Extends trigonometry to triangles with no right angle. States the law of sines, applies it to the AAS and ASA cases, works carefully through the ambiguous SSA case where two triangles may satisfy the data, and computes areas from two sides and their included angle.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 8 — Further Applications of Trigonometry
§8.1 Non-right Triangles: Law of Sines, pp. 904-922
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 904-922 — the pages these objectives are drawn from
Warm-up
Section 5.4's ratios all assumed a right angle. Most triangles do not have one.
Discussion prompt
Why can the sine, cosine and tangent ratios not be used directly on a triangle with no right angle?
Hint: What did opposite over hypotenuse require?
Answer:
Those ratios were defined using a hypotenuse, and only a right triangle has one. Without a right angle there is no hypotenuse to divide by.
So the definitions do not apply as they stand. Something more general is needed for a triangle with three arbitrary angles.
This chapter supplies two such tools. The law of sines relates each angle to its opposite side, and it works for any triangle at all — including right ones, which turn out to be a special case.
Concept
In any triangle, the ratio of an angle's sine to the length of the side opposite it is the same for all three angles.
\[ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \]
The proportionality makes intuitive sense: a bigger angle opens onto a longer opposite side. The law makes that precise and turns it into a computational tool, since knowing one complete pair fixes the common ratio.
Figure (svg): A card stating the law of sines as three equal ratios of a sine to its opposite side
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 904-909
Section
Section 1
Concept
The whole chapter depends on a labelling convention: each vertex takes a capital letter and the side opposite it takes the same letter in lower case.
Mislabelling is the commonest cause of a wrong answer here, and it produces a plausible-looking number rather than an obvious error. Drawing the triangle and marking the pairs before writing any equation is worth the thirty seconds.
Figure (svg): A general triangle with vertices labelled with capital letters and each opposite side labelled with the matching lower-case letter
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 904-910
Picture it
Each lower-case side faces its capital angle across the triangle.
Figure (svg): A general triangle with vertices labelled with capital letters and each opposite side labelled with the matching lower-case letter
The pairing is what the law uses. A side and the angle at one of its endpoints are not a pair, and substituting them gives a wrong answer with no warning.
Worked example
One complete pair fixes the ratio.
\[ \text{In a triangle, } A=40^\circ, \; B=75^\circ, \; a=12. \text{ Find } b. \]
Identify the known pair
Why: A and its opposite side a.
\[ 40 ^\circ\text{ with } 12 \]
Write the law for A and B
Why: Two of the three ratios.
\[ \sin A / a = \sin B / b \]
Solve for b
Why: Cross-multiply.
\[ b = a \sin B / \sin A \]
Compute
Why: Substitute and evaluate.
\[ 12(0.966) / (0.643) \]
Figure (svg): A card stating the law of sines as three equal ratios of a sine to its opposite side
\[ b\approx 18.0 \]
Verify: check the ordering
Why: Angle B is larger than angle A, so its opposite side must be longer — and 18 is longer than 12. That ordering check catches an inverted ratio immediately, which is the commonest slip in applying the law.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 905-908
Prediction
A triangle has angles of 30, 60 and 90 degrees.
Predict first
Which side is longest?
Correct: The one opposite the 90 degree angle.
Why: The law of sines makes each side proportional to the sine of its opposite angle, and among angles in a triangle a larger angle has a larger sine. So the largest angle always faces the longest side, which is a useful check on any computed answer.
