How To Measure A Peak You Can't Reach
Two surveyors at either end of a baseline can pin a distant peak without walking to it. Each measures the angle from their own position to the peak; the baseline between them is known. That is all the law of sines needs to compute how far the peak stands from either surveyor — a triangle solved from two angles and one side.
The law of sines connects each side of any triangle to the sine of the angle opposite it. It is the natural partner to the law of cosines: where the cosine law handles sides and enclosed angles, the sine law handles matched side-and-opposite-angle pairs.
What Is The Law Of Sines?
The law of sines is a proportion that holds for every triangle. For sides $a$, $b$, $c$ opposite angles $\angle A$, $\angle B$, $\angle C$:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
The single most useful reading of it: the ratio of any side to the sine of its opposite angle is the same for all three pairs in the triangle. The sine here is the ordinary sine from the trigonometric ratios, evaluated at each angle. Because the law holds for every triangle, it applies across all types of triangles - acute, right, and obtuse.
What Is That Common Ratio Equal To?
The shared ratio is not just any number. It equals the diameter of the triangle's circumscribed circle:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$
where $R$ is the circumradius - the radius of the unique circle passing through all three vertices. That single fact turns the law of sines into a way to find the circumradius from one side and its opposite angle, a link most competitor explainers leave out.
How Do You Prove The Law Of Sines?
The altitude proof is the shortest. Drop a perpendicular of height $h$ from vertex $C$ to side $c$, splitting the triangle into two right triangles. In each, the height relates to a side and a sine:
$$\sin A = \frac{h}{b} \quad\Rightarrow\quad h = b\sin A$$
$$\sin B = \frac{h}{a} \quad\Rightarrow\quad h = a\sin B$$
Both expressions equal the same height $h$, so set them equal:
$$b\sin A = a\sin B$$
Divide both sides by $\sin A \sin B$:
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Repeating the argument with an altitude from a different vertex brings in the third ratio $\dfrac{c}{\sin C}$. The whole proof rests on the sine definition inside a right-angled triangle, applied twice.
When Do You Use The Law Of Sines?
The law of sines fits the triangle cases where a side is paired with its opposite angle.
AAS / ASA - two angles and a side. Once two angles are known, the third follows from the $180^\circ$ sum, and every side can be found from the one given side.
SSA - two sides and a non-included angle. You know two sides and an angle opposite one of them, and you want the other angle. This case can misbehave - it is the ambiguous case below.
If you instead know two sides and the angle between them (SAS), or all three sides (SSS), the law of sines has no matched pair to start from - switch to the law of cosines.
What Is The Ambiguous Case Of The Law Of Sines?
The SSA case can produce two valid triangles, one, or none - which is why it is called ambiguous. The cause is simple: an angle and its supplement have the same sine, since $\sin\theta = \sin(180^\circ - \theta)$. So when you solve for an angle with the sine law, both an acute and an obtuse value may satisfy the equation.
The test after finding a first angle $\angle B$ is to check its supplement $180^\circ - \angle B$:
Two triangles if the supplement, added to the known angle, still totals less than $180^\circ$.
One triangle if the supplement pushes the angle sum to $180^\circ$ or beyond, ruling the second solution out.
No triangle if the sine value comes out greater than $1$, an impossible sine.
The supplement idea leans on the same reference angle logic that governs sines in the second quadrant.
Examples Of The Law Of Sines
Example 1
In a triangle, $\angle A = 40^\circ$, $\angle B = 75^\circ$, and side $a = 10$. Find side $b$.
$$\frac{a}{\sin A} = \frac{b}{\sin B} \quad\Rightarrow\quad b = \frac{a\sin B}{\sin A} = \frac{10\sin 75^\circ}{\sin 40^\circ}$$
$$b = \frac{10(0.9659)}{0.6428} \approx 15.03$$
Final answer: $b \approx 15.0$.
Example 2
A triangle has $a = 7$, $b = 9$, and $\angle A = 35^\circ$. Find $\angle B$. First instinct, then the correct route.
Set up the sine law and solve for $\sin B$:
$$\sin B = \frac{b\sin A}{a} = \frac{9\sin 35^\circ}{7} = \frac{9(0.5736)}{7} \approx 0.7375$$
The rushed move is to take the calculator's first output and stop: $\angle B = \sin^{-1}(0.7375) \approx 47.5^\circ$, done.
That misses half the answer. Because $\sin B = \sin(180^\circ - B)$, the supplement $180^\circ - 47.5^\circ = 132.5^\circ$ is also a candidate. Check whether it fits: $35^\circ + 132.5^\circ = 167.5^\circ$, which is still under $180^\circ$, so a second triangle genuinely exists.
Both are valid:
$\angle B \approx 47.5^\circ$, giving $\angle C \approx 97.5^\circ$
$\angle B \approx 132.5^\circ$, giving $\angle C \approx 12.5^\circ$
Final answer: two triangles: $\angle B \approx 47.5^\circ$ or $\angle B \approx 132.5^\circ$.
Example 3
In a triangle, $\angle A = 50^\circ$, $\angle C = 60^\circ$, and the included side $b = 8$. Find side $a$.
First, $\angle B = 180^\circ - 50^\circ - 60^\circ = 70^\circ$. Then:
$$a = \frac{b\sin A}{\sin B} = \frac{8\sin 50^\circ}{\sin 70^\circ} = \frac{8(0.766)}{0.9397} \approx 6.52$$
Final answer: $a \approx 6.52$.
