Chapter 5 Project: Determining the Number of Triangles in the Ambiguous Case (SSA): A Construction Exploration

Question 1:

Sin C is the same in both triangles in the ambiguous case because Sin A is the same as Sin C.

Question 2:

The thing that I discovered about the number of triangles you can make given SSA (two sides and an angle is that if BC is < or =" AB,">

Obtuse Triangle

a. BC > AB makes one triangle

b. BC = AB makes zero triangle

c. BC <>

Acute Triangle

a. AC > BC > h makes two triangles

b. BC > AC > h makes one triangle

c. BC = h makes one right triangle

d. BC <>

Ch. 5 - Determining Number of Triangles

Exploration 1. Case #1
Exploration 1. Case #2, 3

Exploration 2. Case #1, 2


Exploration 2. Case #3

Exploration 3.


Question 1: Explain why sin C is the same in both triangles in the ambiguous case (exploration 3). This is why the Law of Sines is also ambiguous in this case.

SinC is the same in both triangle because SinA remains the same since according to the law of sin, sinA= sinC.

Question 2: What did you discover about the number of triangles you can make given SSA (two sides and an angle)? Summarize the "rules" for all cases.

If angel A is obtuse and...

• BC > AB....1 triangle
  
• BC = AB....0 triangle
• BC <>

If angel A is acute and....

• BC <>
• BC = h....1 triangle
• BC > AB > h....1 triangle
• AB > BC > h....2 triangles
  

# CH 5 How to Do Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration

In each of the three different cases, I constructed three triangles with a 120 degree angle but different relationships between the lengths of the sides.

In Case 1, where line BC was longer than line AB, I discovered that one triangle could be formed while keeping angle A at 120 degrees.
In Case 2, where line BC was equal to line AB, it was impossible for line segment AC to form since line segment BC was not long enough to connect with the dashed line. Therefore, no triangles were formed.
In Case 3, where line BC was shorter than line AB, the same problem in Case 2 arose. Line segment BC was not long enough to connect with the dashed line. Again, no triangles were formed.

In conclusion, I found out that in order for an obtuse triangle to be created, the following rules apply*:

• If BC > AB, then one triangle is possible.
• If BC = AB, then no triangles are possible.
• If BC < AB, then no triangles are possible.
*Note: The lettering scheme I use may be different than the lettering schemes used byothers. However, the same concepts I mention here still apply.
In my constructions:
• Line segment BC refers to the side opposite of the obtuse angle measuring 120 degrees
• Line segment AB refers to the leg of the angle measuring 120 degrees which touches line segment BC as mentioned above





These three cases here were based on acute triangles, and the relationships between the lengths of the sides and the height of the triangle. In each of the cases, in order for the triangle to still have an angle A measuring 30 degrees, point C of line segment BC would have had to connect with the line segment.

In Case 1, line segment BC was shorter than the triangle’s “minimum” height, so therefore no triangles were constructed. The dashed red circle I drew represents the fact that no matter where point C is placed, line BC will never connect with the dashed line.
In Case 2, line segment BC was equal to the height, so one triangle was constructed.
In Case 3, line segment BC was greater than line segment AB which was greater than the height. One triangle was possible.

In conclusion, in an acute triangle:
• If BC < h, then no triangles are possible.
• If BC = h, then one triangle is possible.
• If BC > AB > h, then one triangle is possible.



In Exploration 3, I drew a triangle where line AB was greater than line BC which was greater than the triangle’s height. I discovered that two triangles were possible while keeping an angle A measuring 60 degrees. The triangle in Exploration 3 is similar to the triangles in Exploration 2. If this triangle were an Exploration 2 triangle, then point C would be able to connect to the dashed line in two places, therefore creating two triangles that we can see here. Basically, there is enough room for line BC to exist in two places.
In conclusion, if in an acute triangle:
• If AB > BC > h, then two triangles are possible.

