SOHCAHTOA only works on right-angled triangles. The Sine Rule works on any triangle — as long as you know an angle and its opposite side. It links every angle to the side directly across from it.
Any triangle (no right angle needed)Angle opposite its sideFind unknown sidesFind unknown anglesAmbiguous case
Angle A is opposite side a | Angle B is opposite side b | Angle C is opposite side c
When to use the Sine Rule
You need one matched pair (an angle AND the side opposite it), plus one more piece of information:
✅ Use Sine Rule when you have:
• Two angles + one side (AAS or ASA)
• Two sides + an angle opposite one of them (SSA)
❌ Do NOT use when you have:
• Three sides only (SSS) → use Cosine Rule
• Two sides + the angle BETWEEN them (SAS) → use Cosine Rule
Why it worksThe key insight
Drop a perpendicular h from C to AB. Then:
sin A = h/b → h = b·sin A
sin B = h/a → h = a·sin B
So: b·sin A = a·sin B
Divide both sides: a/sin A = b/sin B ✓
💡 The ratio side ÷ sin(opposite angle) is the same for all three sides of any triangle. That’s the whole Sine Rule!
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The Sine Rule Formula
Two equivalent forms — one for finding sides, one for finding angles. Use the form that puts the unknown on top.
Reading the formulaYou only ever use TWO of the three fractions at once
a/sin A = b/sin B → use when finding a or b
a/sin A = c/sin C → use when finding a or c
b/sin B = c/sin C → use when finding b or c
💡 Pick the pair that contains your unknown. The third fraction is irrelevant — ignore it!
Labelling ruleAlways label the triangle first
Label the three vertices A, B, C
Label the side opposite each vertex with the lowercase letter: side a is opposite angle A, etc.
Identify which angle-side pair you already know (both the angle AND its opposite side)
Set up the equation using that pair plus the unknown
⚠️ You must have one complete pair (angle + opposite side) before you can use the Sine Rule. Without it, the formula has two unknowns.
📏
Finding an Unknown Side
Use a/sin A = b/sin B. Put the unknown side on the left, cross-multiply, then divide.
Strategy: Set up a/sin A = b/sin B → unknown = known side × sin(unknown angle) ÷ sin(known angle)
Step-by-step method
Identify the known pair: an angle AND the side opposite it
Identify the unknown side and its opposite angle
Write: unknown / sin(its angle) = known side / sin(its angle)
Multiply both sides by sin(unknown angle)
Calculate: unknown = known side × sin(unknown angle) / sin(known angle)
Check using the third side or angles
More worked examples
Find c: A=55°, C=72°, a=8 cm
Known pair: (A=55°, a=8)
Write: c/sin72° = 8/sin55°
Rearrange: c = 8×sin72°/sin55°
Calculate: c = 8×0.9511/0.8192
Answer:c = 9.28 cm ✅
Find b: B=38°, C=104°, c=15 m
Known pair: (C=104°, c=15)
Write: b/sin38° = 15/sin104°
Rearrange: b = 15×sin38°/sin104°
Calculate: b = 15×0.6157/0.9703
Answer:b = 9.52 m ✅
🧭
Finding an Unknown Angle
Use sin A / a = sin B / b. Cross-multiply, divide, then use sin⁻¹ to get the angle.
A roof truss has two rafters meeting at the ridge. The left rafter makes 42° with the horizontal, the right rafter makes 38° with the horizontal. The horizontal span between the two wall plates is 9.6 m.
Find the length of each rafter.