Example 4
A triangle has side $a = 10$ opposite $\angle A = 30^\circ$. Find the circumradius $R$.
$$2R = \frac{a}{\sin A} = \frac{10}{\sin 30^\circ} = \frac{10}{0.5} = 20$$
$$R = 10$$
Final answer: the circumradius is $R = 10$.
Example 5
Two observers $100$ m apart both sight a landmark. From one end the angle to the landmark is $65^\circ$; from the other it is $75^\circ$. How far is the landmark from the first observer?
The third angle is $180^\circ - 65^\circ - 75^\circ = 40^\circ$, opposite the $100$ m baseline. The distance $d$ is opposite the $75^\circ$ angle:
$$d = \frac{100\sin 75^\circ}{\sin 40^\circ} = \frac{100(0.9659)}{0.6428} \approx 150.3 \text{ m}$$
Final answer: about $150.3$ m — a triangulation solved without reaching the landmark.
Example 6
A triangle has $a = 5$, $b = 12$, and $\angle A = 40^\circ$. Find $\angle B$.
$$\sin B = \frac{b\sin A}{a} = \frac{12\sin 40^\circ}{5} = \frac{12(0.6428)}{5} \approx 1.543$$
A sine cannot exceed $1$, so no angle $\angle B$ exists.
Final answer: no triangle is possible with these measurements.
Where Does The Law Of Sines Earn Its Keep? - "Fixing a point from two angles"
The law of sines exists to solve a triangle you can only partly measure - specifically, when you have angles and one matched side. That is the exact situation of anyone measuring across a distance they cannot cross.
Surveying and triangulation. A known baseline plus two sighting angles fixes any distant point, which is how land is mapped and how early positioning systems worked.
Astronomy and navigation. Angles to the same star from two locations, or from one location at two times, give distances by the same proportion.
Heights and inaccessible lengths. An angle of elevation taken from two spots turns into the height of a tower or cliff through a single sine-law triangle.
Why does the ratio equal $2R$ and not just some triangle-specific constant? Because every side of the triangle is a chord of its circumscribed circle, and a chord's length is the diameter times the sine of its inscribed angle. The proportion is really a statement about that circle, documented at Wolfram MathWorld.
The Mathematicians Behind The Law Of Sines
Naṣīr al-Dīn al-Ṭūsī (1201–1274, Persia) gave the first systematic statement and proof of the plane law of sines, treating trigonometry as a subject in its own right rather than a tool of astronomy; his biography is at MacTutor. Earlier, Brahmagupta (598–668, India) worked with equivalent chord-and-circumradius relationships that anticipated the circumscribed-circle form.
Common Mistakes With The Law Of Sines
Mistake 1: Ignoring The Second Angle In The Ambiguous Case
Where it slips in: Any SSA problem where you solve for an angle using the sine law.
Don't do this: Reporting only the acute angle the calculator returns.
The correct way: Test the supplement $180^\circ - \angle B$. If it still leaves the angle sum under $180^\circ$, a second triangle exists and both answers count. The rusher who trusts a single calculator readout loses the second triangle here - the most flagged error in the whole topic.
Mistake 2: Pairing A Side With The Wrong Angle
Where it slips in: Setting up the proportion when the diagram is not labelled cleanly.
Don't do this: Writing $\dfrac{a}{\sin B}$, matching a side with an angle it is not opposite.
The correct way: Each side pairs with the angle directly across from it: $a$ with $\angle A$, $b$ with $\angle B$, $c$ with $\angle C$. The second-guesser who re-labels the triangle halfway through often crosses these pairs and gets a plausible but wrong ratio.
Mistake 3: Reaching For The Sine Law In An SAS Or SSS Case
Where it slips in: Choosing a method before checking what is actually given.
Don't do this: Applying the law of sines when you have two sides and the angle between them.
The correct way: The sine law needs a side paired with its opposite angle to start. With SAS or SSS you have no such pair - use the law of cosines instead. Choosing the wrong law is the triangulation equivalent of a surveyor using the wrong instrument for the sighting: the numbers come out, but they describe a triangle the data never specified, and the position error rides along unnoticed.
Key Takeaways
The law of sines is $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$, valid for every triangle.
The common ratio equals $2R$, twice the circumradius.
Use it for AAS, ASA, and SSA; use the law of cosines for SAS and SSS.
The SSA case is ambiguous — always test the supplementary angle for a second triangle.
A required sine greater than $1$ means no triangle fits the data.
To work through the law of sines with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for triangle-solving drills, or live math tutoring with peers from 20+ countries.
A Practical Next Step
Practice these to solidify your understanding: solve the SSA triangle $a = 8$, $b = 11$, $\angle A = 30^\circ$ and decide how many triangles fit, then find a circumradius from any side and its opposite angle. If the ambiguous case trips you up, come back to the ambiguous-case section above. Want a live Bhanzu trainer to solve triangles with you? Book a free demo class.
Read More
Cosine formula — the companion law for SAS and SSS triangles.
Sin cos tan — the three primary ratios the sine law is built on.
Similar triangles — the proportional-reasoning ideas behind triangle-solving.
Angle of depression — a common source of the angles used in sine-law problems.
Trigonometric table — the standard-angle sine values used in these examples.
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