Swinging Gate Slides Swinging Gate
Visualizing the Ambiguous Case
GEOMETRY UNIT 4
The Geometry Challenge
If I give you two sticks (side lengths) and one angle , can you always build the exact same triangle as your neighbor?
Recall:
Why did we say "SSA" was NOT a congruence postulate?
A B a c
The "Swinging Gate" Anatomy
Fixed Angle (A)
The angle at vertex A is locked. It determines the direction of the unknown side.
Side 'c' (Adjacent)
The side between the fixed angle and the "pivot" point (Vertex B).
Side 'a' (Opposite)
The "Swinging Gate." It can rotate around Vertex B to find the base.
Today's Mission: Explore how the length of the Swinging Side (a) changes everything.
Hands-On Investigation
1
Draw a ray to represent the unknown base of your triangle.
2
Construct a fixed angle of \( 30^\circ \) at the start of the ray.
3
Mark point B on the side of the angle such that side \( c = 10 \text{ cm} \).
Challenge: If side \( a \) is 3 cm, 5 cm, 7 cm, and 12 cm, how many triangles can you make for each?
Construction View
Pivot B
Use your compass to "swing" the arc of side 'a'.
Swinging Gate Worksheet Swinging Gate Investigation
Geometry: Ambiguous Case Exploration
NAME:
DATE:
Objective
In this investigation, you will use physical or digital construction tools to explore how many triangles can be formed when you are given two side lengths and one non-included angle (SSA) . Pay close attention to the "swinging" side.
Part 1: Setting the Parameters
For all constructions today, we will use the following "fixed" pieces of information:
Fixed Angle (A)
\( 30^\circ \)
Adjacent Side (c)
10 cm
Opposite Side (a)
Variable (The Swinging Side)
Instructions:
Draw a horizontal line (ray) across your page. Mark the start point as A .
Measure and draw a \( 30^\circ \) angle at point A.
Along the new side of the angle, measure and mark point B at exactly 10 cm from A.
Vertex B is now your "pivot" point. You will use a compass to swing an arc of length a .
Part 2: The Swinging Side Scenarios
Use your compass to test each of the following lengths for side a . For each length, record how many times your compass arc intersects the base ray.
Length of Side 'a' Visual Description (Does it reach? Does it hit twice?) # of Triangles 3 cm 5 cm 7 cm 12 cm
Part 3: Critical Thinking
1. At what exact length does side a touch the base ray exactly once (forming a right triangle)? How can you calculate this length using trigonometry?
2. Explain why the 7 cm case produced two triangles. What was unique about that length compared to 3 cm or 12 cm?
3. Based on your 12 cm trial, if side a is longer than side c , can you ever form two triangles? Why or why not?
Swinging Gate Teacher Guide Teacher Guide
Swinging Gate Visualization
Lesson 1: The Geometry of SSA Ambiguity
Duration
50-60 min
Learning Objectives
Construct triangles using SSA parameters.
Identify the "height" of a triangle as the critical threshold for construction.
Visually distinguish between 0, 1, and 2 solution cases.
Materials Needed
Compasses, Protractors, Rulers.
Large "Swinging Gate" Slides.
Investigation Worksheet.
Optional: Digital Geometry Software (GeoGebra/Desmos).
The Mathematical Thresholds
1. The "Too Short" Case
When \( a < c \sin A \), the side is shorter than the height. No triangle exists.
2. The "Evil Twin" Case
When \( c \sin A < a < c \), the side can swing inside and outside, creating 2 triangles.
3. The "Overlap" Case
When \( a \geq c \), the side only hits the ray once in a positive direction. 1 triangle.
Lesson Flow
0-10 min
The Hook & Slides
Use the Swinging Gate Slides to present the "Sticks and Angles" challenge. Emphasize why SSA was not a congruence postulate in previous geometry lessons.
10-35 min
The Investigation
Students work on Part 2 of the worksheet. Circulate to ensure they are keeping the \( 30^\circ \) angle fixed while swinging the compass. Many students will struggle to see the "inside" triangle for the 7 cm case—show them how to swing the arc until it hits the ray twice.
35-50 min
Synthesis & Math Connection
Discuss Part 3 . Guide students to see that \( 10 \sin 30^\circ = 5 \text{ cm} \). This is the height. If \( a < 5 \), it can't reach. If \( a = 5 \), it's a right triangle. If \( 5 < a < 10 \), we get two solutions.
