△ The Cosine Rule

The Cosine Rule works on any triangle when you have two sides and the angle between them (SAS), or all three sides (SSS). It extends Pythagoras’ theorem to non-right-angled triangles.

SAS — find the third side SSS — find any angle Extends Pythagoras Works on any triangle

When to use the Cosine Rule

✅ Use Cosine Rule when you have:
• Two sides + the angle between them (SAS) → find the third side
• All three sides (SSS) → find any angle

❌ Cannot use when you have:
• Two angles + one side (AAS/ASA) → use Sine Rule
• Two sides + angle opposite one of them (SSA) → use Sine Rule

📌 The Sine Rule is usually simpler. Only switch to the Cosine Rule when the Sine Rule doesn’t work.

Link to PythagorasWhy the formula has that shape

Drop a perpendicular from C to AB (length h). Then:

h² = b² − x² (from the left sub-triangle)

a² = h² + (c−x)² (from the right sub-triangle)

Substituting h²: a² = b² − x² + c² − 2cx + x²

Since x = b·cosA: a² = b² + c² − 2bc·cosA ✔

💡 When A=90°: cos90°=0, so a²=b²+c² — this IS Pythagoras!

📐

The Cosine Rule Formulae

Two forms: one for finding a side (SAS), one for finding an angle (SSS). Choose by what you already know.

All three versionsOne for each side/angle

a² = b² + c² − 2bc·cosA

b² = a² + c² − 2ac·cosB

c² = a² + b² − 2ab·cosC

💡 The pattern is always: opposite² = other two² + other two² − 2(other two)(other two)·cos(angle between them)

Rearranged for angles

cos A = (b²+c²−a²) / 2bc

cos B = (a²+c²−b²) / 2ac

cos C = (a²+b²−c²) / 2ab

💡 Once you have cos A = (decimal), press cos⁻¹ (SHIFT + COS) on your calculator to get angle A.

⚠️ If cosA is negative, angle A is obtuse (between 90° and 180°). Your calculator handles this automatically — don’t panic!

📏

Finding an Unknown Side (SAS)

You know two sides and the angle between them. Use a² = b² + c² − 2bc·cosA, then take the square root.

Strategy: Substitute b, c, A into the formula → calculate a² → take √ → answer

Step by step

Identify the two sides (b and c) and the angle between them (A)
Write: a² = b² + c² − 2bc·cosA
Substitute the numbers and calculate 2bc·cosA carefully
Subtract: a² = b² + c² − 2bc·cosA
Take the square root: a = √(answer)

More worked examples

Find b: a=7, c=10, B=42°

Write: b²=a²+c²−2ac·cosB

Substitute: b²=49+100−2(7)(10)cos42°

Calculate: b²=149−140×0.7431=149−104.0=45.0

Answer: b=√45.0=6.71 ✅

Find c: a=15, b=11, C=105°

Write: c²=a²+b²−2ab·cosC

Substitute: c²=225+121−2(15)(11)cos105°

cos105°=−0.2588 (obtuse!)

Calculate: c²=346−330×(−0.2588)=346+85.4=431.4

Answer: c=√431.4=20.77 ✅

🧭

Finding an Unknown Angle (SSS)

You know all three sides. Use cosA = (b²+c²−a²) / 2bc, then apply cos⁻¹.

Strategy: Substitute all three sides → calculate cosA as a decimal → press cos⁻¹ on calculator

Step by step

Label sides a, b, c. Decide which angle to find.
Write cosA = (b²+c²−a²) / 2bc (or equivalent for B or C)
Substitute all three side values and calculate the fraction
Apply cos⁻¹ (SHIFT + COS) to get the angle
Find remaining angles using the Sine Rule (simpler)

⚠️ If cosA is negative, the angle is obtuse (more than 90°). cos⁻¹ gives the correct obtuse angle automatically.

