A Tale of Three Intersecting Lines · Lesson 6 of 10
Two Angles and the Included Side
“Construct from a side and its endpoint angles, then discover when the two rays meet.”
• Identify the included side between two specified angles. • Construct a triangle with two angles and their included side. • Use parallel lines to recognise the boundary between possible and impossible angle pairs. • Apply the condition that two positive angles must sum to less than 180°. • Explain why changing the base length does not change whether the angle pair is possible.
Two Angles and the Included Side
Now the given information consists of a side and the angles at its two endpoints. After drawing that side, we draw one ray from each end. If those rays meet on the chosen side of the base, their meeting point supplies the third vertex. This is a different way to locate C: we use two directions rather than two distances.
The side joining the vertices of the two given angles. For given angles at A and B, the included side is AB.
For example, let AB = 5 cm, ∠A = 45°, and ∠B = 80°. These angles use AB as one arm at A and BA as one arm at B. The second ray at each endpoint must be drawn on the same side of the base to describe the intended triangle.
Constructing the Two Rays
At A, measure 45° from the ray AB. At B, measure 80° from the ray BA, which points back toward A. Because the starting directions are opposite, take particular care with the protractor at B. Extend the new rays sufficiently far to find their intersection rather than assuming short segments have already failed to meet.
- Draw AB = 5 cm.
- At A, construct a 45° ray above AB, measured from AB.
- At B, construct an 80° ray above AB, measured from BA.
- Name their intersection C. The segments AC and BC complete the triangle.
Problem
Construct ΔABC with AB = 5 cm, ∠A = 45°, and ∠B = 80°.
- 1.Draw the 5 cm included side AB.
- 2.Draw the 45° ray from A and the 80° ray from B on the same side of AB.
- 3.Extend them to their intersection C, then check both angles and AB.
- 4.C belongs to both required directions, so the completed triangle has the specified information.
Problem
Construct a triangle from angles 120° and 30° and included side 6 cm.
- 1.Draw PQ = 6 cm. Construct a 120° ray at P from PQ.
- 2.Construct a 30° ray at Q from QP, on the same side of the base.
- 3.Extend the rays until they meet at R. A very wide angle at one endpoint is allowed when the other is sufficiently small.
- 4.The angles add to 150°, leaving room for a positive third angle, as the following reasoning explains.
Do Triangles Always Exist?
The rays do not always meet. If both base angles are right angles, their new rays are parallel. If both are greater than a right angle, the rays lean away from each other. Even one acute angle and one obtuse angle may be too large together. We need a rule that includes all these cases.
Fix ∠A at 40°. At B, imagine rotating the other ray. The boundary position is the ray parallel to the one from A. AB crosses both parallel lines as a transversal, so the interior angles on the same side add to 180°. The angle at B in this boundary position is therefore 180° − 40° = 140°.
If ∠B is a positive angle smaller than 140°, its ray bends toward the first ray and they meet. At 140° the rays are parallel. Beyond 140°, they move apart on the chosen side of AB. This explains why 40° + ∠B must be less than 180° for a triangle to exist.
The reasoning does not depend on whether AB is 5 cm, 7 cm, or another positive length. A longer base can move the intersection farther away, but it does not change whether the rays approach each other, stay parallel, or move apart. The angles decide existence; the given side sets the size of the construction.
Problem
Decide whether pairs 35°, 150°; 70°, 30°; 90°, 85°; and 50°, 150° can occur in a triangle.
- 1.35 + 150 = 185, which exceeds 180, so the first pair is impossible.
- 2.70 + 30 = 100, which is less than 180, so the second pair is possible.
- 3.90 + 85 = 175, which is less than 180, so the third pair is possible, although it leaves only a small third angle.
- 4.50 + 150 = 200, so the final pair is impossible.
| First angle | Other angle must satisfy | Two possible examples | Two impossible examples |
|---|---|---|---|
| 30° | 0° < other < 150° | 50°, 90° | 150°, 170° |
| 70° | 0° < other < 110° | 60°, 80° | 110°, 120° |
| 54° | 0° < other < 126° | 64°, 90° | 126°, 154° |
| 144° | 0° < other < 36° | 20°, 35° | 36°, 90° |
A sum exactly equal to 180° does not work. The boundary rays are parallel, so they never supply a third vertex. Also, a pair with one zero or negative angle is invalid even if its sum is less than 180°.
Construct a triangle with angles 60° and 70° using a 5 cm included side, then repeat with a 7 cm included side. Measure the third angle in each. You should find the same value, about 50°; small differences come from drawing and measurement. The next lesson explains the exact reason.
Try several positive angle pairs whose sums are less than 180°. For each construction, record the first angle, second angle, included side length, measured third angle, and sum of all three measured angles in a table. Repeat some pairs with a different side length. Look for a relationship that stays the same despite changes in size; allow for small measurement errors.
Quiz
For given angles at A and B, which is the included side?
Which pair of angles can occur in a triangle?
With ∠A = 40°, which second angle makes the boundary rays parallel?
What condition must two positive angles satisfy?
If two fixed positive angles sum to less than 180°, does changing the included side from 5 cm to 7 cm make the pair impossible?
Practice Problems
- Construct triangles for 75°, 5 cm, 75°; 25°, 3 cm, 60°; and 120°, 6 cm, 30°, using the length as the included side.
- For a first angle of 30°, 70°, 54°, and 144°, give two possible and two impossible second angles in each case. Explain the boundary.
- Check angle pairs 35°, 150°; 70°, 30°; 90°, 85°; and 50°, 150° using the sum condition.
- Explain with a diagram why 40° and 140° do not make a triangle, however long the included side is.
- Construct the pair 60°, 70° with included sides 5 cm and 7 cm. Compare the measured third angles.
- A learner draws the angle at B from the rightward extension of AB instead of from BA. Explain why this changes the intended interior angle.
Key Takeaways
• The included side joins the two given angle vertices. • Construct each angle from its base ray and draw the new rays on the same side of the base. • Two positive angles can occur in a triangle exactly when their sum is less than 180°. • A sum of 180° gives parallel boundary rays and no triangle. • Changing a positive included side affects the size, but not the possibility, of a fixed valid angle pair.