Discussion Questions
"If we made the initial angle \( 100^\circ \) (obtuse), how many solutions could we have?"
Answer: Only 0 or 1. Two solutions are only possible when the fixed angle is acute.
"What happens to the two triangles as side 'a' gets closer and closer to 5 cm?"
Answer: They 'merge' into a single right triangle.
No Solution Slides Impossible Triangles
Analyzing the No-Solution Case
The Geometric "Gap"
Yesterday, we saw that if the swinging side \( a \) is too short, it never touches the base.
Threshold:
\( a < \text{Height} \)
\( a < c \sin A \)
h a GAP
The Algebra of Impossibility
What happens if we use the Law of Sines on a "gap" triangle?
\[ \frac{\sin A}{a} = \frac{\sin B}{c} \]
\[ \sin B = \frac{c \sin A}{a} \]
The Logic Check:
If \( a < c \sin A \), then the fraction on the right is GREATER THAN 1 .
But \(\sin B\) can NEVER be greater than 1!
The Calculator Verdict
Error: Domain
"Why did it crash?"
When you press sin⁻¹, your calculator is asking:
Geometry's Answer:
No Solution Worksheet Impossible Triangles
Detecting the No-Solution Case
NAME:
DATE:
The Height Threshold
\( h = c \sin A \)
If side a is less than height h , the swinging side is too short to reach the base ray. In the Law of Sines, this results in \( \sin B > 1 \), which is mathematically impossible.
Part 1: Detecting the "Gap" Algebraically
Attempt to solve the following triangles using the Law of Sines. Show your work until you encounter the error.
Triangle 1
\( \angle A = 40^\circ \)
\( c = 15 \)
\( a = 8 \)
SHOW WORK:
Result:
Triangle 2
\( \angle A = 75^\circ \)
\( c = 20 \)
\( a = 12 \)
SHOW WORK:
Result:
Part 2: Visualizing the "Crash"
1. Calculate the actual height \( h \) for Triangle 1 above. Is side \( a \) shorter or longer than \( h \)?
2. Explain in your own words why your calculator gives a "Domain Error" when you try to find \( \angle B \) for these triangles.
3. Sketch a diagram for Triangle 2. Label side \( c \), angle \( A \), height \( h \), and side \( a \). Clearly indicate the "gap."
Diagram Workspace
No Solution Answer Key Answer Key: No Solution Analysis
Geometry Unit 4 | Lesson 2
Part 1: Problem 1
\[ \frac{\sin 40^\circ}{8} = \frac{\sin B}{15} \]
\[ \sin B = \frac{15 \cdot \sin 40^\circ}{8} \]
\[ \sin B \approx 1.205 \]
Result: NO SOLUTION
The value of \(\sin B\) is greater than 1, which is outside the domain of the inverse sine function.
Part 1: Problem 2
\[ \frac{\sin 75^\circ}{12} = \frac{\sin B}{20} \]
\[ \sin B = \frac{20 \cdot \sin 75^\circ}{12} \]
\[ \sin B \approx 1.610 \]
Result: NO SOLUTION
Part 2: Visualizing the "Crash"
1. Calculate the actual height for Triangle 1.
\( h = 15 \cdot \sin 40^\circ \approx 9.64 \)
Conclusion: Side \( a = 8 \) is shorter than the height \( h = 9.64 \). It cannot reach the base.
2. Explain why the calculator gives a "Domain Error."
"The sine of an angle represents the ratio of the opposite side to the hypotenuse in a right triangle. Since the hypotenuse is always the longest side, the sine value can never exceed 1. When we calculate a ratio like 1.2, we are asking the calculator to find an angle where the opposite side is longer than the hypotenuse, which is impossible."
3. Sketch for Triangle 2.
c=20 75° h ≈ 19.3 a=12 GAP
Two Triangle Slides The Evil Twin
Solving for Two Triangle Solutions
The Double Hit
When the swinging side \( a \) is longer than the height but shorter than side \( c \) , it hits the base twice.
The Sweet Spot:
\( h < a < c \)
Two valid triangles exist!
C₁ C₂ B
The Secret Supplement
Your calculator will only give you the acute angle (\( B_1 \)).
Triangle 1
"The Calculator Answer"
\( B_1 = \sin^{-1}(\text{ratio}) \)
Triangle 2
"The Evil Twin"
\( B_2 = 180^\circ - B_1 \)
Why? Because \(\sin(180-\theta) = \sin(\theta)\). Both satisfy the math!
The Reality Check
How do we know if the "Evil Twin" triangle actually works?
