The first tool for a triangle with no right angle. Covers the Law of Sines and its proof by dropping an altitude, the two configurations it handles unambiguously, and the Angle-Side-Side case where the same data may describe no triangle, exactly one, or two genuinely different ones — decided by comparing the given opposite side against the altitude c sine alpha. Closes with the area formula in terms of two sides and their included angle.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.2 The Law of Sines, pp. 896-904
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 896-904 — the pages these objectives are drawn from
Warm-up
You can solve any right triangle. This lesson is about the ones with no right angle, and it starts by noticing what breaks.
Discussion prompt
A triangle has angles 40, 60 and 80 degrees, with the side opposite the 40 degree angle measuring 5. Try to find another side using only right-triangle trigonometry. What goes wrong?
Hint: Where would you put the adjacent and opposite sides?
Answer:
Nothing works directly, because there is no right angle and therefore no hypotenuse, no adjacent side and no opposite side in the sense those words were defined. The ratios from Lesson 10.3b simply do not apply.
The way out is to create a right angle by dropping a perpendicular from one vertex to the opposite side. That splits the triangle into two right triangles, both of which the old tools handle — and the shared perpendicular is what links them.
That single idea proves both of the laws in this chapter. Everything else is bookkeeping.
Concept
Drop a perpendicular from a vertex to the opposite side. It is a leg of two right triangles at once, so it can be written in terms of two different angle-side pairs. Setting those two expressions equal gives the Law of Sines.
angle-side opposite pair — An angle of a triangle together with the side opposite it, conventionally labelled with matching Greek and Latin letters. The Law of Sines relates such pairs, and at least one complete pair must be known before it can be used.
\[ \frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b} = \frac{\sin(\gamma)}{c} \]
The equal ratios can be written either way up. The form with the sines on top is easier when solving for an angle; the form with the sides on top is easier when solving for a side.
Figure (svg): An acute triangle with an altitude dropped from one vertex, splitting it into two right triangles which share the altitude, so that the altitude can be expressed two ways and the two expressions equated
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 897-898
Section
Section 1
Concept
In an acute triangle, an altitude from a vertex is a leg of both right triangles it creates. Writing its length from each side and equating gives one instance of the law; a second altitude gives another.
That last point is not a technicality to skim. The fact that an angle and its supplement share a sine makes the proof work in the obtuse case, and it is the very same fact that makes the Angle-Side-Side case ambiguous later.
Figure (svg): An acute triangle with an altitude dropped from one vertex, splitting it into two right triangles which share the altitude, so that the altitude can be expressed two ways and the two expressions equated
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 897-898
Picture it
The law relates pairs, so it needs at least one complete pair before it can be applied.
Figure (svg): Two columns separating the information the Law of Sines can use from the information it cannot
Three sides with no angle, or two sides with the angle between them, supply no complete pair. Those two cases are exactly what the Law of Cosines exists to handle.
Worked example
The whole proof for the acute case, in four lines.
\[ \text{Prove } \frac{\sin(\alpha)}{a} = \frac{\sin(\gamma)}{c} \text{ for an acute triangle.} \]
Drop an altitude from the vertex between a and c
Why: It meets the opposite side and creates two right triangles.
Write h from the first right triangle
Why: The sine of alpha is h over c.
\[ h = c \sin \alpha \]
Write h from the second
Why: The sine of gamma is h over a.
\[ h = a \sin \gamma \]
Equate and rearrange
Why: Divide both sides by a c.
\[ \sin \alpha / a = \sin \gamma / c \]
Figure (svg): The solution to Worked example prove one ratio shown as a ladder of expressions, one row per legal move
\[ c\sin(\alpha) = a\sin(\gamma) \;\Longrightarrow\; \frac{\sin(\alpha)}{a} = \frac{\sin(\gamma)}{c} \]
Verify: test on a known triangle
Why: In a 30-60-90 triangle with hypotenuse 2, the side opposite 30 degrees is 1 and the side opposite 90 degrees is 2. The ratios are sine 30 over 1, which is 0.5, and sine 90 over 2, which is also 0.5. They agree.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 897-897
Sorting
It needs at least one complete angle-side opposite pair.
Sort into buckets
Sort each set of given data.
Worked example
The altitude falls outside the triangle, and one identity rescues the argument.
\[ \text{Explain why the proof survives when } \alpha \text{ is obtuse.} \]
Note where the altitude falls
Why: For an obtuse angle at that vertex the foot lies outside the base.