Triangle with angles at each end of the base
Setup: Triangle. Base (wall plate) = 9.6 m. Left angle = 42°, right angle = 38°
Ridge angle: C = 180°−42°−38° = 100°
Known pair: (C=100°, c=9.6 m)
Left rafter a: a/sin42° = 9.6/sin100° → a = 9.6×0.6691/0.9848
Left rafter:a = 6.53 m ✅
Right rafter b: b/sin38° = 9.6/sin100° → b = 9.6×0.6157/0.9848
Right rafter:b = 6.01 m ✅
🚚 Problem 3 — Navigation Bearing
A ship leaves port P and travels to lighthouse A, which is 15 km away on a bearing of N 40° E. From A, the ship heads to dock B. The angle at A (PAB) is 112° and AB = 22 km. Find the distance PB (straight line back to port).
Find the angle at B first, then use Sine Rule for PB
Triangle PAB: PA=15 km, AB=22 km, angle A=112°
Using cosine for PB²: PB²=15²+22²−2(15)(22)cos112°
PB²: 225+484−660×(−0.3746) = 709+247 = 956
PB:PB = √956 = 30.9 km ✅
Note: This needed the Cosine Rule — a reminder to check which tool fits!
⚙
Word Problems — Engineering Contexts
Force triangles, linkage mechanisms, structural frames — all use the Sine Rule when angles are not 90°.
⚖️ Problem 1 — Force Triangle
Three forces act at a point and are in equilibrium. The force triangle has angles of 55°, 68° and 57°. The largest force is 480 N. Find the other two forces.
Force triangle — each side represents a force magnitude
Check angles: 55+68+57 = 180° ✓
Largest force is opposite the largest angle (68°): F₁=480N, angle=68°
A crank arm AB is 80 mm long and makes 35° with the horizontal. The connecting rod BC is 200 mm long. Find the angle the connecting rod makes with the horizontal (angle at C with the slider line).
Triangle ABC — find angle at C
Given: AB=80mm (crank), BC=200mm (rod), angle at A=35°
Where this topic is used in engineering, manufacturing, maintenance and daily life.
🏗️Construction:Roof truss rafter lengths in non-right triangles.
🚢Navigation:Triangulation: fix position from two known bearings.
📡Electrical:Locate transmitter from two receiving stations.
🌍Surveying:Measure inaccessible distances across terrain.
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Engineering Applications — Sine Rule
Real-world uses across construction, mechanical, electrical and structural engineering. Each application uses exactly the same method: identify the known pair, write the formula, calculate.
🧱 Hip Roof Rafter Length
A hip roof has a common rafter angle of 34° at the wall plate and a hip rafter angle of 27°. The common rafter span is 4.2 m. Find the hip rafter length.
Known pair: common rafter angle 34°, span = 4.2 m opposite the ridge angle Ridge angle = 180°−34°−27° = 119°
Hip/sin34° = 4.2/sin119° Hip = 4.2 × sin34°/sin119° Hip = 4.2 × 0.5592/0.8746 Hip rafter = 2.68 m
📐 Traverse Survey — Missing Line
A surveying traverse has a closing line. At station A, the bearing deflection is 48°. At station B, the deflection is 61°. The measured line AB = 85 m. Find the closing line AC.
Angle at C = 180°−48°−61° = 71° Known pair: (C=71°, c=AB=85m)
AC/sin61° = 85/sin71° AC = 85 × sin61°/sin71° AC = 85 × 0.8746/0.9455 AC = 78.6 m
🔧 Slider-Crank Phase Angle
A crank AB = 60 mm, connecting rod BC = 180 mm. At a crank angle of 40°, find the angle the connecting rod makes with the cylinder axis (line AC).
Check: 180°−40°−12.4°=127.6° at B Obtuse check: 40°+167.6°>180° ✓ one solution
⚡ Fillet Weld Throat — Joint Geometry
A fillet weld has a leg length of 8 mm and the included angle between the joined plates is 75°. Find the throat thickness (shortest path across the weld).
The weld triangle: two equal legs = 8mm, angle between them = 75° The remaining angles = (180°−75°)/2 = 52.5°
Standard approx: throat ≈ 0.7 × leg = 5.6mm (assumes 90° joint — real geometry matters!)
