01 · Exact side
A = 45°, B = 60° and a = 4. Find b exactly.
Hint
Use sin 45° = √2/2 and sin 60° = √3/2.
Worked solution
b = 4(√3/2)/(√2/2) = 2√6
Understand · explore · practise
Sine rule and ambiguous case
Use the sine rule to find sides and angles. Understand why two triangles may fit the same data, and check zero, one or two possible solutions.
Before you startOpposite sides and angles, inverse sine and triangle angle sums
01 / Match opposite pairs
The sine rule needs a complete side–angle pair.
a / sin A = b / sin B = c / sin C
sin A / a = sin B / b = sin C / c
Choose the form that puts your unknown on top. Use one known side and its opposite angle, plus the relevant second side or angle. If only two angles are given, find the third using A + B + C = 180°.
Why the rule works
Drop a perpendicular from C to the line AB. Its height is h = b sin A = a sin B, including when the foot lies outside the side. Divide by sin A sin B to get a/sin A = b/sin B. Repeating with another altitude includes c/sin C.
02 / Find a side
The same ratio connects each side with its sine.
In a triangle, A = 40°, B = 65° and a = 9 cm. The third angle is C = 75°.
Find b and cWorked example
b/sin 65° = 9/sin 40°
b = 9 sin 65° / sin 40° ≈ 12.7 cm
The 9 cm side is opposite 40°, not 65°.
c = 9 sin 75° / sin 40° ≈ 13.5 cm
The biggest side is opposite 75°, the biggest angle.
03 / Find an angle
Inverse sine gives a candidate, not always the whole answer.
Suppose A = 30°, a = 7 and b = 10. Then sin B = b sin A / a = 5/7.
Acute B = sin⁻¹(5/7) ≈ 45.6°
Obtuse B = 180° − sin⁻¹(5/7) ≈ 134°
Both are between 0° and 180°, and both leave a positive third angle when combined with A = 30°. They give different triangles. Use unrounded angles to calculate the corresponding third side.
Acute case: B ≈ 45.5847°, C ≈ 104.4153°, c ≈ 13.6
Obtuse case: B ≈ 134.4153°, C ≈ 15.5847°, c ≈ 3.76
04 / Zero, one or two?
A circle can meet the base ray twice.
Fix A = 30° and AC = b = 10. Point B must lie on the positive horizontal ray from A and be distance a from C. Those are the intersections of the ray with a circle centred at C.
The perpendicular height of C is h = 10 sin 30° = 5. Compare the circle radius a with 5 and 10. Change a below to see no triangle, a right triangle, two triangles, or one triangle.
One set of data, two possible trianglesChoose and compare
a = 7 gives two triangles. The possible base lengths AB are about 3.761 and 13.559. The circle crosses the positive ray twice.
Watch the same data produce two triangles
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Test every candidate
A triangle needs three strictly positive angles.
- Calculate k = b sin A / a. If k > 1, there is no triangle.
- For 0 < k ≤ 1, find B₁ = sin⁻¹ k and B₂ = 180° − B₁.
- Keep only distinct candidates with A + B < 180°.
- Find C = 180° − A − B and c = a sin C / sin A for each surviving case.
At k = 1 the two candidates are both 90°, so count that triangle once. An angle sum of exactly 180° leaves C = 0° and is not a triangle.
An obtuse given angle
If A = 120°, a = 10 and b = 6, then sin B = 3√3/10 and B₁ ≈ 31.3°. The supplementary candidate is about 149°, which cannot fit with 120°. Only B₁ is valid. The side opposite an obtuse angle must be the longest.
06 / Exact ratios and algebra
Use a given sine directly.
You do not always need to find the angle first. If a = 3√2, sin A = √3/4 and B = 30°, then:
b = a sin B / sin A
= 3√2(1/2)/(√3/4) = 2√6
If side lengths contain an unknown, set up the ratio before solving. For A = 30°, B = 90°, a = x and b = x + 4, the sine rule gives x/(1/2) = (x + 4)/1, so x = 4.
Any alternative angle or additional condition still needs checking; using an exact sine does not remove ambiguity by itself.
07 / Your turn
Check the third angle before accepting an answer.
Lengths use one consistent unit. Give non-exact answers to three significant figures.
02 · Find the missing pair
A = 30°, B = 105° and c = 6√2. Find a.
Hint
First find C.
Worked solution
C = 45°
a = 6√2 sin 30° / sin 45° = 6
03 · Two triangles
A = 30°, a = 6 and b = 6√2. Find both possible pairs (B,C).
Hint
sin B = √2/2.
Worked solution
(B,C) = (45°,105°) or (135°,15°)
Both pairs leave a positive third angle.
04 · Impossible data
A = 30°, a = 4 and b = 10. How many triangles fit?
Hint
Calculate sin B.
Worked solution
sin B = 10(1/2)/4 = 5/4 > 1
No triangle fits.
05 · A repeated candidate
A = 30°, a = 5 and b = 10. Find B and C.
Hint
Count a 90° candidate only once.
Worked solution
sin B = 1 ⇒ B = 90°, C = 60°
Exactly one triangle.
06 · A degenerate alternative
A = 30° and a = b = 10. Which B values are valid?
Hint
Test 30° and 150° against the angle sum.
Worked solution
B = 30° ⇒ C = 120°
B = 150° ⇒ C = 0°
Only B = 30° gives a triangle.
07 · Unknown length
A = 30°, B = 90°, a = x + 1 and b = 3x − 2. Find x.
Hint
b must be twice a.
Worked solution
3x − 2 = 2(x + 1) ⇒ x = 4
The sides a = 5 and b = 10 are positive; C = 60°.
08 · An extra condition
A = 30°, a = 7 and b = 10, and B is known to be obtuse. Find B and C.
Hint
Use the supplementary inverse-sine candidate.
Worked solution
B = 180° − sin⁻¹(5/7) ≈ 134°
C = sin⁻¹(5/7) − 30° ≈ 15.6°
08 / Recap
An opposite pair opens the door; the angle sum decides.
- Pair each side with its opposite angle.
- For an angle, inspect both inverse-sine candidates.
- Reject a candidate that leaves no positive third angle.
- Count coincident 90° candidates once.
- Use exact sine values directly when they are supplied.
Section 1 of 8 · Match opposite pairs