Identify the angle in the right triangle
Why: It is the supplement of alpha rather than alpha itself.
\[ \alpha' = 180 - \alpha \]
Use the supplement identity
Why: An angle and its supplement have the same sine.
\[ \sin(\alpha') = \sin(\alpha) \]
Conclude
Why: The expression for h is unchanged, so the rest of the proof is too.
\[ h = c \sin \alpha \]
Figure (svg): The solution to Worked example why the obtuse case still works shown as a ladder of expressions, one row per legal move
\[ \sin(180^\circ - \alpha) = \sin(\alpha) \;\Longrightarrow\; h = c\sin(\alpha) \text{ either way} \]
Verify: check the identity
Why: The sine of 150 degrees is one half and the sine of 30 degrees is also one half. The two are supplements and their sines agree, which is exactly what the proof needed and exactly what will cause trouble in the third section.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 898-898
Trap
\[ a = 5, \; b = 7, \; \gamma = 40^\circ \;\Longrightarrow\; \frac{\sin\alpha}{5} = \frac{\sin\beta}{7} \]
Write down the law and hope
Why: Two sides and an angle are given, which feels like enough information.
But every ratio in that equation has an unknown in it. Alpha and beta are both unknown, so the equation relates two unknowns and cannot be solved.
The given data is two sides and the angle between them, which supplies no complete angle-side pair — gamma's opposite side c is not given.
The Law of Sines cannot start here. This is the Side-Angle-Side case and it needs the Law of Cosines, which is the next lesson.
The check to run before writing anything: is there an angle whose opposite side is also known? If not, this law is the wrong tool, and recognising that immediately saves a page of futile algebra.
Prediction
In a triangle, side a is longer than side b.
Predict first
What follows about the angles opposite them?
Correct: Alpha is larger than beta.
Why: The Law of Sines says sine alpha over a equals sine beta over b, so sine alpha equals a over b times sine beta. Since a exceeds b, the sine of alpha exceeds the sine of beta — and within a triangle the larger side is always opposite the larger angle. This is worth carrying as a check on any solved triangle: if the largest side is not opposite the largest angle, something is wrong.
Faded example
A triangle has alpha equal to 40 degrees with a equal to 8, and beta equal to 65 degrees. Find b.
Fill in the blanks
\frac40 deg11.28 = \frac______}} \;\Longrightarrow\; b = \frac______ \approx ___
Why: The known pair is alpha with a, so it goes on one side and the wanted side with its opposite angle on the other. The sine of 65 degrees is about 0.906 and the sine of 40 about 0.643, giving about 11.28. Since beta exceeds alpha, b must exceed a — and 11.28 exceeds 8, which checks.
Socratic
The law is stated both with the sines on top and with the sides on top.
Discussion prompt
Explain when each form is easier, and why they are the same statement.
Hint: What do you want isolated in each case?
Answer:
They are the same statement because taking reciprocals of equal quantities gives equal quantities. Nothing mathematical distinguishes them.
Practically, the sides on top form isolates a side with one multiplication, which is what an AAS or ASA problem wants. The sines on top form isolates a sine with one multiplication, which is what an Angle-Side-Side problem wants when hunting for an angle.
Choosing the form that puts your unknown in a numerator saves a rearrangement and, more importantly, avoids the fraction-inversion errors that a rearrangement invites. Set the equation up so the unknown is already on top.
Section
Section 2
Concept
If two angles are known, the third follows from the angle sum, so the triangle's shape is completely fixed. One side then fixes the size, and exactly one triangle results.
A practical habit worth adopting: always compute from the originally given data rather than from a value you computed a step earlier, so that a rounding error in one answer does not propagate into the next.
Figure (svg): The three configurations of given information that the Law of Sines can use, labelled Angle-Angle-Side, Angle-Side-Angle, and Angle-Side-Side, with the last one flagged as ambiguous
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 901-901
Picture it
Two of them are routine and one is not, and telling them apart takes one glance at what was given.
Figure (svg): The three configurations of given information that the Law of Sines can use, labelled Angle-Angle-Side, Angle-Side-Angle, and Angle-Side-Side, with the last one flagged as ambiguous
The distinguishing question is whether two angles were given. If they were, the case is safe; if only one was, it is the ambiguous case and needs the comparison test.
Worked example
Example 11.2.2, part 1. Alpha is 120 degrees, a is 7, and beta is 45 degrees.
\[ \text{Solve the triangle with } \alpha = 120^\circ, \; a = 7, \; \beta = 45^\circ. \]
Find the third angle
Why: The three angles sum to 180.
\[ \gamma = 15 ^\circ \]
Use the known pair to find b
Why: Set b over sine beta equal to a over sine alpha.
\[ b = 7 \sin 45 / \sin 120 \]
Evaluate exactly then approximate
Why: The result simplifies to seven root six over three.
\[ b = 5.72 \]
Find c the same way, from the ORIGINAL pair
Why: Using alpha and a again rather than the computed b.
\[ c = 7 \sin 15 / \sin 120 = 2.09 \]
Figure (svg): The solution to Worked example an AAS triangle shown as a ladder of expressions, one row per legal move
\[ \gamma = 15^\circ, \quad b = \frac{7\sqrt{6}}{3} \approx 5.72, \quad c \approx 2.09 \]
Verify: check the side ordering
Why: The angles in increasing order are 15, 45 and 120 degrees, and the sides opposite them are 2.09, 5.72 and 7. The largest side is opposite the largest angle throughout, which is the standard check on any solved triangle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 898-899
Matching
Two of the three are decided before any arithmetic.
