Geometric Twins · Lesson 5 of 10
Measuring Two Sides and a Non-included Angle
“See how the same two side lengths and a non-included angle can produce two different triangles.”
• Distinguish included and non-included angles for a pair of sides. • Construct the two possible triangles in the given SSA example. • Explain why generic SSA is not a congruence condition. • Decide when measurement information is insufficient for a proof.
Moving the angle changes the information
SAS worked because the given angle fixed the opening between the two measured sides. Now keep two measured sides but give an angle at a different vertex. One measured side lies opposite this angle rather than along one of its arms. This information is called SSA, or Side–Side–Angle. It may leave more than one possible triangle.
For sides PQ and QR, the included angle is angle PQR at Q. An angle at P, such as angle QPR, is non-included. The distinction depends on which two sides have been measured: the same angle can be included for one pair of sides and non-included for another pair.
| Measured sides | Included angle | A non-included angle |
|---|---|---|
| AB and AC | Angle BAC at A | Angle ABC at B |
| PQ and QR | Angle PQR at Q | Angle QPR at P |
| LM and MN | Angle LMN at M | Angle MLN at L |
Problem
AB = 6 cm, AC = 4 cm, and angle ABC = 30° are given. Is this SAS?
- 1.The measured sides AB and AC meet at A, so their included angle would be angle BAC.
- 2.The given angle ABC is at B, not A. It is non-included for the given sides.
- 3.The information has the SSA arrangement. We cannot call it SAS.
One circle meets the ray twice
Try the same arrangement with fresh labels: PQ = 6 cm, QR = 4 cm, and angle QPR = 30°. Draw PQ and make a 30° ray at P. The third vertex must lie on that ray to satisfy the angle, and it must be 4 cm from Q to satisfy the other side. A circle centred at Q with radius 4 cm shows all such locations.
In this construction the circle crosses the ray at two points, S and R. Joining Q to either point gives a triangle with the required measurements: PQ = 6 cm, QS = QR = 4 cm, and angle QPS = angle QPR = 30°. Yet S is nearer to P than R, so PS and PR differ. The two triangles have different shapes and cannot be superimposed exactly.
Problem
Why are triangles PQR and PQS above not congruent under the given pairing?
- 1.The base PQ is common and both QR and QS are radii of the same 4 cm circle.
- 2.R and S lie on the same ray from P, so both triangles have the same 30° angle at P.
- 3.But PS is shorter than PR. The remaining corresponding sides are unequal.
- 4.The construction gives a counterexample: two equal sides and an equal non-included angle do not always force congruence.
Insufficient information is a valid conclusion
A counterexample is one case that disproves a statement claimed to be always true. We have found such a case for generic SSA. This does not mean that every SSA construction makes two triangles. Depending on the measurements, the circle and ray may have no intersection, one suitable intersection, or two. It means that SSA by itself cannot be used as a general guarantee.
Compare this with SSS, SAS, ASA, AAS, and the right-triangle condition you will learn later. Those conditions state precisely when the required information forces exact copies. If a question gives generic SSA and no extra facts, the careful answer is that congruence is not established. Do not assume an unmarked right angle or another equality from the appearance of a sketch.
Problem
A student knows AB = DE, AC = DF, and angle B = angle E and writes ABC ≅ DEF by SAS. Is the reason valid?
- 1.In the first triangle, AB and AC meet at A. In the second, DE and DF meet at D.
- 2.The given angles are at B and E, so they are not between those pairs of sides.
- 3.This is generic SSA, not SAS. The given facts alone do not establish congruence.
- 4.An additional suitable fact, such as angle A = angle D, would allow SAS. It must be given or proved rather than guessed.
SSA being insufficient does not prove that the particular triangles are unequal. It means the stated facts do not guarantee that they are equal in shape and size.
Quiz
For sides AB and AC, which angle is non-included?
Why can the construction above produce two triangles?
Which sides differ in the displayed counterexample?
What is the careful conclusion from generic SSA alone?
Which extra fact would complete SAS for AB = DE and AC = DF?
Practice Problems
- Construct PQ = 6 cm and a ray at P making 30° with PQ. Draw the circle centred at Q with radius 4 cm and locate both intersections.
- Name the two triangles from your construction and list the three pieces of information they share.
- Measure their distances from P to the third vertex. Explain what the difference tells you.
- For sides RS and ST, name one included angle and one non-included angle.
- Explain the difference between “not enough information to prove congruence” and “proved not congruent”.
- A proof uses AB = DF, AC = DE, and angle B = angle F. Identify the arrangement and explain why SAS has not been established.
Key Takeaways
• The angle’s location matters as much as its value. • SSA gives two sides and a non-included angle. • A circle can meet the angle ray at two third-vertex positions. • The resulting triangles can have unequal remaining sides. • Generic SSA is not a sufficient congruence condition.