Two more examples

Find A: a=12, b=8, c=7

Write: cosA=(64+49−144)/(2×8×7)

Numerator: 64+49−144=−31

cosA: −31/112=−0.2768

Answer: A=cos⁻¹(−0.2768)=106.1° (obtuse) ✅

Find C: a=5, b=6, c=4

Write: cosC=(25+36−16)/(2×5×6)

cosC: 45/60=0.75

Answer: C=cos⁻¹(0.75)=41.4° ✅

🏗

Word Problems — Construction & Surveying

Draw the triangle first. Label the sides a, b, c. Identify SAS or SSS. Pick the right formula.

🧱 Problem 1 — Roof Truss Diagonal

A triangular roof truss has two rafters of 6.8 m and 7.2 m. The angle between them at the ridge is 40°. Find the length of the horizontal tie (the base of the triangle).

SAS — two sides, included angle, find third side

Label: b=6.8m, c=7.2m, A=40° (at ridge). Find a (the tie).

Write: a²=b²+c²−2bc·cosA

Substitute: a²=46.24+51.84−2(6.8)(7.2)cos40°

Calculate: a²=98.08−97.92×0.7660=98.08−75.02=23.06

Answer: a=√23.06=4.80 m ✅

📍 Problem 2 — Surveying a Triangular Plot

A triangular plot has sides of 120 m, 85 m and 95 m. Find the largest angle of the plot.

SSS — find the angle opposite the longest side

Label: a=120m (longest), b=85m, c=95m. Find A (largest angle).

Write: cosA=(b²+c²−a²)/2bc

Substitute: cosA=(7225+9025−14400)/(2×85×95)

Calculate: cosA=1850/16150=0.1145

Answer: A=cos⁻¹(0.1145)=83.4° ✅

🚧 Problem 3 — Setting Out a Site

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From a site control point P, peg A is 45 m away and peg B is 60 m away. The angle APB (at P) is 72°. Find the distance AB.

SAS — PA=45, PB=60, angle P=72°

Write: AB²=PA²+PB²−2(PA)(PB)cosP

Substitute: AB²=2025+3600−2(45)(60)cos72°

Calculate: AB²=5625−5400×0.3090=5625−1668.6=3956.4

Answer: AB=√3956.4=62.9 m ✅

⚙

Word Problems — Engineering

Force triangles, linkage geometry, structural analysis — all use the Cosine Rule when you have SAS or SSS.

⚖️ Problem 1 — Resultant Force

Two forces of 350 N and 280 N act at a point with an angle of 65° between them. Find the magnitude of the resultant force.

Force triangle — SAS

Note: Angle in force triangle = 180°−65°=115° (supplementary angle)

Write: R²=350²+280²−2(350)(280)cos115°

Substitute: R²=122500+78400−196000×(−0.4226)

Calculate: R²=200900+82829=283729

Answer: R=√283729=532.7 N ✅

🔧 Problem 2 — Four-Bar Linkage

In a four-bar mechanism, crank AB=40 mm, coupler BC=90 mm. Angle ABC (at the pin) is 130°. Find the distance AC.

SAS — crank AB=40, coupler BC=90, angle=130°

Write: AC²=AB²+BC²−2(AB)(BC)cosB

Substitute: AC²=1600+8100−2(40)(90)cos130°

cos130°=−0.6428: AC²=9700−7200×(−0.6428)=9700+4628=14328

Answer: AC=√14328=119.7 mm ✅

🏗️ Problem 3 — Steel Frame Analysis

Steel Frame: a=5.1m, b=4.6m, c=3.2m A=79.4° B=62.2° C=38.4° a=5.1m b=4.6m c=3.2m

A triangular steel frame has members of 3.2 m, 4.6 m and 5.1 m. Find all three angles.

SSS — find all angles using Cosine Rule then Sine Rule

Label: a=5.1m (longest), b=4.6m, c=3.2m

Find A (largest): cosA=(21.16+10.24−26.01)/29.44=5.39/29.44=0.1832 → A=cos⁻¹(0.1832)

A: 79.4°

Find B: sinB/4.6=sin79.4°/5.1 → sinB=4.6×0.9823/5.1=0.8860 → B=62.2°

C: 180°−79.4°−62.2°=38.4° ✅

⚙️

Applications

Where this topic is used in engineering, manufacturing, maintenance and daily life.

