The law of sines, solving triangles in the AAS and ASA cases, the ambiguous SSA case and how to count the possible triangles, SSA with none, one or two solutions, and the area of a triangle from two sides and the included angle.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 13 — Trigonometric Ratios and Functions
Apply the Law of Sines
Objectives
Five outcomes. Triangles with no right angle at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-887 — the lesson these objectives are drawn from
Warm-up
Lesson 13.1 solved right triangles, where the 90 degree angle was a free third piece of information.
Discussion prompt
A triangle has angles 25 and 107 degrees and the side opposite the 25 degree angle is 15. There is no right angle. Can you still find everything?
Hint: The three angles still sum to 180.
Answer:
The third angle is 180 minus 107 minus 25, which is 48 degrees. So the shape is fully determined, and the single given side sets its size.
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
That proportion is the law of sines, and it is what replaces the right-triangle ratios once the right angle is gone. Each side is proportional to the sine of the angle opposite it.
Concept
In any triangle, each side divided by the sine of the angle opposite it gives the same value. This solves any triangle where two angles and one side are known, and any where two sides and an angle opposite one of them is known, though that last case may give none, one or two triangles.
law of sines — For any triangle ABC with sides a, b and c opposite angles A, B and C respectively, a over sine A equals b over sine B equals c over sine C. It may equally be written with the sines on top.
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \]
To solve any triangle you need one side and two other parts. Three angles alone fix the shape but not the size.
Figure (svg): Two columns comparing right-triangle methods with the law of sines
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-882
Section
Section 1
Concept
When two angles and any side are given, find the third angle by subtraction, then set up one equation per unknown side, each pairing that side with the complete ratio.
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \;\Longrightarrow\; a = \frac{b\sin A}{\sin B} \]
The two angle-side pairings that are both known form the ratio you keep; each unknown gets its own equation against it.
Figure (svg): A general triangle with the law of sines written beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-882 — Law of Sines
Picture it
Each side against the sine of the angle across from it.
Figure (svg): A general triangle with the law of sines written beside it
Three equal ratios give three equations, but only two are ever needed: the complete pair, and the pair holding the unknown.
Worked example
Example 1.
\[ \text{Solve } \triangle ABC \text{ with } C = 107^\circ, \; B = 25^\circ, \; b = 15. \]
Find the third angle
Why: One eighty minus 107 minus 25.
\[ A = 48 ^\circ \]
Write the complete ratio
Why: Side b against sine of B.
\[ 15 / \sin 25 \]
Solve for a
Why: Fifteen times sine 48, over sine 25.
\[ a = 26.4 \]
Solve for c
Why: Fifteen times sine 107, over sine 25.
\[ c = 33.9 \]
Figure (svg): A triangle solved from two angles and one side
\[ A = 48^\circ, \; a \approx 26.4, \; c \approx 33.9 \]
Verify: check the sides rank like their angles
Why: The angles are 25, 48 and 107, and the sides opposite them are 15, 26.4 and 33.9 — same order. In every triangle the longest side faces the largest angle, which is a free check on any law-of-sines answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-882
Fill the middle
Example 1.
Fill in the blanks
A = 180^\circ-107^\circ-25^\circ = 48^\circ
Why: Forty-eight degrees. Two angles always determine the third, which is why AAS and ASA are really the same case.
Worked example
Guided Practice 1 and 2.
\[ \text{Solve } \triangle ABC \text{ with } B = 34^\circ, C = 100^\circ, b = 8; \text{ and with } A = 51^\circ, B = 44^\circ, c = 11. \]
First: find A
Why: One eighty minus 34 minus 100.
\[ A = 46 ^\circ \]
First: solve for a and c
Why: Against the ratio 8 over sine 34.
\[ a = 10.3, c = 14.1 \]
Second: find C
Why: One eighty minus 51 minus 44.
\[ C = 85 ^\circ \]
Second: solve for a and b
Why: Against the ratio 11 over sine 85.
\[ a = 8.6, b = 7.7 \]
Figure (svg): The solution to Worked example two more triangles shown as a ladder of expressions, one row per algebraic move
\[ 46^\circ, 10.3, 14.1; \qquad 85^\circ, 8.6, 7.7 \]
Verify: check the second triangle's ordering
Why: Its angles are 44, 51 and 85 with opposite sides 7.7, 8.6 and 11 — again in the same order. Notice too that the three angles are close together, and so are the three sides; a nearly equilateral triangle in both respects.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-882
Trap
\[ B = 25^\circ, \; b = 15, \; A = 48^\circ \]
Pair each side with the nearest angle in the diagram
Why: The labels are read off the picture rather than from the opposite relationship.
