Law of Sines and Law of Cosines

When to Use These Laws

Right triangles: Use SOH-CAH-TOA

Non-right (oblique) triangles: Use Law of Sines or Law of Cosines

Standard notation:

• Angles: A, B, C (capital letters)
• Opposite sides: a, b, c (lowercase)
• Side a is opposite angle A, etc.

Law of Sines

Formula:

sin(A)/a = sin(B)/b = sin(C)/c

Or equivalently:

a/sin(A) = b/sin(B) = c/sin(C)

Use when you have:

• Two angles and one side (AAS or ASA)
• Two sides and an angle opposite one of them (SSA)

Example 1: AAS (Two Angles and Non-Included Side)

Given: A = 30°, B = 45°, a = 10 Find: b

Find angle C first:

C = 180° - 30° - 45° = 105°

Use Law of Sines:

a/sin(A) = b/sin(B)
10/sin(30°) = b/sin(45°)
10/0.5 = b/0.707
20 = b/0.707
b = 20(0.707) ≈ 14.14

Answer: b ≈ 14.14

Example 2: ASA (Two Angles and Included Side)

Given: A = 50°, B = 60°, c = 20 Find: a

Find C:

C = 180° - 50° - 60° = 70°

Use Law of Sines:

a/sin(50°) = 20/sin(70°)
a = 20 · sin(50°)/sin(70°)
a = 20(0.766/0.940)
a ≈ 16.3

Answer: a ≈ 16.3

Example 3: Find Angle

Given: a = 15, b = 20, A = 40° Find: B

Use Law of Sines:

sin(A)/a = sin(B)/b
sin(40°)/15 = sin(B)/20
sin(B) = 20 · sin(40°)/15
sin(B) = 20(0.643)/15
sin(B) ≈ 0.857
B = sin⁻¹(0.857)
B ≈ 59°

Answer: B ≈ 59°

The Ambiguous Case (SSA)

SSA can have:

• No triangle
• One triangle
• Two triangles

Depends on relationship between given angle, sides

Example: Two Triangles

Given: a = 10, b = 15, A = 30° Find: B (two possible values)

Use Law of Sines:

sin(B) = b · sin(A)/a
sin(B) = 15 · sin(30°)/10
sin(B) = 15(0.5)/10 = 0.75

Two angles with sin(B) = 0.75:

• B₁ = sin⁻¹(0.75) ≈ 48.6°
• B₂ = 180° - 48.6° ≈ 131.4°

Both create valid triangles (check A + B < 180°)

Law of Cosines

Three formulas (use appropriate one):

a² = b² + c² - 2bc·cos(A)

b² = a² + c² - 2ac·cos(B)

c² = a² + b² - 2ab·cos(C)

Use when you have:

• Three sides (SSS)
• Two sides and the included angle (SAS)

Note: Generalizes Pythagorean Theorem (when angle = 90°, cos = 0)

Example 1: SAS (Two Sides and Included Angle)

Given: a = 8, b = 10, C = 60° Find: c

Use Law of Cosines:

c² = a² + b² - 2ab·cos(C)
c² = 8² + 10² - 2(8)(10)cos(60°)
c² = 64 + 100 - 160(0.5)
c² = 164 - 80
c² = 84
c ≈ 9.17

Answer: c ≈ 9.17

Example 2: SSS (Three Sides)

Given: a = 7, b = 8, c = 9 Find: A

Rearrange for cos(A):

a² = b² + c² - 2bc·cos(A)
2bc·cos(A) = b² + c² - a²
cos(A) = (b² + c² - a²)/(2bc)
cos(A) = (64 + 81 - 49)/(2·8·9)
cos(A) = 96/144
cos(A) = 2/3
A = cos⁻¹(2/3)
A ≈ 48.2°

Answer: A ≈ 48.2°

Example 3: Find All Angles (SSS)

Given: a = 5, b = 7, c = 10 Find: All angles

Find angle C (opposite longest side):

cos(C) = (25 + 49 - 100)/(2·5·7)
cos(C) = -26/70
cos(C) ≈ -0.371
C = cos⁻¹(-0.371)
C ≈ 111.8°

Use Law of Sines for A:

sin(A)/5 = sin(111.8°)/10
sin(A) = 5·sin(111.8°)/10
A ≈ 27.7°

Find B:

B = 180° - 111.8° - 27.7° ≈ 40.5°

Choosing Which Law to Use

Law of Sines:

• Know two angles and one side (AAS, ASA)
• Know two sides and angle opposite one (SSA) - watch for ambiguous case

Law of Cosines:

• Know three sides (SSS)
• Know two sides and included angle (SAS)

Summary table:

Area of Triangle

With angle between two sides:

Area = (1/2)ab·sin(C)

Where a and b are sides, C is included angle

Example: Find Area

Triangle with sides 6 and 8, included angle 30°

Calculate:

Area = (1/2)(6)(8)sin(30°)
     = (1/2)(48)(0.5)
     = 12 square units

Real-World Applications

Surveying: Measuring distances when direct measurement impossible

Navigation: Finding distances and bearings

Engineering: Bridge construction, roof angles

Astronomy: Calculating distances to stars

Physics: Resolving force vectors

Example: Surveying

Two surveyors 100 m apart both sight a tree. From point A, angle to tree is 40°. From point B, angle to tree is 55°.

Find distance from A to tree.

Set up:

• AB = 100 m (base)
• Angle at A = 40°
• Angle at B = 55°
• Angle at tree C = 180° - 40° - 55° = 85°

Use Law of Sines:

c/sin(85°) = 100/sin(C at tree looking back)

Actually, let's be clearer:

• Angle CAB = 40°
• Angle CBA = 55°
• Angle ACB = 85°
• We want side a (opposite A, which is BC)

Wait, let me reconsider. If we want distance from A to tree (call it b):

b/sin(55°) = 100/sin(85°)
b = 100·sin(55°)/sin(85°)
b ≈ 82.2 m

Answer: About 82.2 meters from point A to tree

Special Cases

Obtuse triangles: Law of Cosines handles these well

When cos < 0: Angle is obtuse (> 90°)

Check validity: Sum of angles = 180°

Practice