Worked example
Find the third angle first, then two sides.
\[ \text{Solve the triangle with } A=35^\circ, \; C=80^\circ, \; a=9. \]
Find the third angle
Why: The three sum to a straight angle.
\[ B = 65 ^\circ \]
Set up the ratio
Why: Using the known pair.
\[ \sin 35 / 9 \]
Find b
Why: Using angle B.
\[ b = 9 \sin 65 / \sin 35 = 14.2 \]
Find c
Why: Using angle C.
\[ c = 9 \sin 80 / \sin 35 = 15.5 \]
Figure (svg): The solution to Worked example solve an AAS triangle completely shown as a ladder of expressions, one row per legal move
\[ B=65^\circ,\; b\approx 14.2,\; c\approx 15.5 \]
Verify: check the ordering of all three
Why: The angles from smallest to largest are 35, 65 and 80, and the opposite sides are 9, 14.2 and 15.5 — in the same order. That correspondence must always hold, and checking it verifies every computed side at once.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 908-911
Trap
\[ \frac{\sin A}{b}=\frac{\sin B}{a} \]
Pair each angle with a side touching it
Why: The sides are matched to angles by proximity in the drawing rather than by being opposite.
The equation is wrong, and it produces a plausible number with no error signal.
Each sine goes over its opposite side, the one across the triangle from that vertex.
An adjacent side touches the angle; the opposite one does not. The law uses only opposite pairs.
Mark the pairs on a drawing before writing the equation. The error is silent, so a structural check beforehand is more reliable than spotting it afterwards.
Faded example
Solving for side c.
Fill in the blanks
\fracac}=\frac______} \;\Longrightarrow\; c=\frac______
Why: Each sine sits over the side opposite it, so A pairs with a and C pairs with c. Cross-multiplying then isolates the unknown side in one step.
Sorting
Each angle pairs only with the side opposite it.
Sort into buckets
Sort each pairing.
Step zero
You are given a triangle to solve.
Discussion prompt
What do you do before writing any equation?
Hint: What does the law require to be available?
Answer:
Draw the triangle and label it, marking which sides are opposite which angles. Thirty seconds of drawing prevents the silent pairing error.
Then check whether a complete side-angle pair is known. If one is, the law of sines applies; if not, it does not and the law of cosines is needed.
That single check determines the whole method. Reaching for the law of sines when no pair is known produces an equation with two unknowns, which is where a lot of wasted work comes from.
Section
Section 2
Concept
The law of sines needs one ratio to be computable, which means knowing a side and the angle opposite it. Three of the standard triangle cases supply that and two do not.
In the ASA case the given side is between the two known angles, so it is not opposite either of them. Finding the third angle first creates a complete pair, after which the law applies normally.
| case | what is known | law of sines? |
|---|---|---|
| AAS | two angles and a non-included side | yes |
| ASA | two angles and the included side | yes, after finding the third angle |
| SSA | two sides and a non-included angle | yes, but ambiguous |
| SAS | two sides and the included angle | no |
| SSS | three sides | no |
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 909-914
Picture it
The dividing line is whether a complete pair is available.
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
The two right-hand cases have no side opposite a known angle, so no ratio can be computed. Section 8.2 supplies the tool for them.
Worked example
The third angle creates the pair.
\[ \text{Solve the triangle with } A=50^\circ, \; B=60^\circ, \; c=10. \]
Notice the side is included
Why: c lies between A and B.
Find the third angle
Why: The sum is a straight angle.
\[ C = 70 ^\circ \]
Now a pair exists
Why: C with c.
\[ \sin 70 / 10 \]
Find the other sides
Why: Using the ratio twice.
\[ a = 8.15, b = 9.21 \]
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
\[ C=70^\circ,\; a\approx 8.15,\; b\approx 9.21 \]
Verify: check the ordering
Why: The angles are 50, 60 and 70 and the opposite sides are 8.15, 9.21 and 10 — matching order. The included side turned out to be the longest, which is right since it faces the largest angle.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 910-913
Sorting
It needs a side and its opposite angle.
Sort into buckets
Sort each case.
Worked example
Two sides and the angle between them.
\[ \text{Given } a=7, \; b=9, \; C=40^\circ, \text{ can the law of sines find } c? \]
Identify what is known
Why: Two sides and the included angle.
Look for a complete pair
Why: C is known but c is not.