\[ \frac{a}{\sin 25^\circ} = \frac{15}{\sin 48^\circ} \quad \text{(wrong)} \]
Side a is opposite angle A, and side b is opposite angle B. The pairing is by OPPOSITE, never by adjacency.
\[ \frac{a}{\sin 48^\circ} = \frac{15}{\sin 25^\circ} \]
Pair each side with the angle across from it
Why: That is what the lowercase and uppercase letters record.
\[ a = \frac{15\sin 48^\circ}{\sin 25^\circ} \approx 26.4 \]
The wrong pairing would have given about 8.5, shorter than the given side despite facing a larger angle — an immediate contradiction.
Fill the middle
Example 1.
Fill in the blanks
a = \frac26.4___ = \frac______ \approx ___
Why: About 26.4. The result exceeds 15 because angle A of 48 degrees is larger than angle B of 25, so its opposite side must be longer.
Sorting
You need one side and two other parts.
Sort into buckets
Sort each set of given information.
Three angles give a whole family of similar triangles of every size, which is exactly the similarity that made the trigonometric ratios well defined in the first place.
Prediction
Commit before reasoning.
Predict first
Why can two angles and a side never produce two different triangles?
Correct: Because two angles fix the shape completely, and the one side then fixes the size.
\[ \text{AAS: shape fixed} + \text{one side} \;\Longrightarrow\; \text{one triangle} \]
Why: Knowing two angles gives the third for free, so all three angles are determined and every triangle with them is similar to every other. Supplying one side then scales that shape to exactly one size. The SSA case is different precisely because it leaves an angle unknown, and an unknown angle can be acute or obtuse with the same sine.
Section
Section 2
Concept
Given two sides and an angle opposite one of them, compute h equal to b sine A. Comparing side a with h and with b tells you whether zero, one or two triangles exist.
\[ h = b\sin A \]
The value h is the shortest that side a could be and still reach the base, which is why a shorter than h gives no triangle at all.
Figure (svg): The possible numbers of triangles in the SSA case
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 883-883 — Possible Triangles in the SSA Case
Picture it
The cases when A is obtuse and when A is acute.
Figure (svg): The possible numbers of triangles in the SSA case
Only one configuration gives two triangles: A acute with a between h and b. Everything else gives one or none.
Worked example
The boxed cases on page 883.
\[ \text{How many triangles when } A = 115^\circ, a = 20, b = 11? \text{ When } A = 40^\circ, a = 13, b = 16? \]
First: is A obtuse?
Why: One fifteen exceeds 90.
First: compare a with b
Why: Twenty exceeds 11.
Second: A is acute, so find h
Why: Sixteen times sine 40.
\[ h = 10.3 \]
Second: compare
Why: Ten point three, then 13, then 16.
\[ h < a < b,\text{ so two triangles} \]
Figure (svg): The possible numbers of triangles in the SSA case
\[ \text{one}; \qquad \text{two} \]
Verify: explain the obtuse rule physically
Why: If A is obtuse then side a faces the largest angle and must be the longest side. When a exceeds b that is possible and the triangle closes exactly one way; when a is at most b it is impossible and no triangle exists. The obtuse case never gives two, because there is no room for a second closing point.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 883-883
Fill the middle
Example 4.
Fill in the blanks
h = b\sin A = 16\sin 40^\circ = 16(0.643) \approx 10.3
Why: About 10.3, the shortest side a could be and still reach the base. Since a is 13, longer than that, the arc does reach and a triangle exists.
Worked example
Guided Practice 3 and 5.
\[ \text{How many triangles when } A = 122^\circ, a = 18, b = 12? \text{ When } A = 50^\circ, a = 2.8, b = 4? \]
First: A is obtuse
Why: One twenty-two exceeds 90.
First: compare
Why: Eighteen exceeds 12.
Second: A is acute, find h
Why: Four times sine 50.
\[ h = 3.06 \]
Second: compare
Why: Three point zero six exceeds 2.8.
\[ h > a,\text{ no triangle} \]
Figure (svg): The solution to Worked example two more counts shown as a ladder of expressions, one row per algebraic move
\[ \text{one}; \qquad \text{none} \]
Verify: picture the second case
Why: Side a of length 2.8 is a compass arc swung from vertex C, and the nearest point of the base is 3.06 away — the arc never reaches it. Drawing the situation with a ruler makes the impossibility obvious in a way the algebra does not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 884-884
Error analysis
A student is given A = 51 degrees, a = 3.5 and b = 5.