Questions
1. Explain why sin C is the same in both triangles in the ambiguous case (exploration 3). This is why the Law of Sines is also ambiguous in this case.
Let’s recall the values for the triangle in Exploration 3. Line AB equals 8 cm (in the Law of Sines formula, this would be side c), angle A equals 60 degrees, line BC equals 7 ½ cm (LoS formula: side a), and the height equals 7 cm.
The Law of Sines is ambiguous in this case since two different triangles will share part of the same Law of Sines formula since they have a congruent angle and two sides (otherwise known as the misleading SSA congruency conjecture). In the triangle in Exploration 3:
  sin 60 = sin C
     7.5           8
The Law of Sines here misleads us into thinking there is only one triangle, when two are possible. If one were to solve this equation they would only get one of the triangles. It would require one to solve the sin B/b part (which is where the ambiguity comes in), to get the full picture.
2. Explain from diagram 4 in Exploration 2 why a unique triangle is determined if BC < AB.
In acute triangles, a triangle can be constructed when BC < AB since point C of line BC has enough length to “connect” to AB. However, two triangles are not possible since unlike in Exploration 3, there is not enough “room” for point C to connect in two places without point C overshooting line AB.
3. What did you discover about the number of triangles you can make given SSA (two sides and an angle)? Summarize the "rules" for all cases.
Given SSA, the case in Exploration 3, two triangles can be made.
 
All in all:

• If in an obtuse triangle:




• If BC > AB, then one triangle is possible.


• If BC = AB, then no triangles are possible.


• If BC < AB, then no triangles are possible.




• If in an acute triangle:
  
  
  
  • If BC < h, then no triangles are possible.
  
  
  • If BC = h, then one triangle is possible.
  
  
  • If BC > AB > h, then one triangle is possible.
  
  
  • If AB > BC > h, then two triangles are possible.
  
  
  
  

Since other people use different lettering schemes, I will list how my line segments correspond to the Law of Sines formula:
• AB = c
• BC = a
• AC = b

Anna Algebra

Question 1: SinC remains at the same point because SinC is the vertex of the triangle, it determines the height of the triangle and the length of the sides.

Question 2: In an obtuse triangle, you would always want the sides add up to more than the hypotenuse for a triangle to be able to form. In an acute triangle, the height would always need to be the smallest, like BC>AC>h. The general rule to be able to make a triangle would be AC>BC>h.

How to do Determining the number of Triangles in the Ambiguous case (SSA): A construction exploration

Exploration 1a:

AC = 120 degrees

AB = 6 cm

BC = 7 cm

1) Draw a 120 degree angle, 7 cm, 6 cm, and 5 cm line (6 and 5 cm line also used for other two diagrams in exploration 1)

2) Copy angle A

3) Fix compass to length 6 cm. Place pointy end on A and pencil end on the intial side. This will be point B.

*Step 1-3 are also used to start the other two diagrams of exploration 1

4) Fix compass to 7 cm length. Place pointy end on B and make an arc that intersects the terminal side (ray AC) The intersection will be point C.

5) construct a perpendicular to form height.

Exploration 1b:

Exploration 1c:

1) Repeat Step 1-3 of first diagram of exploration 1
2) Fix compass to 5 cm length. Place pointy end on B and make an arc that intersects ray AC. The arc did not intersect AC, it intersected AB. Therefore creating a line, not a triangle.

Exploration 2a:

1)Draw 30 degree angle with a protractor

2) Copy 30 degree angle

3) Draw ray AC and AB

4) AC can be any length, in this case, I fixed the compass to length 7 cm

5) Construct a perpendicular from AC to ray AB to form a altitude

*These 5 steps are the same for the other two triangles in exploration 2

6) First Case, let BC be less then height, which eventually, it won't form a triangle, because if BC is less than h then BC won't touch ray AB (point B is nonexistence)

Exploration 2b:

Exploration 2C:

Exploration 3:

1) Draw a 30 degree angle, 7 cm line segment and 6 cm line segment

2) Copy the 30 degree angle
3) Fix compass to 7 cm length. Place pointy end of compass on angle A, then place pencil end somewhere along the terminal side of the angle. That point will be AC.
4) Fix compass to 6 cm length. Place pointy end on C, then place pencil end to the left of C and right of C. Make arcs. These will be the two B points.

5) Construct altitude by constructing a perpendicular bisector.
6) Measure h.

Questions:

1) Because sin A remains the same in the ambiguous case, sin c also remain the same

2) If angle A is obtuse, then side BC (long side) has to be greater than side AB in order to form a triangle. A triangle cannot form if BC is less than or equal to AB.

In a right triangle, side BC has to equal the height of the triangle. BC cannot be less than h.