Match the pairs
Why: AAS and ASA both give two angles, which fixes the shape, and one side, which fixes the size — so each determines a unique triangle. ASS gives only one angle and the outcome depends on a comparison. SSS gives no angle-side pair at all and requires the Law of Cosines.
Worked example
Example 11.2.2, part 2. The given side lies between the two given angles.
\[ \text{Solve the triangle with } \alpha = 85^\circ, \; \beta = 30^\circ, \; c = 5.25. \]
Find the third angle
Why: One hundred eighty minus 85 minus 30.
\[ \gamma = 65 ^\circ \]
Now a complete pair exists
Why: Gamma with c, since c was given and gamma just computed.
Find a from that pair
Why: Set a over sine alpha equal to c over sine gamma.
\[ a = 5.25 \sin 85 / \sin 65 = 5.77 \]
Find b from the same pair
Why: Using the original pair again rather than the computed a.
\[ b = 5.25 \sin 30 / \sin 65 = 2.90 \]
Figure (svg): The solution to Worked example an ASA triangle shown as a ladder of expressions, one row per legal move
\[ \gamma = 65^\circ, \quad a \approx 5.77, \quad b \approx 2.90 \]
Verify: check the ordering again
Why: Angles 30, 65 and 85 degrees with opposite sides 2.90, 5.25 and 5.77. Increasing angles pair with increasing sides, as they must. Note that the given side 5.25 sits between the two computed ones, which is consistent with its angle sitting between the other two.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 899-899
Error analysis
A student solves an AAS triangle and finds the third side from the second.
Annotate
On: \( b \approx 5.72 \;\Longrightarrow\; c = \frac{5.72\sin(15^\circ)}{\sin(45^\circ)} \approx 2.09 \)
The book makes this point twice, and it is worth taking seriously in any multi-step numerical problem. Compute every unknown from the original data, not from your own earlier answers.
Faded example
A triangle has beta equal to 50 degrees, gamma equal to 70 degrees, and a equal to 12. Find b.
Fill in the blanks
\alpha = 180^\circ - 50^\circ - 70^\circ = 60 deg \;\Longrightarrow\; b = \frac10.61___}} \approx ___
Why: The third angle is 60 degrees, which pairs with the given side a. Then b is 12 times the sine of 50 over the sine of 60, about 10.61. Since beta is less than alpha, b should be less than a — and 10.61 is less than 12, confirming the arrangement.
Prediction
You are given two angles of a triangle and one side.
Predict first
How many triangles satisfy the data?
Correct: Exactly one.
Why: Two angles determine the third and therefore the shape of the triangle completely. Any two triangles with those three angles are similar, differing only in size, and the one given side fixes the size. So the triangle is unique. This holds whether the given side is between the two angles or not, which is why AAS and ASA behave identically despite having different names.
Edge cases
Two angles determine the third by the angle sum.
Discussion prompt
What if the two given angles sum to 180 degrees or more? What does that mean for the problem?
Hint: What would the third angle have to be?
Answer:
The third angle would be zero or negative, which no triangle has. So no triangle exists with those two angles, and the problem is inconsistent rather than hard.
Geometrically, two angles summing to 180 degrees would mean the two other sides are parallel and never meet, so no third vertex exists to close the triangle.
This is worth checking first in any AAS or ASA problem, because it costs one addition and identifies an impossible problem before any trigonometry is attempted. Two angles of a triangle must sum to strictly less than 180 degrees, and that is a genuine constraint rather than a formality.
Section
Section 3
Concept
Given an angle, its opposite side, and one other side, the number of possible triangles depends on how the opposite side compares with the altitude — the shortest it could be and still reach.
Theorem 11.3 — With alpha, a and c given: if a is less than h there is no triangle; if a equals h exactly one right triangle; if h is less than a which is less than c there are two triangles; and if a is at least c there is exactly one.
\[ h = c\sin(\alpha) \]
The picture to hold is a compass: fix the angle and the adjacent side, then swing an arc of radius a from the far vertex. How many times that arc meets the base line is how many triangles there are.
Figure (svg): The four possible outcomes in the Angle-Side-Side case, drawn as a fixed angle with a fixed adjacent side and a swinging opposite side of four different lengths
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 901-902
Picture it
One length decides all four cases, and it is computed before anything else.