Annotate
On: \( \sin B = \frac{5\sin 51^\circ}{3.5} \approx 1.11 \)
A sine above 1 is not an arithmetic slip here; it is the correct answer to the question of whether the triangle can be built. It cannot.
Sorting
Compute h, then compare.
Sort into buckets
Sort each SSA case by how many triangles it gives.
Both two-triangle cases have an acute given angle with a strictly between h and b. That single configuration is the entire ambiguous case.
Comparison
Fill the blanks. The count depends on which.
Comparison matrix
| Question | A obtuse | A acute |
|---|---|---|
| What to compare | a against b | a against h and against b |
| Can give two triangles? | never | yes, when h < a < b |
| No triangle when | a is at most b | a is less than h |
| Why | a must face the largest angle | the arc may miss, touch or cross twice |
When A is obtuse, side a faces the biggest angle and so must be the biggest side. That single requirement collapses all the possibilities to one comparison.
Prediction
Commit before reasoning.
Predict first
Solving an SSA case, you obtain sin B = 1.11. What does that tell you?
Correct: That no such triangle exists, since no angle has a sine above 1.
\[ \sin B > 1 \;\Longrightarrow\; \text{no triangle} \]
Why: The law of sines was applied correctly; the impossible output is the algebra's way of reporting an impossible triangle. This is a genuinely useful behaviour: rather than needing to test h against a first, you can simply solve and let a sine above 1 signal the empty case. The other two counts still need the comparison, since one and two triangles both produce perfectly ordinary sines.
Section
Section 3
Concept
When the count is one, use the law of sines to find the second angle, subtract for the third, and use the law again for the remaining side. When the count is zero, say so and show why.
\[ \frac{\sin B}{b} = \frac{\sin A}{a} \;\Longrightarrow\; \sin B = \frac{b\sin A}{a} \]
The inverse sine returns the acute answer. When only one triangle exists, that acute answer is the right one.
Figure (svg): The possible numbers of triangles in the SSA case
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 883-884 — Solve the SSA case with one solution
Picture it
The six SSA configurations again.
Figure (svg): The possible numbers of triangles in the SSA case
Two of the six give no triangle at all, three give one, and only the middle acute case gives two.
Worked example
Example 2.
\[ \text{Solve } \triangle ABC \text{ with } A = 115^\circ, \; a = 20, \; b = 11. \]
Count the triangles
Why: A obtuse and a exceeds b.
Find sine of B
Why: Eleven times sine 115, over 20.
\[ 0.4985 \]
Find B and then C
Why: Inverse sine, then subtract.
\[ B = 29.9, C = 35.1 \]
Find c
Why: Twenty times sine 35.1, over sine 115.
\[ c = 12.7 \]
Figure (svg): The solution to Worked example SSA with one triangle shown as a ladder of expressions, one row per algebraic move
\[ B \approx 29.9^\circ, \; C \approx 35.1^\circ, \; c \approx 12.7 \]
Verify: check that B could not be obtuse
Why: The obtuse partner of 29.9 is 150.1, and 115 plus 150.1 already exceeds 180. So the obtuse option is impossible here, which is another way of seeing why this case gives only one triangle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 883-883
Fill the middle
Example 2.
Fill in the blanks
\sin B = \frac29.9___ \approx 0.4985 \;\Longrightarrow\; B \approx ___^\circ
Why: About 29.9 degrees. A sine near one half gives an angle near 30, which is a useful landmark for checking the arithmetic.
Worked example
Example 3, and Guided Practice 6.
\[ \text{Solve } \triangle ABC \text{ with } A = 51^\circ, a = 3.5, b = 5; \text{ and with } B = 105^\circ, b = 13, a = 6. \]
First: compute h
Why: Five times sine 51.
\[ h = 3.9 \]
First: compare with a
Why: Three point nine exceeds 3.5.
Second: count
Why: B obtuse and b exceeds a.