BC could also be greater than h, but BC also has to be greater than AC as well and angle A has to be acute. Though this is not a right triangle, one obtuse triangle and two acute

Exploration 1:

Exploration 2:

Exploration 3:

Exploration 1

1.First we construct a 120 degree angle three times2.In the first triangle, we make line AC

3.On the second triangle, we make line AC>BC and this makes no triangles because the point C isn't long enough to reach ray A so a triangle could not be formed

4.On the third triangle, we make line BC=AC and this also does not make a triangle because point A is also point C therefore it cannot connect to ray A

The rule is on an obtuse angle, when AC

Exploration 2

1.First we construct three 30 degree triangles and construct a perpendicular bisector on each triangle

2.On the first triangle, we measure the height and draw line BC, making it longer than the height which makes one triangle

3.On the second triangle, we also measure the height and make line BC same is the height which makes one triangle

4.On the third triangle, we make line BC shorter than the height which forms no triangle

The rule is in an acute triangle, when BC>AC then 1 triangle, also when BC=AC then it is also one triangle

Exploration 3

1. First, we draw a 6cm line, 7cm line, 8cm line and a 30 degree angle.

2. Then we construct a 30 triangle, making line AC 7cm (copying the line), and making 2 BC lines that are both 6m long.

3. Then we construct a perpendicular bisector as the height

4. Measure the height

Then we see that we can have 2 triangles when line AC is greater than line BC and this is all greater than the height (AC>BC>h)

Chapter 5 Project: How to Do Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration

Exploration 1:

1. First, I drew a line that is over 6 cm and drew a point on that line and labeled it A.
2. Then, I constructed a 120 degree angle.
3. Afterwards, i made a 7 cm arch using a compass. Then the arch touched the 120 degree angle line, so I label it point C. *Note: in my Exploration 1, 1 the 7cm line is suppose to touch point B but it accidently got cut off.
4. I did the same for the other lengths using 5 and 6 cm.

Questions:

1. For 1, I discovered that there was one triangle that could have been made and for 2 and 3, i learned that zero triangles could be made.
3. The rule: BC > AB to make a triangle.



1. First, I drew a 30 degree angle with a protractor.
2. Then, i constructed a perpendicular bisector through point C and measured my height (3.5 cm).
3. Next, i copied a 30 degree angle and drew ray AC and AB.
4. Afterwards, I constructed a perpendicular bisector from AC to ray AB to form a right triangle.
5. I then found a value less than 3.5 cm (2 cm) and drew an arch from point C to see if a triangle could be formed. A triangle couldn't be formed.
6. I repeated steps 3-4 but for the next two triangles, I used a segment equal to the height of my triangle (3.5 cm) and a segment where BC>AC>h (7.3 cm).

Questions-

1. I discovered that i could make zero triangles in 6a, 1 right triangle in 6b, and 2 triangles in 6 c.
2. Rule: Triangles can only be made if its length is equal or greater than its height.




Exploration 3:

1. First, i drew a 30 degree angle with a protractor.
2. Then, I drew a 7 cm and 6 cm segment.
3. Next, I drew a line through the page and made a point and labeled it A.
4. I constructed a 30 degree angle with the line segment of 7 cm.
5. After that, I connected the points and labeled it Line segment AC= 7 cm.
6. Then, I used a 6 cm segment to draw arches on line AB-which should make two archs.
7. I connected these arches to point C, which then formed 2 triangles.
8. Finally, i created a perpendicular bisector through point C and measured the height.

The rule I got was AC>B>h

Questions:

1. SinA is the same as SinC
2. If BC is < or = AB, it can cause a right triangle or obtuse triangle to form.
3. BC = AB makes zero triangles
BC > AB makes one triangle
BC < AB makes zero triangles
BC < h makes zero triangles
BC = h makes one triangle
BC > AC > h makes two triangles
AC > B > makes two triangles

Chapter 5 Construction Exploration Project

1) Angle A stays the same in both triangles and sin (pi - C) = sin C so sin C stays the same in both triangles
2) A unique triangle is determined when BC = AB. BC and AB are the sides of the triangle and when we constructed a perpendicular from ray AC to ray AB, we made a 90° angle, which made it a right triangle. If we have BC, angle B and AB, we have a SAS case which determines only one unique triangle and the unique triangle is a isosceles right triangle.
3) IF angle A is obtuse then:
When BC > AB, 1 triangle can be formed
When BC = AB, no triangles can be formed
When BC < AB, no triangles can be formed IF angle A is acute then: When AC > BC > H, 2 triangles can be formed
When BC > AC > H, 1 triangle can be formed
When BC = H, 1 right t

Chapter 5 Project: How to Do Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration

In this exploration we copied angles using a compass and protratcor and tried froming triangles with acute and obtuse angles and different sides our conclusion are:

Angle A is obtuse:

BC>AB one triangle

BC=AB zero triangles

Angle A is Acute:

AC>BC>h two triangles

BC>AC>h one triangle

BC=h one triangle

BC

This is my exploration:

CH 5 Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration.