When you are given two sides and an angle that is not between them, the triangle may have two possible shapes, one, or none. Explain which case applies before you calculate, then check every angle you find against the angle sum of 180°.
This lesson extends choosing the sine or cosine rule. The idea that two angles share one sine also appeared in area from two sides and an included angle.
Why can there be two triangles?
The sine rule gives sin C = c × sin A ÷ a, and the inverse sine returns the acute angle. But an obtuse angle of 180° minus that value has the same sine. Both might fit.
To picture it, fix angle A and side c, and swing side a around like a gate.
If a is long enough, it touches the third side once. If it is in the middle range, it can touch in two places. If it is too short, it never touches.
A test you can do before calculating
Take the given angle A (acute), its opposite side a, and the other given side c.
- Find the height c × sin A.
- If a is less than that height, no triangle exists.
- If a equals the height, there is one right-angled triangle.
- If a is between the height and c, there are two triangles.
- If a is at least c, there is one triangle.
Worked example
In triangle ABC, angle A = 35°, BC = 6 cm and AB = 9 cm. Find angle C, and explain the situation.
Step 1, test: a = 6, c = 9. The height is 9 × sin 35° = 5.16. Since 5.16 < 6 < 9, there are two possible triangles.
Step 2, sine rule: sin C = 9 × sin 35° ÷ 6 = 0.8604.
Step 3, two angles: C = 59.4° or C = 180° − 59.4° = 120.6°.
Step 4, check the angle sum: 35° + 59.4° = 94.4°, which is less than 180°, so valid. 35° + 120.6° = 155.6°, which is less than 180°, so also valid.
Step 5, finish each triangle:
- If C = 59.4°, then B = 85.6° and AC = 6 × sin 85.6° ÷ sin 35° = 10.4 cm.
- If C = 120.6°, then B = 24.4° and AC = 6 × sin 24.4° ÷ sin 35° = 4.31 cm.
Answer: angle C is 59.4° or 120.6°, and the two triangles have AC = 10.4 cm or 4.31 cm.
The mistake to watch for
The usual slip is to stop at the calculator answer.
Mistaken answer: “C = 59.4°.”
The student did not test the situation, so the second valid triangle was never found.
The correction is to make the test in step 1 a habit: compare a with c sin A and c. When the test shows one triangle, the acute angle is enough. When it shows two, both need to be written and checked.
Check yourself
1. In a triangle, angle A = 40°, a = 5 cm and c = 10 cm. How many triangles are possible?
Show answer
c sin A = 10 × 0.6428 = 6.43. Since a = 5 is less than 6.43, the side is too short to reach. No triangle exists. (The sine rule would give sin C = 1.29, which is impossible.)
2. Angle A = 50°, a = 8 cm and c = 6 cm. How many triangles are possible, and what is angle C?
Show answer
Since a = 8 is greater than c = 6, there is one triangle. sin C = 6 × 0.7660 ÷ 8 = 0.5745, so C = 35.1°. The obtuse option 144.9° would make an angle sum of 194.9°, which is over 180°, so it is rejected.
3. Angle A = 30°, a = 7 cm and c = 10 cm. How many triangles are possible?
Show answer
c sin A = 10 × 0.5 = 5. Since 5 < 7 < 10, there are two triangles.
Where this leads next
Try a full set of mixed questions in the module practice set, where two of the questions have a second triangle. The triangle and bearings reasoning board asks for this kind of explanation before calculating, and the non-calculator working trainer supports the sine values.
Understanding why a second answer exists is different from remembering a rule about it. A teacher in online one-to-one Mathematics tuition can test that understanding with a diagram you draw yourself.