🏗️Structural: Diagonal brace in a truss with no right angles.
⚡Electrical: Phasor resultant of two AC voltages at an angle.
🚢Navigation: Distance between two GPS points on a course change.
🔧Mechanical: Resultant when two forces act at an angle.

🔨

Engineering Applications — Cosine Rule

The Cosine Rule solves SAS and SSS problems across all engineering disciplines. These are the contexts you will meet in T Level / Level 3 assessments and professional practice.

🏗️ Diagonal Brace in a Structural Frame

A rectangular frame is 2.4 m wide and 1.8 m tall. A diagonal brace is fitted. The brace meets the bottom chord at 52° and a vertical member at 38°. A second diagonal is 3.1 m long. Find the length of the first diagonal.

SAS: vertical=1.8m, horizontal=2.4m, angle=90°
But with Pythagoras: d=√(2.4²+1.8²)=√(5.76+3.24)=√9=3.0m

For non-rectangular frames (angle≠90°), use Cosine Rule.
Example: sides 2.4m, 1.8m, included angle 105°:
d²=5.76+3.24−2(2.4)(1.8)cos105°
d²=9+8.64×0.2588=9+2.237=11.24
d = √11.24 = 3.35 m

🛠️ Weld Inspection — Overlap Joint Geometry

Two plates overlap and are welded. The upper plate is 6 mm thick, the lower 10 mm. The weld toe is 14 mm from the overlap root. The angle between the plate faces at the root is 72°. Find the weld face length.

SSS triangle at the joint:
a=6mm (upper plate thickness)
b=10mm (lower plate thickness)
c=14mm (root-to-toe distance)

cosA=(100+196−36)/(2×10×14)=260/280=0.9286
A=cos⁻¹(0.9286)=21.8°

Weld face/sinA=14/sin(check angle)
Joint angle verified = 21.8°

⚡ Three-Phase Power — Phasor Triangle

In a three-phase circuit, two voltage phasors V₁=230 V and V₂=230 V are separated by 120°. Find the resultant (line voltage).

SAS: V₁=230, V₂=230, angle=120°

V_L²=230²+230²−2(230)(230)cos120°
V_L²=52900+52900−105800×(−0.5)
V_L²=105800+52900=158700
V_L=√158700
V_L = 398.4 V ≈ 230√3 ✓

This confirms the √3 ratio between phase
and line voltage in three-phase systems.

🔧 Connecting Rod — Minimum Bearing Clearance

A connecting rod has big-end radius 35 mm, small-end radius 22 mm, and centre-to-centre length 180 mm. The crank pin is 40 mm from the crankshaft axis. At a crank angle of 120°, find the distance between piston pin and crankshaft axis.

Triangle: crank=40mm, rod=180mm, angle at crank=120°
SAS → Cosine Rule:
d²=40²+180²−2(40)(180)cos120°
d²=1600+32400−14400×(−0.5)
d²=34000+7200=41200
d = √41200 = 202.9 mm

📡 Satellite Dish Alignment

Two ground stations A and B are 850 km apart. A satellite S is tracked from both. AS = 1200 km, BS = 980 km. Find the angle at the satellite (angle ASB).

SSS: a=AB=850, b=AS=1200, c=BS=980
Find angle S (at satellite):

cosS=(AS²+BS²−AB²)/(2·AS·BS)
cosS=(1440000+960400−722500)/(2×1200×980)
cosS=1677900/2352000=0.7134
S = cos⁻¹(0.7134) = 44.5°

This is the angular separation of the two
stations as seen from the satellite.

🧲 Pipe Rack — Offset Calculation

A pipe must travel from point A to point B, offset 1.2 m horizontally and 0.8 m vertically using two equal-length angled sections meeting at point C. The bend angle at C is 150°. Find each section length.