Figure (svg): The altitude h equals c times the sine of alpha, shown as the perpendicular distance from the far vertex down to the line containing the base
Computing h first turns a case analysis into a single comparison, and it costs one multiplication.
Worked example
Example 11.2.2, parts 3 and 4. The same angle and adjacent side, two different opposite sides.
\[ \text{With } \alpha = 30^\circ \text{ and } c = 4, \text{ analyse } a = 1 \text{ and } a = 2. \]
Compute the altitude
Why: Four times the sine of 30 degrees.
\[ h = 4(0.5) = 2 \]
Compare the first case
Why: a equals 1, which is less than h.
\[ 1 < 2,\text{ too short} \]
Conclude for the first
Why: The side cannot reach the base.
Compare the second case
Why: a equals 2, exactly h.
Figure (svg): The solution to Worked example no triangle, and one right triangle shown as a ladder of expressions, one row per legal move
\[ h = 2: \quad a = 1 < h \;\Rightarrow\; \text{none}; \qquad a = 2 = h \;\Rightarrow\; \text{one, with } \gamma = 90^\circ \]
Verify: check the second with the law
Why: Sine gamma over 4 equals sine 30 over 2, so sine gamma equals 4 times one half over 2, which is 1. The only angle in a triangle with sine 1 is 90 degrees, confirming the right triangle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 899-900
Sorting
In each case alpha is 40 degrees and c is 10, so h is about 6.43.
Sort into buckets
Sort each value of a by how many triangles it admits.
Worked example
Example 11.2.2, part 5. The interesting case, with a between h and c.
\[ \text{With } \alpha = 30^\circ, \; a = 3, \; c = 4, \text{ solve the triangle.} \]
Compare against h and c
Why: The altitude is 2, and a is 3, between 2 and 4.
\[ h < a < c:\text{ two triangles} \]
Find sine gamma from the law
Why: Sine gamma over 4 equals sine 30 over 3.
\[ \sin \gamma = \frac{2}{3} \]
Take both angles with that sine
Why: An acute one and its obtuse supplement.
\[ 41.81 ^\circ\text{ or } 138.19 ^\circ \]
Complete each triangle
Why: The third angle and then the third side, for each.
Figure (svg): The solution to Worked example two triangles shown as a ladder of expressions, one row per legal move
\[ \gamma \approx 41.81^\circ, \; \beta \approx 108.19^\circ, \; b \approx 5.70 \qquad \text{or} \qquad \gamma \approx 138.19^\circ, \; \beta \approx 11.81^\circ, \; b \approx 1.23 \]
Verify: check both are genuine triangles
Why: In the first, the angles sum to 30 plus 108.19 plus 41.81, which is 180. In the second, 30 plus 11.81 plus 138.19 is also 180. Both are valid, and both use exactly the given alpha, a and c — so a complete answer must give both.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 900-900
Trap
\[ \sin(\gamma) = \tfrac{2}{3} \;\Longrightarrow\; \gamma = \arcsin\left(\tfrac{2}{3}\right) \approx 41.81^\circ \]
Apply the arcsine and take the single answer it returns
Why: The arcsine is a function and returns exactly one value, so it looks complete.
But the arcsine's range stops at 90 degrees, so it can never return an obtuse angle. Here the obtuse supplement, about 138.19 degrees, also has sine two thirds and also fits in a triangle with alpha.
\[ \gamma \approx 41.81^\circ \quad\text{or}\quad \gamma \approx 180^\circ - 41.81^\circ = 138.19^\circ \]
Always consider the supplement, then test whether it fits
Why: The supplement fits whenever it plus the given angle is still under 180 degrees.
This is the restriction from Lesson 10.6a causing a genuine problem in an application. The arcsine cannot see the obtuse candidate, so the calculator will never volunteer it and the second triangle has to be looked for deliberately.
Fill the middle
With alpha equal to 25 degrees and c equal to 12, decide the case for a equal to 6.
Fill in the blanks
h = 12\sin 25^\circ \approx 5.07 \;\Longrightarrow\; a = 6 \textgreater than ___ \text___ h
Why: The altitude is about 5.07, and 6 lies between that and c equal to 12. So the case is h less than a less than c, which gives two triangles. Computing h first is what makes this a one-line decision rather than a hunt through the algebra.
Prediction
In an Angle-Side-Side problem, the given angle is obtuse.
Predict first
How many triangles can there be?
Correct: At most one.
Why: The two-triangle case requires both a candidate angle and its obtuse supplement to fit alongside the given angle. If the given angle is already obtuse, no second obtuse angle can join it, since two obtuse angles exceed 180 degrees. So the supplement is always rejected and at most one triangle results — and none at all if the given side is too short. The book invites the reader to see this before reading on, and it is worth noticing.
Explain it to yourself
The four cases are usually presented as a list to memorise.