Try writing a ratio
Why: Every equation has two unknowns.
Conclude
Why: A different tool is needed.
Figure (svg): The solution to Worked example recognise when it fails shown as a ladder of expressions, one row per legal move
\[ \text{use the law of cosines instead} \]
Verify: confirm by counting unknowns
Why: Writing the ratio for A and C gives sine A over 7 equals sine 40 over c, with both A and c unknown — one equation, two unknowns. Recognising this before computing saves the effort of discovering it algebraically.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 913-914
Error analysis
A student has three sides and tries the law of sines.
Annotate
On: \( a=5,\;b=7,\;c=9 \;\Longrightarrow\; \frac{\sin A}{5}=\frac{\sin B}{7} \)
The equation is true but useless, which is a distinct kind of dead end from an equation that is wrong. Checking for a complete pair before writing anything identifies these cases in one glance.
Prediction
Two angles and the side between them are known.
Predict first
What do you do first?
Correct: Find the third angle, which creates a complete pair.
Why: The given side is between the two known angles, so it is opposite neither. The third angle is opposite it, and finding it takes one subtraction — after which the law of sines applies normally.
Faded example
Two angles are given.
Fill in the blanks
C=180^\circ-50^\circ-60^\circ=70^\circ \quad(A=50,\;B=60)
Why: The three angles of a triangle sum to a straight angle, so the third is what remains after subtracting the two known ones. This one line converts an ASA problem into an AAS one.
Explain it to yourself
The law needs a complete side-angle pair.
Discussion prompt
Explain why, in terms of what the law says.
Hint: What is the common ratio?
Answer:
The law says three ratios are equal, but it does not say what they equal. That common value is unknown until one ratio can actually be computed.
Computing a ratio needs both its parts: an angle's sine and the length of the opposite side. One complete pair fixes the common value for all three.
Without such a pair every equation you can write has two unknowns, so the law is true but gives no information. That is why the case analysis matters — it identifies in advance which problems this tool can start.
Section
Section 3
Concept
In the SSA case the given data can describe no triangle, exactly one, or two different ones. The inverse sine returns only one candidate, so the second must be sought deliberately.
The geometric picture is a side swinging from a fixed vertex: it may miss the base entirely, touch it once, or cross it twice. The algebraic test — does the angle sum still work — corresponds exactly to those three possibilities.
Figure (svg): A diagram showing the ambiguous SSA case, where a given side length can reach a baseline at two different points, producing two possible triangles
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 914-919
Picture it
The same side length reaches the base at two different points.
Figure (svg): A diagram showing the ambiguous SSA case, where a given side length can reach a baseline at two different points, producing two possible triangles
Both configurations have the same two side lengths and the same non-included angle, so the data cannot distinguish them. The angle sum is what rules one out when only one is valid.
Worked example
Check the supplement as well.
\[ \text{Given } a=8, \; b=10, \; A=35^\circ, \text{ solve the triangle(s).} \]
Apply the law of sines
Why: Solve for the sine of B.
\[ \sin B = 10 \sin 35 / 8 = 0.717 \]
Take the inverse
Why: The acute candidate.
\[ B = 45.8 ^\circ \]
Take the supplement
Why: The obtuse candidate.
\[ B = 134.2 ^\circ \]
Test the angle sums
Why: Both leave room for a third angle.
Figure (svg): A diagram showing the ambiguous SSA case, where a given side length can reach a baseline at two different points, producing two possible triangles
\[ B\approx 45.8^\circ \text{ or } 134.2^\circ \]
Verify: check both angle sums
Why: With B at 45.8 the third angle is 99.2 degrees; with B at 134.2 it is 10.8. Both are positive, so both triangles genuinely exist. Had the second sum exceeded 180 degrees, that candidate would have been impossible.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 915-917
Prediction
In an SSA case, the supplement of the computed angle would make the angle sum exceed 180 degrees.
Predict first
How many triangles exist?