The straight-line distance AB:
AB=√(1.2²+0.8²)=√(1.44+0.64)=√2.08=1.442m

Triangle ABC: AC=BC=L (equal sections),
angle ACB=150° → angles at A and B = 15° each

AB/sin150°=L/sin15°
L=1.442×sin15°/sin150°
L=1.442×0.2588/0.5
Each section = 0.747 m

🏗️ Bridge Truss — Panel Point Load

A Warren truss has inclined members of 2.5 m at 60° to the horizontal. A vertical load creates a force triangle. The horizontal reaction is 18 kN and the load acts at 90° to the chord. Find the inclined member force and the diagonal force.

Force triangle: horizontal=18kN, angle=60° (between
horizontal chord and inclined member)
Third angle = 180°−90°−60°=30°

F_inclined/sin90°=18/sin30°
F_inclined=18×1.0/0.5
Inclined member = 36 kN

F_diagonal/sin60°=18/sin30°
F_diagonal=18×0.866/0.5
Diagonal = 31.2 kN

🌡️ Heat Exchanger — Tube Pitch Angle

Tubes in a heat exchanger shell are on a triangular pitch. The tube outer diameter is 25 mm and the pitch (centre-to-centre) is 32 mm. Find the angle between adjacent tube centres as seen from a third tube centre.

Equilateral triangle (equal pitch = 32mm):
a=b=c=32mm

cosA=(32²+32²−32²)/(2×32×32)
cosA=1024/2048=0.5
A = cos⁻¹(0.5) = 60° ✓

This confirms equilateral triangle pitch gives
60° between centres — the most compact
tube arrangement in shell-and-tube exchangers.

🧩

Which Rule Should I Use?

The hardest part is choosing between Sine Rule and Cosine Rule. Use this flowchart every time.

Quick decision table

Given	Find	Use
2 angles + 1 side	other sides	Sine Rule
2 sides + opposite angle	angle or side	Sine Rule
2 sides + included angle	third side	Cosine Rule
3 sides	any angle	Cosine Rule

Typical exam question types

Roof truss — usually SAS or SSS → Cosine Rule
Surveying bearings — angle between two known distances → Cosine Rule (SAS)
Force triangles — often AAS or SSA → Sine Rule
Finding all angles of a triangular frame → Cosine Rule first, then Sine Rule for the remaining two
Navigation problems — depends on what’s given; draw first!

💡 Always draw the triangle before choosing. Label every known side and angle. The choice becomes obvious from the diagram.

🧮

Cosine Rule Solver

Enter any three known values to find all missing sides and angles — with every step of the working shown clearly.

📐 Cosine Rule Calculatora² = b² + c² − 2bc cos A

ℹ️ How to use: Use the dropdown to choose what you want to find, then type the known values into the boxes. The full step-by-step cosine rule working and a labelled triangle appear instantly.

I want to find

Select what you want to find and enter the known values above.

👆 Try: b=7, c=5, A=60° → a · a=3, b=4, c=5 → A · b=8, c=6, A=90° → all · a=5, b=7, c=9 → all

The Cosine Rule

a² = b² + c² − 2bc cos A (finding side)

cos A = (b² + c² − a²) / 2bc (finding angle)

Similarly: b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

💡 Use the cosine rule when you know: (1) all three sides (SSS), or (2) two sides and the included angle (SAS).

Which rule to use?

Two angles + any side → Sine Rule

Two sides + opposite angle → Sine Rule

Three sides (SSS) → Cosine Rule

Two sides + included angle (SAS) → Cosine Rule

✅ Right-angled triangle? Use SOHCAHTOA or Pythagoras first — they are simpler.

🎮

Quick Fire Quiz

Test your knowledge — 10 questions, instant feedback.

⚡ Cosine Rule Blitz

ℹ️ How to play: A cosine rule question appears. Click the correct answer from the four options. Score and streak update after each — click Next to continue.

Score: 0 Streak: 0 🔥 Q 1/10

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✏

Practice Questions

State which form you’re using (SAS or SSS) and show every step.

🟢 Tier 1 — Foundations

ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer for the full worked solution.