Discussion prompt
Explain them instead using a compass, and say which single quantity the whole classification turns on.
Hint: Fix the angle and the adjacent side, then swing an arc.
Answer:
Draw the angle and mark off the adjacent side c, giving one fixed vertex. Now put a compass point there and swing an arc of radius a. Every place that arc meets the base line is a possible third vertex, so the number of triangles is the number of intersections.
If a is shorter than the perpendicular distance h, the arc never reaches the line — none. If it equals h, the arc just touches — one, and it touches perpendicularly, hence the right angle. If a is between h and c, the arc cuts the line twice on the correct side — two. And if a is at least c, one of those two intersections falls on the wrong side of the original vertex and is discarded — one.
The whole classification turns on h equals c sine alpha, and once that number is computed the case is a single comparison rather than four rules.
Section
Section 4
Concept
The classic use of the Law of Sines is finding an inaccessible distance from two angle measurements and one measured baseline — the same structure as the two-sighting problems of Lesson 10.3b, now without any right angle required.
Because the case is always ASA, this method never runs into the ambiguity. That is one practical reason surveyors measure two angles and one length rather than two lengths and one angle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 902-903
Picture it
When the ambiguous case does arise, the two answers can be very different, and reporting only one would be badly wrong.
Figure (svg): The two triangles arising from the same Angle-Side-Side data, one with an acute gamma and one with its obtuse supplement, drawn side by side
The third sides here are 5.70 and 1.23 — a factor of more than four apart. In a navigation problem that is the difference between a safe course and a grounding.
Worked example
Example 11.2.3. Two sightings 5 miles apart along a shoreline.
\[ \text{From } P \text{ the island bears } 30^\circ \text{ from the shoreline; from } Q, \; 5 \text{ miles on, it bears } 45^\circ \text{ back. Find the distance from } Q. \]
Identify the triangle and its angles
Why: The angle at Q inside the triangle is 180 minus 45, and the third angle follows.
Find the third angle
Why: One hundred eighty minus the two known angles.
\[ \gamma = 15 ^\circ \]
Apply the Law of Sines
Why: The baseline pairs with the angle opposite it.
\[ d = 5 \sin 30 / \sin 15 \]
Evaluate
Why: The sine of 15 degrees is about 0.2588.
\[ d = 9.66\text{ miles} \]
Figure (svg): The solution to Worked example distance to an island shown as a ladder of expressions, one row per legal move
\[ d = \frac{5\sin(30^\circ)}{\sin(15^\circ)} \approx 9.66 \text{ miles} \]
Verify: sanity-check the magnitude
Why: The two bearings differ by only 15 degrees, which means the island is far away relative to the 5 mile baseline — a small angular difference implies a large distance. Nearly ten miles for a five mile baseline is consistent with that.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 902-903
Prediction
A surveyor uses a 100 metre baseline and measures the two angles to a distant target as 89 and 89.5 degrees.
Predict first
What does that imply about the target's distance?
Correct: It is very far away.
Why: The two angles sum to 178.5 degrees, leaving only 1.5 degrees for the angle at the target. A tiny angle at the far vertex means the two sight lines are nearly parallel, which means the target is at a great distance — about 3800 metres here. This is the fundamental limitation of the method: as the target gets further, the third angle shrinks and small errors in the measured angles produce large errors in the distance.
Worked example
Example 11.2.3 continued. Once one distance is known, a right triangle finishes it.
\[ \text{With } d \approx 9.66 \text{ miles at } 45^\circ \text{ to the shoreline, find the island's distance from the coast.} \]
Identify the right triangle
Why: The perpendicular from the island to the shoreline creates one.
Choose the relation
Why: The perpendicular distance is opposite the 45 degree angle, and d is the hypotenuse.
\[ \sin 45 = \frac{y}{d} \]
Solve
Why: Multiply by d.
\[ y = 9.66 \sin 45 \]
Evaluate
Why: The sine of 45 degrees is about 0.7071.
\[ y = 6.83\text{ miles} \]
Figure (svg): The solution to Worked example the perpendicular distance to shore shown as a ladder of expressions, one row per legal move
\[ y = d\sin(45^\circ) \approx 9.66\left(\frac{\sqrt{2}}{2}\right) \approx 6.83 \text{ miles} \]
Verify: check with the other leg
Why: The remaining angle is also 45 degrees, so the triangle is isosceles and the along-shore distance equals the perpendicular one, about 6.83 miles. Checking with Pythagoras, 6.83 squared twice is about 93.3, and 9.66 squared is about 93.3 as well.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 903-903
Error analysis
A surveyor sights a target at 45 degrees from the baseline, measured on the far side.
Annotate
On: \( \text{interior angle at } Q = 45^\circ \)
Bearings and interior angles are frequently supplements of one another, and which you have depends on which way the measurement was taken. Drawing the triangle and checking that the three angles sum to 180 catches this every time.
Faded example
A 60 m baseline has angles of 72 degrees and 63 degrees to a target from its two ends.
Fill in the blanks
\gamma = 180^\circ - 72^\circ - 63^\circ = 45 deg \;\Longrightarrow\; d = \frac75.6___}} \approx ___ \text___
Why: The third angle is 45 degrees, opposite the baseline. The distance from the first endpoint is 60 times the sine of 63 over the sine of 45, about 75.6 metres. Since 63 degrees exceeds 45, that distance exceeds the baseline, which the answer confirms.
Sorting
Some field methods can land in the ambiguous case and some cannot.
Sort into buckets
Sort each scheme.
Real world
A ship's radar reports a contact at a bearing, and ten minutes later reports it again at a different bearing, with the ship having travelled 3 nautical miles in between on a steady course.
Discussion prompt
Explain how the two bearings and the run give the contact's distance, and identify which case this is and why that matters.
Hint: What are the three pieces of information, and what shape do they form?
Answer:
The 3 mile run is a baseline and the two bearings give the angles at each end of it, so the third angle follows from the angle sum. This is ASA, which always determines exactly one triangle.
The Law of Sines then gives the distance from the current position to the contact directly, as the run times the sine of the first angle over the sine of the third.
The case matters enormously. ASA is unambiguous, so the answer is unique and the navigator can act on it. Had the method produced Angle-Side-Side data instead — one bearing and two ranges, say — there could have been two possible positions for the contact, and acting on the wrong one is exactly the kind of error that causes collisions. This is one concrete reason the running fix uses two bearings and a known run rather than any other combination.
Section
Section 5
Concept
The familiar area formula needs a perpendicular height, which is rarely measured directly. Trigonometry supplies it: the height is a side times the sine of the angle at the base, giving a formula in two sides and their included angle.
Theorem 11.4 — The area enclosed by a triangle is half the product of any two sides with the sine of the angle between them.
\[ A = \tfrac{1}{2}bc\sin(\alpha) = \tfrac{1}{2}ac\sin(\beta) = \tfrac{1}{2}ab\sin(\gamma) \]
The proof uses exactly the same altitude picture as the Law of Sines, which is why the two theorems sit together.
Figure (svg): The area formula for a triangle in terms of two sides and the included angle, shown with the altitude that derives it
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 903-903
Picture it
One altitude again, this time substituted into a formula rather than equated with another.
Figure (svg): The area formula for a triangle in terms of two sides and the included angle, shown with the altitude that derives it
Notice the angle must be the one between the two sides used. Choosing an angle not between them gives a number that is not the area.
Worked example
Example 11.2.4. The triangle from the first worked example, using the most given data.
\[ \text{Find the area of the triangle with } \alpha = 120^\circ, \; a = 7, \; \beta = 45^\circ. \]
Choose the version using the most given data
Why: Beta and a were given, and c was computed; the version with a, c and beta uses two of the three.
\[ A = (\frac{1}{2}) a c \sin \beta \]
Recall the computed side
Why: c is seven sine 15 over sine 120.
\[ c = 2.09 \]
Substitute
Why: Half of 7 times c times the sine of 45.
\[ A = 0.5(7) (2.09) (0.7071) \]
Evaluate
Why: About 5.18 square units.
\[ A = 5.18 \]
Figure (svg): The solution to Worked example the area of a solved triangle shown as a ladder of expressions, one row per legal move
\[ A = \tfrac{1}{2}(7)\left(\frac{7\sin 15^\circ}{\sin 120^\circ}\right)\sin(45^\circ) \approx 5.18 \]
Verify: cross-check with another version
Why: Using a and b with gamma: half of 7 times 5.72 times the sine of 15 degrees is about 5.18 as well. All three versions must agree, and computing two of them is a genuine check on the arithmetic.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 903-903
Matching
The angle used is always the one between the two sides.
Match the pairs
Why: In each case the angle is the one whose opposite side is missing from the product. Sides b and c meet at the vertex where alpha sits, so alpha is between them; and a, which is opposite alpha, is the side not used. The pattern is completely regular and is the fastest way to remember which angle goes with which pair.
Worked example
The formula's natural use, needing no solving at all.
\[ \text{A triangle has sides } 9 \text{ and } 14 \text{ with an angle of } 62^\circ \text{ between them. Find its area.} \]
Check the angle is between the two sides
Why: It is, so the formula applies directly.