Second: solve
Why: Sine A is 6 sin 105 over 13.
\[ A = 26.5, C = 48.5, c = 10.1 \]
Figure (svg): The solution to Worked example SSA with no triangle shown as a ladder of expressions, one row per algebraic move
\[ \text{none}; \qquad A \approx 26.5^\circ, \; C \approx 48.5^\circ, \; c \approx 10.1 \]
Verify: notice the second case has the given angle at B
Why: The letters do not have to be A. What matters is that the given angle is OPPOSITE one of the two given sides, and here 105 degrees sits opposite the side of length 13. The whole analysis transfers with the letters relabelled.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 884-884
Trap
\[ A = 51^\circ, \; a = 3.5, \; b = 5 \]
Apply the law of sines immediately
Why: The count is never checked.
\[ \sin B \approx 1.11 \;\Longrightarrow\; B = ? \]
The calculator returns an error, and a student who was not expecting it may assume they mis-keyed something.
\[ h = 5\sin 51^\circ \approx 3.9 > 3.5 = a \]
Count the triangles first
Why: Then solve only if there is something to solve.
\[ \text{no triangle exists} \]
Counting first turns a confusing calculator error into an expected and reportable conclusion.
Fill the middle
Example 3.
Fill in the blanks
h = 5\sin 51^\circ \approx 3.9 > 3.5 \;\Longrightarrow\; no \text___
Why: No triangle. Side a of 3.5 is shorter than the 3.9 needed to reach the base, so the two ends never meet.
Matching
Count first, then solve.
Match the pairs
Why: Three of the four have an obtuse given angle with its opposite side the longest, which is the reliable one-triangle configuration. The remaining case fails the reach test entirely.
Prediction
Commit before reasoning.
Predict first
When the given angle A is obtuse and a triangle exists, why is there exactly one?
Correct: Because the second angle B would have to be acute, since a triangle cannot hold two obtuse angles.
\[ A > 90^\circ \;\Longrightarrow\; B < 90^\circ \text{ forced} \]
Why: The two-triangle ambiguity comes from an angle whose sine is known being either acute or obtuse. When A is already obtuse, B has no choice but to be acute, so the ambiguity vanishes. That is why only an acute given angle can ever produce two triangles, and it explains the whole shape of the case table.
Section
Section 4
Concept
When h is less than a and a is less than b with A acute, two triangles exist. The inverse sine gives the acute value of B; the obtuse one is 180 minus it. Each choice leads to its own third angle and third side.
\[ B_1 = \sin^{-1}k; \qquad B_2 = 180^\circ-B_1 \]
Every remaining part must be computed separately for each triangle, so the work is done twice from that point on.
Figure (svg): One set of measurements producing two different triangles
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 884-884 — Solve the SSA case with two solutions
Picture it
Example 4: A is 40 degrees, a is 13 and b is 16.
Figure (svg): One set of measurements producing two different triangles
Both have the same angle A and the same two sides, yet one is tall and one is squat, and both are perfectly valid answers.
Worked example
Example 4.
\[ \text{Solve } \triangle ABC \text{ with } A = 40^\circ, \; a = 13, \; b = 16. \]
Count the triangles
Why: H is 10.3, and 10.3 < 13 < 16.
Find both values of B
Why: Sine B is 0.7911.
\[ 52.3\text{ and } 127.7 ^\circ \]
Triangle 1: C and c
Why: One eighty minus 40 minus 52.3.
\[ C = 87.7, c = 20.2 \]
Triangle 2: C and c
Why: One eighty minus 40 minus 127.7.
\[ C = 12.3, c = 4.3 \]
Figure (svg): One set of measurements producing two different triangles
\[ B \approx 52.3^\circ \text{ or } 127.7^\circ \]
Verify: check both third angles are positive
Why: Forty plus 52.3 is 92.3, leaving 87.7; forty plus 127.7 is 167.7, leaving 12.3. Both are positive, so both triangles are genuine. When the obtuse option makes the first two angles exceed 180 it must be discarded, which is exactly what rules out the second triangle in the one-solution cases.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 884-884
Fill the middle
Example 4.
Fill in the blanks
B_1 \approx 52.3^\circ \;\Longrightarrow\; B_2 = 180^\circ-52.3^\circ = 127.7^\circ
Why: One hundred twenty-seven point seven degrees. Supplementary angles have equal sines, which is the whole source of the ambiguity.
Worked example
Guided Practice 4.
\[ \text{Solve } \triangle ABC \text{ with } A = 36^\circ, \; a = 9, \; b = 12. \]
Count
Why: H is 12 sine 36, about 7.05, and 7.05 < 9 < 12.
Find both values of B
Why: Sine B is 12 sine 36 over 9, about 0.7837.
\[ 51.6\text{ and } 128.4 ^\circ \]
Triangle 1
Why: C is 92.4; c is 9 sine 92.4 over sine 36.
\[ c = 15.3 \]
Triangle 2
Why: C is 15.6; c is 9 sine 15.6 over sine 36.
\[ c = 4.1 \]
Figure (svg): The solution to Worked example another two-triangle case shown as a ladder of expressions, one row per algebraic move
\[ 51.6^\circ/92.4^\circ/15.3; \qquad 128.4^\circ/15.6^\circ/4.1 \]
Verify: compare the two third sides
Why: Fifteen point three against 4.1 — the two triangles are not remotely similar in size, even though they share an angle and two sides. That is worth noticing, because it shows the ambiguity is not a minor rounding matter but two genuinely different answers.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 884-884
Error analysis
A student solves the case A = 40 degrees, a = 13, b = 16.
Annotate
On: \( \sin B \approx 0.7911 \;\Longrightarrow\; B \approx 52.3^\circ \text{ only} \)
Half the answer is missing. The inverse sine returns only its range's value, so the obtuse partner must always be considered when the count says two.
Fill the middle
Example 4, Triangle 2.
Fill in the blanks
C = 180^\circ-40^\circ-127.7^\circ = 12.3^\circ
Why: Twelve point three degrees. It is positive, so this triangle is genuine; a negative result here would have ruled the obtuse option out.
Ranking
Smallest first.
Put in order
Why: The values are about 4.1, 4.3, 12.7, 15.3 and 20.2. Within each ambiguous pair the obtuse option always gives much the shorter third side, because it leaves much the smaller third angle.
Prediction
Commit before reasoning.
Predict first
Why can the same angle A and sides a and b produce two different triangles?
Correct: Because side a swung as an arc from vertex C crosses the base line at two different points.
\[ h < a < b \;\Longrightarrow\; \text{the arc crosses twice} \]
Why: Fix angle A and side b, so vertex C is pinned. Now swing side a from C like a compass: if it is longer than the perpendicular distance h but shorter than b, the arc cuts the base twice, once on each side of the foot of the perpendicular. Each crossing is a legitimate vertex B. Once a is at least as long as b the near crossing falls on the wrong side of A and stops being a triangle, which is why the upper bound is b.
Section
Section 5
Concept
The area of any triangle is one half the product of two side lengths times the sine of their included angle. Any of the three pairs may be used.
\[ \text{Area} = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B = \tfrac{1}{2}ab\sin C \]
The angle must be the one BETWEEN the two chosen sides. Using a different angle gives a wrong answer.
Figure (svg): A triangular region with two sides and the included angle
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 885-885 — Area of a Triangle
Picture it
Example 5: sides of 125 and 223 miles with a 54.2 degree angle between them.
Figure (svg): A triangular region with two sides and the included angle
The familiar one half base times height is hiding inside: b times sine A is exactly the height dropped onto side c.
Worked example
Example 5.
\[ \text{Find the area of a triangle with } b = 125, \; c = 223 \text{ and } A = 54.2^\circ. \]
Check the angle is included
Why: A sits between sides b and c.
Write the formula
Why: Half of b times c times sine A.
\[ (\frac{1}{2}) (125) (223) \sin 54.2 \]
Multiply the sides
Why: One twenty-five times 223, halved.
\[ 13, 937.5 \]
Multiply by the sine
Why: Sine of 54.2 is about 0.811.
\[ \text{about } 11, 300 \]
Figure (svg): A triangular region with two sides and the included angle
\[ \text{Area} \approx 11{,}300 \text{ mi}^2 \]
Verify: compare with the right-angle case
Why: Had the angle been 90 degrees the sine would be 1 and the area 13,937.5 — the largest it could be for those two sides. The actual angle of 54.2 gives about 81 percent of that, which is exactly what a sine of 0.811 should do.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 885-885
Fill the middle
Guided Practice 7.
Fill in the blanks
\text50.4 = \tfrac______(10)(14)\sin 46^\circ = 70(0.719) \approx ___
Why: About 50.4 square units. Since sine 46 is close to 0.72, the triangle has about 72 percent of the area a right angle would give.