Correct: One.
Why: The supplement is geometrically impossible because there is no room left for a third angle. The acute candidate survives, giving exactly one triangle. Testing the angle sum is what distinguishes the cases.
Worked example
The supplement fails the angle sum.
\[ \text{Given } a=12, \; b=9, \; A=60^\circ, \text{ solve the triangle(s).} \]
Apply the law of sines
Why: Solve for the sine of B.
\[ \sin B = 9 \sin 60 / 12 = 0.6495 \]
Take the inverse
Why: The acute candidate.
\[ B = 40.5 ^\circ \]
Take the supplement
Why: The obtuse candidate.
\[ B = 139.5 ^\circ \]
Test the angle sums
Why: The second exceeds 180 with A.
Figure (svg): The solution to Worked example only one triangle shown as a ladder of expressions, one row per legal move
\[ B\approx 40.5^\circ \text{ only} \]
Verify: check why the supplement fails
Why: Sixty plus 139.5 is 199.5, already past a straight angle before the third angle is counted. So that configuration is geometrically impossible. Notice also that a is longer than b here, which guarantees a single triangle — the longer side faces the larger angle, so B cannot be obtuse when A is 60.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 917-919
Trap
\[ \sin B=0.717 \;\Longrightarrow\; B=45.8^\circ \text{ and stop} \]
Report the calculator's angle as the answer
Why: The obtuse possibility is never considered.
A second valid triangle is missed entirely.
Always compute the supplement too in the SSA case, and test both against the angle sum.
The inverse sine returns only acute angles for positive inputs, by its restricted range — but a triangle's angle may perfectly well be obtuse.
The angle sum is the test, not the calculator. If both candidates leave a positive third angle, both triangles exist and both must be reported.
Sorting
Only SSA can produce two triangles.
Sort into buckets
Sort each case.
Faded example
The inverse sine gave 42 degrees.
Fill in the blanks
B_2=180^\circ-42^\circ=138^\circ
Why: The supplement has the same sine, so it satisfies the same equation. Whether it gives a real triangle depends on the angle sum, which must still leave room for a positive third angle.
Explain it
SSA does not determine a triangle.
Discussion prompt
Explain to a classmate why two sides and a non-included angle can fit two triangles.
Hint: Picture the third side swinging.
Answer:
Fix the angle and one side, then the other given side swings from its far end looking for the base. It may reach the base at two different points.
Both landing points give a triangle with the same two side lengths and the same angle — so the data cannot tell them apart. One has an acute angle at that vertex and the other an obtuse one.
Algebraically this is the inverse sine returning only one of two angles with the same sine. The angle sum is the test: compute both candidates and keep whichever leave room for a positive third angle. A good explanation stresses that SSA is missing from the congruence criteria for exactly this reason.
Section
Section 4
Concept
The familiar half base times height becomes half the product of two sides times the sine of the angle between them, since that sine supplies the height.
The derivation is one line: the height dropped from one vertex equals a side times the sine of the included angle, by the right-triangle definition. Substituting that into half base times height gives the formula directly.
Figure (svg): A general triangle with vertices labelled with capital letters and each opposite side labelled with the matching lower-case letter
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 919-922
Picture it
Any two sides and the angle between them determine the area.
Figure (svg): A general triangle with vertices labelled with capital letters and each opposite side labelled with the matching lower-case letter
Which pair of sides you use makes no difference to the answer, which is a useful check when more than one pair is known.
Worked example
Two sides and the angle between.
\[ \text{Find the area with } a=8, \; b=11, \; C=42^\circ. \]
Check the angle is included
Why: C lies between sides a and b.
