The idea, step by step
Match each side to its opposite angle
Lower-case a names the side opposite angle A, and similarly for b/B and c/C. A known complete side-angle pair supplies the reference for the sine rule.
Rearrange to put an unknown side alone before calculating. If an angle is unknown, isolate its sine and apply inverse sine.
SSA can be ambiguous
Sine has the same value at an acute angle and its supplement. When two sides and a non-included angle are known, inverse sine may therefore miss a second possible triangle.
Test both candidates using the angle sum. A candidate with A+B at least 180° is impossible. If the calculated sine exceeds 1, there is no triangle; do not round it down to force an answer.
A common trap
Accepting only the inverse-sine angle.
Instead: For SSA, test the supplementary candidate and reject it only if the geometry rules it out.
See the thinking
Worked example 1
A=40°, B=65° and b=12 m. Find a to one decimal place.
- The complete pair is b=12 opposite B=65°.
- a/sin40°=12/sin65°.
- a=12sin40°/sin65°≈8.5 m.
Worked example 2
A=30°, a=8 cm and b=10 cm. Find every possible B.
- sinB=10sin30°/8=0.625.
- The first candidate is 38.6822°; its supplement is 141.3178°.
- Both candidates leave a positive third angle, so both triangles fit.
Check your understanding
A=70° and B=45°. Find C.
Show answer
65°
C=180°−70°−45°=65°.
Your turn
6 questions · 17 marks · Work at your own pace.
Read the questions online
- A=50°, B=70° and a=9 cm. Find b to one decimal place.2 marks
- A=70°, a=20 cm and b=15 cm. Find B to one decimal place and explain whether its supplement fits.3 marks
- A=45°, B=60° and a=10 m. Find C and c.3 marks
- A=30°, a=10 cm and b=14 cm. Find both B angles to one decimal place.3 marks
- A=50°, a=8 cm and b=13 cm. Does a triangle exist? Explain.3 marks
- A=35°, B=85° and c=18 m. Find a to one decimal place.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- A=50°, B=70° and a=9 cm. Find b to one decimal place.2 marks
b=9sin70°/sin50°.
b≈11.0 cm.
11.0 cm - A=70°, a=20 cm and b=15 cm. Find B to one decimal place and explain whether its supplement fits.3 marks
sinB=15sin70°/20.
The supplementary candidate plus A=70° exceeds 180°.
44.8°; the supplementary angle does not fit. - A=45°, B=60° and a=10 m. Find C and c.3 marks
C=180°−45°−60°=75°.
c=10sin75°/sin45°.
C=75°; c≈13.7 m - A=30°, a=10 cm and b=14 cm. Find both B angles to one decimal place.3 marks
sinB=14sin30°/10=0.7.
Both inverse-sine and supplementary candidates leave a positive third angle.
44.4° or 135.6° - A=50°, a=8 cm and b=13 cm. Does a triangle exist? Explain.3 marks
sinB=13sin50°/8≈1.245>1.
No angle can have this sine.
No triangle exists. - A=35°, B=85° and c=18 m. Find a to one decimal place.3 marks
C=60°.
a=18sin35°/sin60°.
11.9 m
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.
Official curriculum source ↗