Part III
1. How many triangles did you draw in I.1? How many in I.2? Why was there a difference?
I.1 has one triangle; there is no possible triangle for I.2.
The lengths for sides I.1 and I.2 are reversed. So, in I.2, the side opposite the 100° angle is shorter than the
other side, making is impossible.
2. How many triangles did you draw in I.3?
1 triangle
3. Complete the table based on your work in Part II.
Situation
Number of
Type of Triangle(s) (i.e., acute, obtuse, or right)
Triangles
Note: If the angle measures very close to 90°, you may
assume that it is a right angle.
II.1.a
2
Acute, obtuse
II.1.b
1
Right
II.1.c
0
II.2.a
2
Acute, obtuse
II.2.b
1
Right
II.2.c
0
4. What do the situations II.1.a and II.2.a have in common?
Each situation produces one acute and one obtuse triangle.
5. What do the situations II.1.b and II.2.b have in common?
Each situation produces one right triangle.
6. What do the situations II.1.c and II.2.c have in common?
Neither situation produces a triangle. The side opposite the angle is too short to reach the remaining side.
7. BONUS Explain the relationship between the various triangles you drew in Part II. How can you
determine the number of triangles that exist when ∠A is acute and a < b? Hint: Compare a and b*sin A.
Compare the value of a to the value of b*sin A.
If a = b*sin A, then there is a right triangle. I.e., side a is exactly long enough to meet side b in a single point.
If a > b*sin A, then there are two possible triangles.
If a < b*sin A, then there are no possible triangles. I.e., side a is too short to reach side b.