The Angle Sum Rule:
\[ \angle A + \angle B_2 < 180^\circ \]
If there is room for a third angle (\( \angle C \)), then two triangles exist.
If the sum is 180 or more, the twin is an imposter. Only one triangle exists.
Two Triangle Protocol Handout The Two-Solution Protocol
A Step-by-Step Guide for the Ambiguous Case
1
Solve for the Reference Angle (\( B_1 \))
Use the Law of Sines to find the first possible angle for vertex B:
\[ \sin B_1 = \frac{c \sin A}{a} \]
Note: Use your calculator's \(\sin^{-1}\) function. This will always give an acute angle.
2
Find the "Evil Twin" (\( B_2 \))
Calculate the supplementary angle. This is your second potential solution for vertex B:
\[ B_2 = 180^\circ - B_1 \]
3
Validate Solution 2
Check: Is there enough room for \(\angle C_2\)?
Add Angles
\( \angle A + B_2 = \text{Sum} \)
If Sum < 180°
Two triangles exist!
If Sum ≥ 180°
Only one triangle exists.
4
Solve Remaining Parts
Triangle 1 Path
\( \angle C_1 = 180 - (A + B_1) \)
Solve for side \( b_1 \) using Law of Sines.
Triangle 2 Path
\( \angle C_2 = 180 - (A + B_2) \)
Solve for side \( b_2 \) using Law of Sines.
Double Trouble Worksheet Double Trouble
Solving for Two Solutions
NAME:
DATE:
Problem 1
Two Triangle Existence
Given: \( \angle A = 35^\circ \), \( c = 12 \), and \( a = 8 \). Find all possible values for \( \angle B \), \( \angle C \), and side \( b \).
Triangle 1 (Acute \(\angle B\))
Find \( \angle B_1 \):
Find \( \angle C_1 \):
Find side \( b_1 \):
Triangle 2 (Obtuse \(\angle B\))
Find \( \angle B_2 \) (Supplement):
Find \( \angle C_2 \):
Find side \( b_2 \):
Problem 2
The Single Solution Case
Given: \( \angle A = 50^\circ \), \( c = 10 \), and \( a = 11 \). Does this SSA case produce two triangles? Prove your answer algebraically.
1. Calculate the first possible angle (\( B_1 \)):
2. Calculate the "potential" second angle (\( B_2 = 180 - B_1 \)):
3. Perform the Validation Check (\( \angle A + B_2 \)):
Final Verdict:
How many triangles exist? Explain why using the results above.
Predicting Triangles Slides Predicting Triangles
A Systematic Approach to SSA
The Power of Prediction
Do you really need to use the Law of Sines to know how many answers you'll get?
The Golden Rule:
Compare the Swinging Side (a) to the Height (h) and the Adjacent Side (c) .
1. Height: \( h = c \sin A \)
2. Sides: \( a \) vs \( c \)
The Decision Matrix
Condition Visual Outcome # Solutions \( a < h \) Side too short to reach 0 \( a = h \) Forms a Right Triangle 1 \( h < a < c \) The Ambiguous Case 2 \( a \geq c \) Swings beyond vertex A 1
Rapid Fire Round
How many triangles? (Quick height check required!)
\( \angle A = 30^\circ \)
?
\( \angle A = 45^\circ \)
?
\( \angle A = 60^\circ \)
?
\( \angle A = 30^\circ \)
?
SSA Decision Tree Handout The SSA Decision Tree
Classifying Triangles with Precision
Variables
A = The given acute anglea = Side opposite the anglec = Side adjacent to the angle
Step 1: Find Height
\( h = c \sin A \)
\( a < h \)
0 Triangles
"The side is shorter than the height."
\( a = h \)
1 Triangle (Right)
"The side is exactly the height."
\( h < a < c \)
2 Triangles
"The side can swing inside or outside."
\( a \geq c \)
1 Triangle (Obtuse/Acute)
"The side is longer than the fixed side."
Pro Tip
Always find height first. If \( a \geq c \), you don't even need to find the height to know there's exactly 1 triangle (provided \( a \) is long enough to exist).
The Predictor Exit Ticket EXIT TICKET: The Predictor
Predict the number of triangles (0, 1, or 2).
NAME: ________________
\( \angle A = 30^\circ, c = 10, a = 4 \)
Solutions:
\( \angle A = 45^\circ, c = 8, a = 12 \)
Solutions:
\( \angle A = 60^\circ, c = 20, a = 18 \)
Solutions:
\( \angle A = 30^\circ, c = 14, a = 7 \)
Solutions:
Briefly explain your reasoning for Problem #3. Why did you choose that number of solutions?
Real World Ambiguity Slides Signal Ambiguity
Real-World Applications of SSA
The Radar Conflict
A radar station tracks a plane. It knows the angle of the signal and the distance to the plane.
But if the plane's flight path isn't fixed, that distance might match two different locations .
POS A POS B
Which One is Real?
In real life, we usually have "context clues" to eliminate one of the two triangles:
Geography
"The plane is reported to be over the ocean, not the city."
History
"The plane was at 10,000 ft last minute, so it must be at the closer point."
Heading
"The pilot is flying North-East, eliminating the Southern path."
The Case Study
A surveyor is measuring a plot of land. He stands at Point A and measures an angle of \( 40^\circ \). He knows the distance to a fence post (Point B) is 50 meters. He measures the distance from Point B to the corner of the house (Point C) as 35 meters.
\( \angle A = 40^\circ \)
Challenge:
Find both possible distances from Point A to Point C.
If the lot is known to be "shallow," which answer is correct?
Ambiguous Signal Worksheet The Signal Conflict
Resolving Real-World Ambiguity
NAME:
DATE:
Scenario 1
Property Line Survey
A surveyor at Point A measures an angle of \( 38^\circ \) between a North-running fence and a landmark at Point C. He knows the distance from Point A to a secondary marker (Point B) is 60 meters. The distance from Point B to the landmark at Point C is 45 meters.
Solution A (Acute Case)
Find \(\angle B_1\), \(\angle C_1\), and distance AC:
Solution B (Obtuse Case)
Find \(\angle B_2\), \(\angle C_2\), and distance AC:
The Context Clue:
"The landmark at Point C is known to be situated more than 70 meters away from the surveyor at Point A."
Which distance is correct? ________________
Scenario 2
Radar Ghosting
An air traffic controller sees a plane at an angle of \( 22^\circ \). The distance to the plane's known flight path entry (Point B) is 100 miles. The plane's transponder says it is currently 50 miles away from Point B.
1. Calculate both possible angles for the plane's current position (\( \angle C_1 \) and \( \angle C_2 \)):
2. Calculate both possible distances from the controller to the plane:
Resolution:
"The plane reported its position 5 minutes ago at 120 miles away. Based on its speed, it is getting closer to the station."
Which solution represents the plane's current position? Why?
Real World Answer Key Answer Key: The Signal Conflict
Geometry Unit 4 | Lesson 5
Scenario 1: Property Line Survey
\[ \frac{\sin 38^\circ}{45} = \frac{\sin B}{60} \implies \sin B \approx 0.8209 \]
Solution A (Acute)
\( \angle B_1 \approx 55.2^\circ \)
\( \angle C_1 = 180 - (38 + 55.2) = 86.8^\circ \)
\( b_1 = \frac{45 \sin 86.8^\circ}{\sin 38^\circ} \approx \mathbf{72.96 \text{ m}} \)
Solution B (Obtuse)
\( \angle B_2 \approx 124.8^\circ \)
\( \angle C_2 = 180 - (38 + 124.8) = 17.2^\circ \)
\( b_2 = \frac{45 \sin 17.2^\circ}{\sin 38^\circ} \approx \mathbf{21.6 \text{ m}} \)
Final Answer: 72.96 meters
Context: The landmark was "more than 70 meters away."
Scenario 2: Radar Ghosting
\[ \frac{\sin 22^\circ}{50} = \frac{\sin C}{100} \implies \sin C \approx 0.7492 \]
Trial 1 (Acute Angle at C)
\( \angle C_1 \approx 48.5^\circ \)
\( \angle B_1 = 180 - (22 + 48.5) = 109.5^\circ \)
\( b_1 = \frac{50 \sin 109.5^\circ}{\sin 22^\circ} \approx \mathbf{125.8 \text{ miles}} \)
Trial 2 (Obtuse Angle at C)
\( \angle C_2 \approx 131.5^\circ \)
\( \angle B_2 = 180 - (22 + 131.5) = 26.5^\circ \)
\( b_2 = \frac{50 \sin 26.5^\circ}{\sin 22^\circ} \approx \mathbf{59.6 \text{ miles}} \)
Final Answer: 59.6 miles
Context: The plane was at 120 miles and is "getting closer." The 125.8 mile solution would mean it moved further away.