🏗️ Truss Member Forces (Method of Sections)
A Pratt truss has a diagonal member at 55° to the horizontal chord. A vertical member makes 90° with the chord. A load creates a force triangle where the known force is 12 kN opposite the 35° angle. Find the diagonal member force.
From instrument station I, two control pegs A and B are visible. IA = 56 m, IB = 44 m, and the angle AIB = 83°. A new peg C lies on line AB such that IC bisects angle AIB. Find IC.
Angle IAC: use triangle IAB first. AB²=56²+44²−2(56)(44)cos83° [Cosine Rule] AB=65.4m
Then in triangle IAC, angle AIC=41.5°, angle IAC found via Sine Rule in IAB: sinA/44=sin83°/65.4 → A=41.3°
IC/sinA = IA/sin(AIC) IC=56×sin41.3°/sin(180°−41.5°−41.3°) IC=56×0.659/sin97.2°=36.9/0.992 IC = 37.2 m
🏎️ Cam Follower — Pressure Angle
A cam has a base circle radius of 25 mm. The follower arm is 80 mm long, pivoted 40 mm from the cam centre. The instantaneous pressure angle (cam to follower) is 22°. Find the follower arm angle.
Two bevel gears mesh with a shaft angle of 90°. The gear ratio is 2:1 (driven:driver). Find the pitch cone angles of each gear.
For bevel gears at 90°: tan(pitch angle driver) = z_driver/z_driven = 1/2 → pitch angle = tan⁻¹(0.5) = 26.6°
Driven pitch angle = 90°−26.6° = 63.4°
Verify with Sine Rule on pitch triangle: sin(26.6°)/1 = sin(63.4°)/2 0.4472/1 ≈ 0.8944/2 = 0.4472 ✓ Driver: 26.6° Driven: 63.4°
🧮
Sine Rule Solver
Enter any two angles and one side (or two sides and one angle) to find all missing values — with every step of the working shown clearly.
📐 Sine Rule Calculatora/sin A = b/sin B = c/sin C
ℹ️ How to use: Use the dropdown to choose what you want to find, then type the known values into the boxes. The full sine rule working appears instantly. Leave unknown values blank.
Select what you want to find and enter the known values above.
👆
Try:
a=6, A=30°, B=90° → b ·
a=8, b=12, A=40° → B ·
a=10, A=45°, B=75° → all
The Sine Rule
a/sin A = b/sin B = c/sin C
To find a side: a = b ⋅ sin A / sin B
To find angle: sin B = b ⋅ sin A / a
Then: B = sin⁻¹(b ⋅ sin A / a)
💡 Use the sine rule when you know: (1) two angles + one side, or (2) two sides + the angle opposite one of them.
Ambiguous case warning
⚠ When finding an angle, there can be two solutions: if sin B = 0.6 then B = 36.9° or B = 143.1° (= 180° − 36.9°). Always check which fits the triangle (angles must sum to 180°).
✅ The solver checks both possibilities automatically and tells you which are valid.
Third angle: C = 180° − A − B
Area = ½ a b sin C
🎮
Quick Fire Quiz
Test your knowledge — 10 questions, instant feedback.
⚡ Sine Rule Blitz
ℹ️ How to play: A sine rule question appears. Click the correct answer from the four options. Score and streak update after each — click Next to continue.
Score: 0Streak: 0 🔥Q 1/10
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Score: 0/10
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Practice Questions
Always draw and label the triangle first. State which pair you’re using. Show every step.
🟢 Tier 1 — Foundations
ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer to reveal the full worked solution and mark scheme.
Q1
In triangle ABC: A=50°, B=70°, a=8 cm. Find b.
[3]
Known pair (A,a). Use a/sinA = b/sinB.
b/sin70° = 8/sin50° b = 8×sin70°/sin50° = 8×0.9397/0.7660 b = 9.81 cm
Q2
In triangle ABC: B=42°, C=68°, c=15 m. Find b.