Q1

Find a: b=10, c=8, A=50°

[3]

SAS: a²=b²+c²−2bc·cosA

a²=100+64−2(10)(8)cos50°=164−160×0.6428=164−102.8=61.2
a=√61.2=7.82

Q2

Find angle A: a=9, b=7, c=6

[3]

cosA=(49+36−81)/(2×7×6)=4/84=0.0476
A=cos⁻¹(0.0476)=87.3°

Q3

Find c: a=14, b=11, C=78°

[3]

c²=196+121−2(14)(11)cos78°=317−308×0.2079=317−64.0=253.0
c=√253.0=15.9

Q4

Find angle B: a=8, b=12, c=9

[3]

cosB=(64+81−144)/(2×8×9)=1/144=0.0069
B=cos⁻¹(0.0069)=89.6°

Q5

Find c: a=6, b=8, C=120°

[3]

cos120°=−0.5 — be careful with the minus sign

c²=36+64−2(6)(8)cos120°=100−96×(−0.5)=100+48=148
c=√148=12.17

🟡 Tier 2 — Word Problems

Q6

Two forces of 420 N and 310 N act with 55° between them. Find the resultant.

[4]

Angle in triangle=180°−55°=125°
R²=420²+310²−2(420)(310)cos125°=176400+96100−260400×(−0.5736)
R²=272500+149305=421805
R=√421805=649.5 N

Q7

A triangular plot has sides 75m, 90m and 110m. Find all three angles.

[6]

a=110 (largest), b=90, c=75
cosA=(8100+5625−12100)/13500=1625/13500=0.1204 → A=83.1°
sinB/90=sin83.1°/110 → sinB=90×0.9927/110=0.8122 → B=54.3°
C=180−83.1−54.3=42.6°

Q8

A steel tie bar of length AB = ? connects two joints. Joint A is 3.8 m from pivot P, joint B is 5.2 m from P, and angle APB = 48°. Find AB.

[3]

AB²=3.8²+5.2²−2(3.8)(5.2)cos48°=14.44+27.04−39.52×0.6691=41.48−26.45=15.03
AB=√15.03=3.88 m

Q9

A surveyor measures triangle PQR: PQ=240m, QR=185m, PR=195m. Find angle Q.

[3]

cosQ=(PQ²+QR²−PR²)/(2·PQ·QR)=(57600+34225−38025)/88800=53800/88800=0.6059
Q=cos⁻¹(0.6059)=52.7°

Q10

Decide which rule to use, then solve: A=35°, B=72°, a=18m. Find b.

[3]

Matched pair (A=35°, a=18m) → SINE RULE
b/sin72°=18/sin35° → b=18×0.9511/0.5736
b=29.8 m

🔴 Tier 3 — T Level Challenge

Q11

A four-bar linkage has: fixed link=120mm, crank=45mm, coupler=110mm, follower=80mm. When the crank makes 60° with the fixed link, find the diagonal d (from crank-coupler pin to follower-fixed pin) and then find the follower angle.

[6]

Step 1: Triangle with crank=45, fixed=120, angle=60°
d²=45²+120²−2(45)(120)cos60°=2025+14400−10800×0.5=16425−5400=11025
d=√11025=105 mm
Step 2: Triangle with follower=80, fixed-remaining=120−d_proj, d=105mm
cosθ=(80²+120²−105²)/(2×80×120)=(6400+14400−11025)/19200=9775/19200=0.5091
θ=cos⁻¹(0.5091)=59.4°

Q12

From a ship S, lighthouse L is 18 km away on a bearing of 040°. A buoy B is 12 km from S on a bearing of 115°. Find the distance LB.

[5]

Angle LSB = 115°−40° = 75° (angle between the two bearings at S)
SL=18km, SB=12km, angle S=75° → SAS → Cosine Rule
LB²=18²+12²−2(18)(12)cos75°=324+144−432×0.2588=468−111.8=356.2
LB=√356.2=18.9 km

🎯 Score: Q1–5 foundations (15 marks) · Q6–10 word problems (19 marks) · Q11–12 challenge (11 marks)

👉 You now have all three trig rules: SOHCAHTOA (right triangles), Sine Rule (AAS/ASA/SSA), Cosine Rule (SAS/SSS).