: $C = 180^\circ - 80^\circ - 60^\circ = 40^\circ$.
We know side $AB = 50$ m (it's side $c$, opposite angle $C$). We need side $AC$ (side $b$, opposite angle $B$) to later find the perpendicular width. Use the Sine Rule: $\frac{b}{\sin B} = \frac{c}{\sin C}$ -> $\frac{b}{\sin 80^\circ} = \frac{50}{\sin 40^\circ}$.
So, $b = \frac{50 \times \sin 80^\circ}{\sin 40^\circ} \approx \frac{50 \times 0.9848}{0.6428} \approx 76.6$ m.
Now, the width $w$ is the altitude from $C$ to $AB$: $\sin A = \frac{w}{b}$ -> $w = b \times \sin A = 76.6 \times \sin 60^\circ \approx 76.6 \times 0.8660 \approx 66.3$ meters. The river is approximately 66.3 meters wide.
Application 2: Navigation (Bearing Problems)
A ship leaves port $P$ and sails 15 km on a bearing of $040^\circ$ to point $Q$. It then changes course and sails 10 km to point $R$, which is due east of $P$. What was the bearing from $Q$ to $R$?
We need to model this as a triangle. In triangle $PQR$, $PQ=15$ km, $QR=10$ km. Bearing $040^\circ$ means angle $NPQ = 40^\circ$ (where $N$ is North). Since $R$ is due east of $P$, angle $NPR = 90^\circ$. Therefore, angle $QPR = 90^\circ - 40^\circ = 50^\circ$. We also know that $PR$ is east, creating a potential right angle. But we don't know if triangle $PQR$ is right-angled. We know two sides and a non-included angle: $PQ=15$, $QR=10$, and angle $QPR=50^\circ$ (SSA). Let's use the Sine Rule.
Label: Side $q = PR$ (opposite angle $Q$), side $r = PQ = 15$ (opposite angle $R$). We have angle $P=50^\circ$. Write: $\frac{\sin R}{r} = \frac{\sin P}{p}$, where $p = QR = 10$.
So, $\frac{\sin R}{15} = \frac{\sin 50^\circ}{10}$ -> $\sin R = \frac{15 \times \sin 50^\circ}{10} = 1.5 \times 0.7660 \approx 1.149$. Since sine cannot exceed 1, this is impossible. This tells us our initial diagram assumption was wrong. Let's re-analyze: If $R$ is east of $P$, and the ship sailed from $Q$ to $R$, point $R$ must be closer. Let's instead find angle $PQR$. Use the Sine Rule with known sides: $\frac{\sin Q}{q} = \frac{\sin P}{p}$. We don't know $q$. Instead, let's find angle $R$ using the correct labeling: side $r = PQ = 15$ (opposite $R$), side $p = QR = 10$ (opposite $P$). So $\frac{\sin R}{15} = \frac{\sin 50^\circ}{10}$ gave $\sin R >1$, error. This means a triangle with these dimensions cannot be drawn where $R$ is east of $P$ with the given distances. The problem illustrates the "no triangle" outcome of the ambiguous case, showing the Sine Rule's value in diagnosing impossible real-world scenarios.
Important Questions
Q1: Why does the Sine Rule use sines of angles, and not cosines or tangents?
The sine function directly relates an angle in a right triangle to the ratio of the opposite side over the hypotenuse. When we drop an altitude in a non-right triangle, we create two right triangles where this opposite/hypotenuse relationship appears naturally for the shared height. This makes sine the natural choice for the derivation. Cosine and tangent rules exist (the Law of Cosines) but they describe different relationships, more suited for situations involving included angles.
Q2: How do I know whether to use the acute or obtuse angle when solving for an angle with the Sine Rule?
This is the crux of the ambiguous case. First, use the conditions in the table above (comparing side $a$ to $b \sin A$ and side $b$) to predict if one or two triangles exist. If two are possible, you must consider both the acute angle ($\theta$) from your calculator and its supplementary obtuse angle ($180^\circ - \theta$). You then check each possibility: add the known angle and the potential angle. If the sum is less than $180^\circ$, the combination is valid for a triangle. Often, the context of the problem (like a diagram or real-world constraint) will rule out one of the possibilities.
Q3: Can the Sine Rule be used on right-angled triangles?
Yes, absolutely. It works for all triangles, including right-angled ones. For a right-angled triangle with $C = 90^\circ$, $\sin C = \sin 90^\circ = 1$. The rule then becomes $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{1} = c$. This is consistent with standard right-triangle trigonometry, where $\sin A = \frac{a}{c}$, so $a = c \sin A$, which rearranges to $\frac{a}{\sin A} = c$.
The Sine Rule is a cornerstone of trigonometry, extending our problem-solving abilities far beyond the confines of right-angled triangles. Its elegant statement—that the ratio of a side to the sine of its opposite angle is constant for all three sides of any triangle—provides a reliable method for finding unknown lengths and angles. While it requires careful attention, especially in the ambiguous SSA case, mastering it opens doors to practical applications in science, engineering, and everyday logic. By connecting the geometry of triangles with the periodic nature of the sine function, this rule beautifully demonstrates the interconnectedness of mathematical concepts.
Footnote
1 AAS: Angle-Angle-Side. A congruence condition where two angles and a non-included side are known.
2 ASA: Angle-Side-Angle. A congruence condition where two angles and the included side are known.
3 SSA: Side-Side-Angle. A condition where two sides and a non-included angle are known, leading to the ambiguous case.
4 SSS: Side-Side-Side. A congruence condition where all three side lengths are known.
5 SAS: Side-Angle-Side. A congruence condition where two sides and the included angle are known.
6 Cosine Rule (Law of Cosines): A related rule stating $c^2 = a^2 + b^2 - 2ab \cos C$, used primarily for SAS and SSS cases.