[3]
b/sin42° = 15/sin68° b = 15×sin42°/sin68° = 15×0.6691/0.9272 b = 10.83 m
C=180°−35°−85°=60°. Known pair (C=60°,c=20). a/sin35°=20/sin60° → a=20×0.5736/0.8660=13.24 m b/sin85°=20/sin60° → b=20×0.9962/0.8660=23.01 m
Q5
Is it possible to have a triangle with a=20, b=14, B=110°? Use the Sine Rule to check.
[2]
sinA/20 = sin110°/14 sinA = 20×0.9397/14 = 1.342 > 1 No — impossible. No such triangle exists.
🟡 Tier 2 — Word Problems
Q6
A triangular plot has angles 62°, 48° and 70°. The side opposite the 70° angle is 85 m. Find the other two sides.
[4]
Known pair (70°,85m). a/sin62°=85/sin70° → a=85×0.8829/0.9397=79.9 m b/sin48°=85/sin70° → b=85×0.7431/0.9397=67.2 m
Q7
A crane jib (arm) is 12 m long. It makes 72° with the vertical mast. A tie rope connects the top of the jib to a point on the mast 8 m below the pivot. Find the angle the rope makes with the mast.
[4]
Triangle: jib=12m, mast segment=8m, angle at pivot=72°. sinθ/12 = sin72°/8 → wait, need rope length first. Actually angle at pivot between mast and jib = 72°. sinB/12 = sin72°/rope. Use cosine for rope first: rope² = 12²+8²−2(12)(8)cos72° = 144+64−192×0.309 = 148.7 → rope=12.19m sinB/12=sin72°/12.19 → sinB=12×0.9511/12.19=0.9357 B = sin⁻¹(0.9357) = 69.4°
Q8
In a steel frame, member AB=6.5 m, member BC=9.0 m, and angle BAC=35°. Find angle ABC.
A force of 650 N acts at 40° to the horizontal. It is resolved into two components along directions 25° and 90° to the horizontal. Find both components using the Sine Rule on the force triangle.
[4]
Angles in force triangle: 40°−25°=15° between force and first component direction. 90°−40°=50° between force and second component direction. Third angle = 180°−15°−50°=115°. Known: resultant=650N opposite 115°. F₁/sin50°=650/sin115° → F₁=650×0.766/0.906=549 N F₂/sin15°=650/sin115° → F₂=650×0.259/0.906=186 N
Q10
Investigate the ambiguous case: a=10, b=12, A=55°. Find all possible values of angle B and side c.
A roof truss forms a triangle. The left rafter is 5.8 m, the right rafter is 6.4 m, and the angle at the ridge (apex) is 36°. Find both base angles and the span (base of the triangle).
[5]
C=36° (ridge), a=6.4m (opp A), b=5.8m (opp B). Need span c. Use Cosine Rule first: c²=5.8²+6.4²−2(5.8)(6.4)cos36°=33.64+40.96−74.24×0.809=33.64+40.96−60.05=14.55 c=√14.55=3.81 m span sinA/6.4=sin36°/3.81 → sinA=6.4×0.5878/3.81=0.9872 → A=80.8° B=180°−36°−80.8°=63.2°
Q12
Two observers at A and B are 500 m apart on level ground. They both measure the angle of elevation to the top of a tower C. Observer A measures 28°, observer B measures 41°. The tower is between them. Find the height of the tower.
[6]
Triangle ABC (horizontal): angle at A=28°, angle at B=(180°−41°)=139° (since observer looks back), AB=500m. Angle at C=180°−28°−139°=13°. BC/sin28°=500/sin13° → BC=500×0.4695/0.2250=1043 m. Height h=BC×sin41°=1043×0.6561=h=684 m. Check: AC/sin139°=500/sin13° → AC=500×0.6561/0.2250=1458m. h=AC×sin28°=1458×0.4695=684m ✓