Substitute
Why: Half times 9 times 14 times the sine of 62 degrees.
\[ A = 0.5(9) (14) \sin 62 \]
Evaluate the sine
Why: The sine of 62 degrees is about 0.8829.
\[ \sin 62 = 0.8829 \]
Compute
Why: Sixty-three times 0.8829.
\[ A = 55.6 \]
Figure (svg): The solution to Worked example area from two sides and the included angle shown as a ladder of expressions, one row per legal move
\[ A = \tfrac{1}{2}(9)(14)\sin(62^\circ) \approx 55.6 \]
Verify: compare against the maximum possible
Why: With sides 9 and 14 the largest possible area is half of 126, which is 63, achieved when the angle is 90 degrees. The answer 55.6 is below that and close to it, which is right for an angle of 62 degrees whose sine is 0.88.
Trap
\[ a = 9, \; b = 14, \; \alpha = 62^\circ \;\Longrightarrow\; A = \tfrac{1}{2}(9)(14)\sin(62^\circ) \]
Substitute the two given sides and the given angle
Why: All three numbers are present and the formula has three slots.
But alpha is opposite side a, not between a and b. The formula requires the angle between the two sides used, which here would be gamma.
\[ A = \tfrac{1}{2}ab\sin(\gamma) \quad \text{with } \gamma \text{ between } a \text{ and } b \]
Match the angle to the two sides it lies between
Why: The subscript pattern is fixed: the angle omitted from the product is the one used in the sine.
The pattern is worth stating explicitly: the sine uses the angle whose opposite side does not appear. In half b c sine alpha the sides are b and c and the angle is alpha, whose opposite side a is the one missing. Checking that one thing prevents the error entirely.
Faded example
A triangle has sides 11 and 6 with an included angle of 150 degrees.
Fill in the blanks
A = \tfrac1/216.5(11)(6)\sin(150^\circ) = 33 \cdot ___ = ___
Why: The sine of 150 degrees is one half, since 150 degrees has reference angle 30 and lies in quadrant two where the sine is positive. So the area is 33 times one half, namely 16.5. Note that an obtuse included angle gives a smaller area than a right angle would, since its sine is smaller — the area peaks at 90 degrees.
Prediction
Two sides of a triangle have fixed lengths, and the angle between them can vary.
Predict first
For what angle is the area largest?
Correct: 90 degrees.
Why: The area is half the product of the two sides times the sine of the included angle, and the sides are fixed, so the area is largest when the sine is largest — which happens at 90 degrees, where the sine equals 1. Making the angle either smaller or larger reduces the sine and therefore the area. This is why a right angle maximises the area enclosed by two fixed sides, and it is a useful design fact well beyond trigonometry.
Counterexample
A student proposes: knowing all three angles of a triangle determines its area.
Discussion prompt
Give two triangles that refute this, and say what does determine the area.
Hint: What do two triangles with the same three angles have in common?
Answer:
Take a 30-60-90 triangle with hypotenuse 2 and another with hypotenuse 20. Their areas are about 0.87 and about 87 — a factor of a hundred apart, with identical angles.
Three angles determine the shape but not the size, since all such triangles are similar. Area is a size quantity and scales with the square of any linear dimension, so it cannot be determined by shape alone.
What does determine the area is two sides and the angle between them, or equivalently any data that pins down the triangle completely. This is the same reason AAA is not a congruence criterion in geometry, and the area formula's need for two side lengths reflects it exactly.
Comparison
Fill the blanks from memory. Everything is compared against h equal to c sine alpha.
Comparison matrix
| Condition on a | Number of triangles | Why |
|---|---|---|
| a < h | none | the side is too short to reach the base |
| a = h | exactly one, right-angled | the arc just touches, perpendicularly |
| h < a < c | two | the arc cuts the base twice |
| a at least c | one | the second crossing falls on the wrong side |
Compute h first and the four rows become a single comparison. Without it, the case analysis has to be reconstructed from the algebra every time.
Pattern
Whatever the triangle problem, the same five moves cover it.
For an area, use the version of the formula that involves the most originally given quantities, and make sure the angle used lies between the two sides used.
OpenStax Algebra and Trigonometry 2e, §10.1 Non-right Triangles: Law of Sines §10.1
Check
Which case is it?
Check your understanding
A triangle has a = 10, b = 14 and the angle between them equal to 50 degrees. Can the Law of Sines start?
Answer: C
Why: The given angle lies between the two given sides, so it is not opposite either of them. Its opposite side is the third one, which was not given. With no complete pair every ratio in the law contains an unknown, and this Side-Angle-Side case needs the Law of Cosines.
Check
The ambiguous case. Compute h first.
Check your understanding
With alpha equal to 30 degrees, c equal to 8 and a equal to 5, how many triangles are there?
Answer: C
Why: The altitude is h equal to 8 times the sine of 30 degrees, which is 4. Since 5 lies strictly between 4 and 8, the case is h less than a less than c and there are two triangles. Both an acute gamma and its obtuse supplement fit alongside the 30 degree angle.
Check
The area formula. Match the angle to the sides.