Worked example
Guided Practice 7 to 10.
\[ \text{Find the area for } a=10,b=14,C=46^\circ; \; a=19,c=8,B=75^\circ; \; b=11,c=7,A=120^\circ; \; a=20,b=24,C=87^\circ. \]
First: half of 10 times 14
Why: Times sine 46.
\[ 70(0.719) = 50.4 \]
Second: half of 19 times 8
Why: Times sine 75.
\[ 76(0.966) = 73.4 \]
Third: half of 11 times 7
Why: Times sine 120.
\[ 38.5(0.866) = 33.3 \]
Fourth: half of 20 times 24
Why: Times sine 87.
\[ 240(0.999) = 239.7 \]
Figure (svg): The solution to Worked example four areas shown as a ladder of expressions, one row per algebraic move
\[ 50.4; \; 73.4; \; 33.3; \; 239.7 \]
Verify: notice the third uses an obtuse angle
Why: Sine of 120 is 0.866, the same as sine of 60, so an obtuse included angle poses no difficulty at all — the formula simply uses the sine, which is positive throughout the second quadrant. An angle of 87 degrees, by contrast, gives a sine of almost exactly 1, so the fourth triangle is nearly right-angled.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 885-885
Trap
\[ b = 125, \; c = 223, \; B = 54.2^\circ \]
Use whichever angle is given
Why: The formula is applied without checking the position.
\[ \text{Area} = \tfrac{1}{2}bc\sin B \quad \text{(wrong)} \]
The formula needs the angle BETWEEN the two sides. Angle B is opposite side b, not between b and c.
\[ \text{Area} = \tfrac{1}{2}bc\sin A \]
Match the missing letter
Why: Sides b and c call for angle A.
\[ \approx 11{,}300 \text{ mi}^2 \]
The pattern is easy to remember: the angle in the formula is the one whose letter does not appear among the sides.
Matching
The angle whose letter is missing.
Match the pairs
Why: The rule is purely a matter of letters: the angle to use is the one whose lowercase letter is not among the sides you have. That works because each angle sits between the two sides not named after it.
Comparison
Fill the blanks. The old one and the new one.
Comparison matrix
| Question | Half base times height | Half two sides times sine |
|---|---|---|
| Needs | a base and a perpendicular height | two sides and the angle between them |
| Formula | (1/2) b h | (1/2) b c sin A |
| Height comes from | measurement or construction | c sin A |
| Works for obtuse angles? | yes, with an external height | yes, directly |
The two are the same formula. Substituting h equal to c sine A into one half b h gives one half b c sine A, so the new version simply computes the height for you from data you are more likely to have.
Prediction
Commit before reasoning.
Predict first
Two sides of a triangle are fixed at 125 and 223. What included angle gives the largest area?
Correct: Ninety degrees, because that is where the sine reaches its maximum of 1.
\[ \sin 30^\circ = \sin 150^\circ = \tfrac{1}{2} \]
Why: The area is one half times a fixed product times the sine, so it rises and falls exactly as the sine does, peaking at 90 degrees with an area of 13,937.5. Notice too that 30 degrees and 150 degrees give identical areas, since their sines are equal — a very skinny triangle and a very flat one can enclose exactly the same space, which is a genuinely counter-intuitive consequence of the formula.
Comparison
Fill the blanks. What each set of givens determines.
Comparison matrix
| Question | AAS or ASA | SSA |
|---|---|---|
| Given | two angles and a side | two sides and an angle opposite one |
| Number of triangles | always exactly one | none, one or two |
| First step | find the third angle by subtraction | count the triangles using h = b sin A |
| Watch out for | pairing sides with opposite angles | the obtuse partner of the inverse sine |
The ambiguity in SSA is not a defect of the law of sines but a fact about triangles: those three measurements genuinely do not always pin one down.
Pattern
Count, then solve.
Pair each side with the angle OPPOSITE it, never with an adjacent one.
OpenStax Algebra and Trigonometry 2e, §10.1 Non-right Triangles: Law of Sines §10.1
Check
Find the third angle first.
Check your understanding
In triangle ABC, C = 107 degrees, B = 25 degrees and b = 15. What is a?
Answer: A
Why: A = 48 degrees, and a = 15 sin 48 / sin 25, about 26.4.
Check
Compute h and compare.
Check your understanding
How many triangles have A = 40 degrees, a = 13 and b = 16?
Answer: A
Why: h = 16 sin 40 is about 10.3, and 10.3 < 13 < 16, so the arc crosses the base twice.