Write the formula
Why: Half the product times the sine.
\[ (\frac{1}{2}) (8) (11) \sin 42 \]
Evaluate the sine
Why: About 0.669.
\[ 0.669 \]
Compute
Why: Multiply out.
\[ 29.4 \]
Figure (svg): A general triangle with vertices labelled with capital letters and each opposite side labelled with the matching lower-case letter
\[ \text{Area}\approx 29.4 \]
Verify: compare with the right-angle case
Why: If C were a right angle the area would be half of 88, which is 44. A 42 degree angle gives a shorter height, so a smaller area is expected — and 29.4 is about two thirds of 44, matching the sine of 42 degrees. The comparison is a quick sanity check.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 920-921
Prediction
You know sides b and c.
Predict first
Which angle does the area formula need?
Correct: Angle A, which lies between them.
Why: By the labelling convention side b is opposite B and side c is opposite C, so the angle those two sides meet at is A. Only the included angle supplies the height that makes the formula correct.
Worked example
Solve first, then use any pair.
\[ \text{With } A=35^\circ, \; C=80^\circ, \; a=9, \text{ find the area.} \]
Find the third angle
Why: The sum is a straight angle.
\[ B = 65 ^\circ \]
Find a second side
Why: By the law of sines.
\[ b = 14.2 \]
Choose an included angle
Why: C lies between a and b.
\[ C = 80 ^\circ \]
Apply the formula
Why: Half the product times the sine.
\[ (\frac{1}{2}) (9) (14.2) \sin 80 \]
Figure (svg): The solution to Worked example area after solving a triangle shown as a ladder of expressions, one row per legal move
\[ \text{Area}\approx 62.9 \]
Verify: recompute with a different pair
Why: Using sides a and c with the included angle B gives half of 9 times 15.5 times the sine of 65, which is also about 63. Any pair with its included angle gives the same area, so agreement between two computations is a genuine check on both.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 921-922
Error analysis
A student computes an area from two sides and an angle.
Annotate
On: \( a=8,\;b=11,\;A=42^\circ \;\Longrightarrow\; \text{Area}=\tfrac{1}{2}(8)(11)\sin 42^\circ \)
The formula's derivation depends on the sine supplying the height between the two named sides, which only works for the included angle. Checking which angle lies between the two sides takes a glance at the labelled drawing.
Faded example
Sides of 6 and 10 with a 30 degree angle between.
Fill in the blanks
\text0.5=\tfrac15___(6)(10)\sin 30^\circ=30\cdot___=___
Why: The sine of thirty degrees is one half, so the area is half of thirty. The formula reduces to half base times height once the sine has supplied the height.
Sorting
Two sides and their included angle are needed.
Sort into buckets
Sort each set of information.
Explain it to yourself
Half a product of two sides and a sine.
Discussion prompt
Explain how it follows from half base times height.
Hint: What supplies the height?
Answer:
Take one of the two sides as the base. The height is the perpendicular distance from the opposite vertex down to it.
That perpendicular forms a right triangle with the other given side as its hypotenuse, so the height equals that side times the sine of the included angle.
Substituting into half base times height gives half the product of the two sides times the sine. It is the familiar formula with the height rewritten, which is why it needs the included angle specifically — no other angle sits in that right triangle.
Section
Section 5
Concept
Most applications of the law of sines involve two known positions observing the same point, which gives two angles and the distance between the observers — an ASA configuration.
Bearing conventions vary, and translating them into a triangle's interior angles is the part of a surveying problem most likely to go wrong. The trigonometry itself is routine once the drawing is right.
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 911-922
Picture it
Applications with two sightings are almost always the ASA case.
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
Recognising which case a word problem has told you is the step that decides everything after it, so it is worth naming explicitly before computing.
Worked example
Two angles and the baseline between the observers.
\[ \text{Two points } 500 \text{ m apart sight a tower at } 42^\circ \text{ and } 58^\circ \text{ from the baseline. Find the nearer distance.} \]
Identify the case
Why: Two angles and the included side.