Check your understanding
A triangle has b = 12, c = 7 and alpha = 40 degrees. What is its area?
Answer: A
Why: Alpha lies between sides b and c, so the formula applies directly: half of 12 times 7 times the sine of 40 degrees. That is 42 times 0.6428, about 27.0 square units.
Real world
An aircraft is 200 km from an airport on a bearing that puts it at 35 degrees from the runway's extended centreline. Air traffic control knows the aircraft's distance to a second beacon is 150 km, and needs its distance from the airport along the centreline.
Discussion prompt
Identify which case this is, determine how many positions are consistent with the data, and say what a controller should do about it.
Hint: One angle and two sides, and check whether the angle lies between them.
Answer:
This is Angle-Side-Side: an angle of 35 degrees, the side opposite it of 150 km, and an adjacent side of 200 km. The altitude is 200 times the sine of 35 degrees, about 114.7 km.
Since 150 lies strictly between 114.7 and 200, the case is h less than a less than c and two positions are consistent with the data. The two solutions place the aircraft at substantially different distances along the centreline.
What a controller should do is obtain one more observation — a second bearing, an altitude, or a radar range from a different station — rather than pick one of the two. This is exactly why air traffic surveillance uses multiple independent sources: a single range-and-bearing measurement of this shape does not determine a position uniquely, and the ambiguity is a property of the geometry rather than of the equipment.
It is also why the phrase ambiguous case is not academic. The mathematics is telling you the measurement is insufficient, and the right response is more data, not a guess.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
You solve for an angle using the Law of Sines and your calculator returns 38 degrees. What should you do next?
Correct: Also consider 142 degrees and test whether it fits.
\[ \sin(38^\circ) = \sin(142^\circ) \approx 0.6157 \]
Why: The arcsine's range stops at 90 degrees, so it can never return an obtuse angle — but an angle and its supplement share a sine, and 142 degrees has the same sine as 38. Whether the supplement is admissible depends on whether it fits alongside the given angle, which is tested by checking that the two sum to less than 180 degrees. The complement is irrelevant here, since complementary angles do not share a sine, and a negative angle cannot occur in a triangle.
Explain it
They have solved an Angle-Side-Side triangle, got one answer from their calculator, and are marked wrong for missing a second triangle. They think the marking is unfair.
Discussion prompt
In no more than five sentences, explain why two answers were expected and how they could have known.
Hint: What does the arcsine refuse to tell them?
Answer:
A usable answer: the calculator's inverse sine only ever gives an angle under 90 degrees, because that is the range it was defined with. But an obtuse angle has exactly the same sine as its supplement, so whenever the calculator says 38 degrees, 142 degrees is an equally valid solution of the same equation — the calculator simply cannot report it.
How to know in advance: work out h, which is the other given side times the sine of the given angle. If the side opposite your angle is longer than h but shorter than the other given side, there are two triangles and you must find both. That check takes one multiplication and settles it before you start.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Whether it can start is fixed by looking for a matched angle-side opposite pair before writing anything. The count is fixed by computing h and comparing. Producing both triangles is fixed by always taking the supplement of an arcsine answer and testing whether it fits. The area angle is fixed by the pattern that the sine uses the angle whose opposite side is missing from the product. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw an acute triangle with an altitude dropped from one vertex, and write out the four-line proof of the Law of Sines beside it, stating the law in both of its forms. Underneath, write the three cases AAS, ASA and ASS with a one-line note on how many triangles each gives, and circle the ambiguous one. In the middle of the page draw the compass picture for the ambiguous case: a fixed angle, a fixed adjacent side, and four arcs of different radii showing none, one, two and one triangle, labelling the condition under each. Write the formula for h beside it. In the bottom left, solve completely the case alpha equal to 30 degrees, a equal to 3, c equal to 4, producing both triangles. In the bottom right, draw a triangle and write all three versions of the area formula, marking on the picture which angle goes with which pair of sides. Finally, circle the single quantity that the whole ambiguous-case classification depends on.
The circled quantity is h, the altitude c sine alpha. Computing it first turns four memorised rules into one comparison, and it is the only new idea the ambiguous case actually contains.
Recap
Five things, and the third is where the marks are usually lost.
| If the question says | Your first move is |
|---|---|
| Two angles are given | Find the third; the triangle is unique |
| One angle and two sides | Compute h = c sin alpha and compare |
| Your calculator gave an angle | Test the supplement as well |
| Three sides, or two sides and the angle between | This law cannot start; use the Law of Cosines |
| Find the area | Use the angle between the two sides you have |
Two configurations remain unsolvable: three sides with no angle, and two sides with the angle between them. Neither supplies an angle-side pair, so neither can start the Law of Sines — and the next lesson supplies the law that handles both.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.2 The Law of Sines §11.2, pp. 896-904 — everything on these slides traces back here
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