Check
The angle must sit between the two sides.
Check your understanding
What is the area of a triangle with b = 125, c = 223 and A = 54.2 degrees?
Answer: A
Why: Area = (1/2)(125)(223) sin 54.2 = 13,937.5 times about 0.811.
Real world
Two fire lookout towers are 18 miles apart along a straight road. From the west tower a plume of smoke bears 42 degrees east of the road; from the east tower it bears 71 degrees west of the road.
Discussion prompt
How far is the smoke from each tower, and what area does the triangle they form cover?
Hint: This is AAS, so start with the third angle.
Answer:
\[ C = 180^\circ-42^\circ-71^\circ = 67^\circ \]
\[ \frac{a}{\sin 42^\circ} = \frac{18}{\sin 67^\circ} \;\Longrightarrow\; a = \frac{18\sin 42^\circ}{\sin 67^\circ} \approx 13.1 \text{ mi} \]
\[ b = \frac{18\sin 71^\circ}{\sin 67^\circ} \approx 18.5 \text{ mi} \]
\[ \text{Area} = \tfrac{1}{2}(13.1)(18.5)\sin 67^\circ \approx 112 \text{ mi}^2 \]
The smoke is about 13.1 miles from the east tower and 18.5 miles from the west one.
This is triangulation, and it is how surveying, navigation and seismology all locate something no one can reach. Notice the two towers alone were never enough: it took the 18 mile baseline plus two bearings to make the AAS case, and had only one bearing been available the position would have been a whole ray rather than a point. Notice also that this is emphatically not SSA — there is no ambiguity here, and a second possible fire location would be a serious operational problem.
Commit first
Answer, then rate your confidence honestly.
Predict first
Solving an SSA case you find sin B = 0.7911 and report B = 52.3 degrees. Could that be only half the answer?
Correct: Yes — 127.7 degrees has the same sine, and if h < a < b both triangles are real.
\[ \sin 52.3^\circ = \sin 127.7^\circ \approx 0.7911 \]
Why: Supplementary angles have equal sines, so every sine value between 0 and 1 comes from one acute and one obtuse angle. The inverse sine returns only the acute one, by the range restriction of Lesson 13.4. Whether the obtuse partner also gives a triangle is settled by the count: if the two known angles would already exceed 180 it must be discarded, and otherwise it is a second genuine answer that a complete solution has to report.
Explain it
They have solved right triangles and think trigonometry needs a right angle.
Discussion prompt
In four sentences or fewer, explain what the law of sines says and why it helps.
Hint: Talk about which side faces which angle.
Answer:
In any triangle, a bigger angle always faces a longer side. The law of sines makes that precise: each side divided by the sine of the angle opposite it gives the same number.
So if you know one side together with the angle across from it, you know that number. Then any other angle gives you its own opposite side, with no right angle needed anywhere.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For pairing, remember that lowercase a always faces uppercase A. For counting, compute h equal to b sine A and compare a with h and with b, or note that an obtuse given angle needs only a compared with b. For the obtuse partner, subtract the acute answer from 180 and check the angles still fit under 180. For area, use the angle whose letter is missing from the two sides.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a law of sines page. Top left: draw a general triangle, label all six parts, and write the law of sines in both forms, marking clearly which side pairs with which angle. Top right: solve one AAS triangle completely, showing the third angle first and one equation per unknown side, then check that the sides rank in the same order as the angles. Middle: draw all six SSA configurations with h marked on each, labelling the count under every one. Bottom left: solve a two-triangle SSA case in full, keeping the two triangles in separate columns and finishing both. Bottom right: compute one area from two sides and their included angle, and beside it show why the formula is the same as one half base times height.
If your SSA work reported only the acute angle when the count said two, you have solved half the problem. The obtuse partner is 180 minus it.
Recap
Five things, and triangles with no right angle at all.
| If you see | Then |
|---|---|
| Two angles and a side | Find the third angle, then one equation per side |
| Two sides and an opposite angle | Count before solving |
| An obtuse given angle | Compare a with b alone |
| h less than a less than b | There are two triangles |
| A sine greater than 1 | No triangle exists |
| Two sides and the angle between | Area is half their product times the sine |
Lesson 13.6 covers the two cases the law of sines cannot touch: three sides, or two sides with the angle between them. Those need the law of cosines.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.5 Apply the Law of Sines §13.5, pp. 882-887 — everything on these slides traces back here
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