Find the third angle
Why: At the tower.
\[ 180 - 42 - 58 = 80 \]
Apply the law of sines
Why: The baseline is opposite the tower's angle.
\[ 500 / \sin 80 \]
Find the nearer side
Why: Opposite the smaller angle.
\[ 500 \sin 42 / \sin 80 \]
Figure (svg): A contrast between the situations calling for the law of sines and those calling for the law of cosines
\[ d\approx 340\text{ m} \]
Verify: check which side is nearer
Why: The side opposite the 42 degree angle is the shorter of the two, at 340 metres, and the other is 500 times the sine of 58 over the sine of 80, about 431 metres. The observer with the larger sighting angle is closer, which matches the geometry.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 912-916
Prediction
Two observers a known distance apart each measure an angle to the same object.
Predict first
Which triangle case is this?
Correct: ASA, since the baseline is between the two angles.
Why: The two measured angles are at the two ends of the known baseline, so the side lies between them. Finding the third angle at the object then creates a complete pair and the law of sines applies.
Worked example
Two angles measured from different points on a hill.
\[ \text{From two points } 60 \text{ m apart on a slope, a peak subtends } 32^\circ \text{ and } 47^\circ. \text{ Find the far distance.} \]
Find the interior angle at the nearer point
Why: It is the supplement.
\[ 180 - 47 = 133 \]
Find the third angle
Why: At the peak.
\[ 180 - 32 - 133 = 15 \]
Apply the law of sines
Why: The baseline faces the peak's angle.
\[ 60 / \sin 15 \]
Find the far distance
Why: Opposite the 133 degree angle.
\[ 60 \sin 133 / \sin 15 \]
Figure (svg): The solution to Worked example a height from a slope shown as a ladder of expressions, one row per legal move
\[ d\approx 170\text{ m} \]
Verify: check the small angle's effect
Why: The peak's angle of 15 degrees is small, so its sine is small and the divided distance is large — which is why a shallow difference between the two sightings gives a distant object. That sensitivity is real: a one-degree measurement error would shift the answer by about twelve metres.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 916-922
Trap
\[ \text{the sighting reads } 47^\circ, \text{ so the triangle's angle is } 47^\circ \]
Copy the measured angle straight into the triangle
Why: The measurement's reference direction is not checked against the triangle's sides.
If the angle was measured on the other side of the baseline, it is the supplement that belongs in the triangle.
Draw the picture and identify the interior angle — the one between the two sides of the triangle at that vertex.
A sighting measured forwards along a line gives the supplement of the interior angle when the object is behind the observer.
The drawing settles it in seconds and nothing else does. This conversion, not the trigonometry, is where surveying problems go wrong.
Faded example
A forward sighting of 47 degrees with the object behind.
Fill in the blanks
\text47=180^\circ-133^\circ=___^\circ
Why: When the measured angle opens away from the triangle, the interior angle is its supplement. Drawing the configuration is the only reliable way to tell which applies, and it takes seconds.
Sorting
Some steps are geometry and some are trigonometry.
Sort into buckets
Sort each step.
Explain it
The trigonometry in these problems is easy and the answers are often wrong.
Discussion prompt
Explain to a classmate where the difficulty actually lies.
Hint: Which step has no formula?
Answer:
The law of sines itself is one equation and takes a line. The hard part is deciding what the triangle is — which measured angle is which interior angle, and which side is the baseline.
Bearings and sightings are measured against reference directions that have nothing to do with the triangle, so a conversion is always required and it has no formula to follow.
So the discipline is: draw first, label every angle and side, and only then write an equation. A good explanation notes that a wrong drawing gives a plausible number with no error signal, which is why the check has to come before the computation rather than after.
Comparison
Fill the blanks from memory. Which tool applies is decided here.
Comparison matrix
| case | law of sines | number of triangles |
|---|---|---|
| AAS | yes, directly | exactly one |
| ASA | yes, after the third angle | exactly one |
| SSA | yes, but check both candidates | zero, one or two |
| SAS and SSS | no, use the law of cosines | exactly one |
The third row is the only one where the data does not determine the answer. That is why SSA is absent from the congruence criteria in geometry.
Pattern
Five steps, and the third is the one people skip.
Step 5 catches an inverted ratio, a mislabelled pair and an impossible SSA candidate, all in one glance at the finished triangle.
OpenStax Algebra and Trigonometry 2e, §10.1 Non-right Triangles: Law of Sines §10.1
Check
When the law applies.
Check your understanding
Which case can the law of sines not start from?
Answer: A
Why: With two sides and the angle between them, no side is opposite a known angle, so no ratio can be computed. Every equation the law offers has two unknowns, and the law of cosines is needed instead.
Check
The ambiguous case.
Check your understanding
In an SSA problem, why must you check the supplement of the computed angle?
Answer: A
Why: The inverse sine returns only an acute angle, but an obtuse angle with the same sine satisfies the same equation. Whether it gives a real triangle depends on the angle sum, so both candidates must be tested.
Check
Area.
Check your understanding
Which angle does the area formula require?
Answer: A
Why: The formula comes from rewriting the height as a side times a sine, and that only works for the included angle. Using a non-included angle gives a number with no geometric meaning.
Real world
Every survey and every triangulated position fix is this computation.
Discussion prompt
A ship's position is fixed by taking bearings on two known landmarks. Why does that determine the position?
Hint: What triangle is formed?
Answer:
The two landmarks and the ship form a triangle. The distance between the landmarks is known from the chart, and the two bearings give the angles at each landmark.
That is the ASA case, so the third angle and both distances follow — which places the ship exactly. The whole fix is one application of the law of sines.
It is worth noticing the sensitivity: when the two landmarks are nearly in line from the ship, the angle at the ship is small and its sine is small, so a small measurement error produces a large positional error. Navigators are taught to choose landmarks well separated in bearing for exactly this reason.
Commit first
State your confidence along with your answer.
Predict first
Why is SSA ambiguous when the other cases are not?
Correct: Two sides and a non-included angle can be arranged into two different triangles.
Why: The unattached side can swing to meet the base at two different points, giving two triangles with identical given data. This is exactly why SSA is missing from the congruence criteria, and it is why both candidates must be tested against the angle sum.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate asks how to know whether to use the law of sines or the law of cosines. What is the test?
Hint: What does the law of sines need to get started?
Answer:
Look for a complete side-angle pair — a side and the angle directly opposite it. If you have one, the law of sines works.
Because that pair is what makes one ratio computable, and the law only says the three ratios are equal without saying what they equal.
If no pair exists — two sides and the angle between them, or three sides — every equation has two unknowns and the law is useless. Then it is the law of cosines. A good explanation adds that this test takes one glance at a labelled drawing, which is why drawing first pays for itself.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is the hardest and the one most likely to cost marks, since the second triangle has to be looked for deliberately. The second is what makes the next section's tool feel necessary rather than arbitrary.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a labelled triangle and write the law of sines beside it, marking the opposite pairs. Underneath, list the five triangle cases and which law each needs. Then sketch the ambiguous case showing both possible triangles, and write the test that decides how many exist.
If your case list explains why SAS and SSS fail — no complete pair — and your ambiguous sketch shows both landing points, the section's two decisions are both on the page.
Recap
Five things, and the second decides which tool you reach for.
| if you remember one thing | it should be this |
|---|---|
| about the law | each sine over its opposite side, never an adjacent one |
| about when it applies | you need one complete side-angle pair |
| about SSA | compute the supplement too and test the angle sum |
| about area | the angle must be the one between the two sides |
Section 8.2 supplies the law of cosines, which handles the two cases this one cannot — and reduces to the Pythagorean theorem when the angle happens to be right.
OpenStax, Precalculus, §8.1 Non-right Triangles: Law of Sines §8.1, pp. 904-922 — everything on these slides traces back here
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