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Law of Sines & Cosines - Interactive Lesson | Study Mondo
Law of Sines & Cosines - Complete Interactive Lesson Part 1: Law of Sines ⚖️ Law of Sines
Part 1 of 7
The Law of Sines lets us solve oblique triangles (no right angle) when we know an angle–side pair.
The Law of Sines
For any triangle with sides a a a , b b b , c c c opposite angles A A A , B B B , C C C :
a sin A = b sin B = c sin C \boxed{\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}} sin A a = sin B
Equivalently: sin A a = sin B b = sin C c \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} a s i n A = b
When to Use It
Known Case Name Law of Sines? Two angles + one side (AAS or ASA) Angle–Angle–Side ✅ Yes Two sides + angle opposite one (SSA) Side–Side–Angle ✅ Yes (watch for ambiguous case!) Two sides + included angle (SAS) Side–Angle–Side ❌ Use Law of Cosines Three sides (SSS) Side-Side-Side ❌ Use Law of Cosines
📝 Worked Examples
Example 1: AAS — Find a Missing Side
In △ A B C \triangle ABC △ A BC : A = 42 ° A = 42° A = 42° , B = 73 ° B = 73° B = 73° , a = 12 a = 12 a = .
🧠 Step-by-Step Strategy
Solving AAS/ASA Triangles
Find the third angle using A + B + C = 180 ° A + B + C = 180° A + B + C = 180°
Set up the proportion using the known angle–side pair
Cross-multiply and solve
Key Formula Rearrangements
To find side b b b : b = a sin B sin A
Solve Triangles 🧮
Round to the nearest integer.
1) △ A B C \triangle ABC △ A BC : A = 40 ° A = 40° A = 40° , B = 60 ° B = 60° B = 60° , a = 10 a = 10 a . Find in degrees. (e.g., if , then )
Part 2: Ambiguous Case ⚠️ The Ambiguous Case (SSA)
Part 2 of 7
When given two sides and an angle opposite one of them (SSA), there might be zero, one, or two possible triangles.
Why It's Ambiguous
Given a a a , b b b , and angle A A A :
We compute sin B = b sin A a \sin B = \frac{b \sin A}{a} sin B = .
Part 3: Law of Cosines 📐 Law of Cosines
Part 3 of 7
The Law of Cosines generalizes the Pythagorean theorem to any triangle — even those without a right angle.
The Law of Cosines
c 2 = a 2 + b 2 − 2 a b cos C \boxed{c^2 = a^2 + b^2 - 2ab\cos C} c 2 = a 2 +
Part 4: Area of Triangles 🔄 Choosing the Right Law & Combined Problems
Part 4 of 7
Knowing when to use Law of Sines vs Law of Cosines is half the battle. This part helps you develop that judgment.
Decision Guide
Do you have a complete angle–side pair (angle and its opposite side)? \boxed{\text{Do you have a complete angle–side pair (angle and its opposite side)?}} Do you have a complete angle–side pair (angle and its opposite side)?
Answer Case Method Yes + need another side/angleAAS, ASA, SSA Law of Sines No + have SASTwo sides + included angle Law of Cosines (find side)
Part 5: Applications 🌍 Applications — Navigation & Surveying
Part 5 of 7
The Laws of Sines and Cosines solve real-world problems involving distances and angles that can't be measured directly.
Application Types
Application Typical Setup Navigation Find distance between two points using bearings Surveying Measure inaccessible distances from two known positions Engineering Force resolution in non-rectangular systems Aviation Distance between waypoints on non-straight routes
The Triangulation Method
To find an inaccessible distance:
Measure a baseline (known distance between two observation points)
Measure the angles from each end of the baseline to the target
Use Law of Sines to compute the unknown distances
📝 Worked Examples
Example 1: Distance Across a River
From points A A A and on one bank, m. Angles to a tree on the other bank: , .
Part 6: Problem-Solving Workshop 📏 Advanced Problem-Solving Techniques
Part 6 of 7
This part covers problems that require creative setups, multi-step approaches, and combining both laws.
Advanced Techniques
Breaking complex shapes into triangles — add diagonals
Using supplementary angles in parallelograms
Computing distances in 3D by projecting onto 2D triangles
Inscribed circle radius : r = Area s r = \frac{\text{Area}}{s} r = s Area
Circumscribed circle radius :
Part 7: Review & Applications 🏆 Law of Sines & Cosines — Full Synthesis
Part 7 of 7
Master review covering all concepts: Law of Sines (including ambiguous case), Law of Cosines, area formulas, and applications.
Complete Decision Tree
Given info → Identify case → Choose law → Solve \boxed{\text{Given info} \to \text{Identify case} \to \text{Choose law} \to \text{Solve}} Given info → Identify case → Choose law → Solve
b
=
sin C c
s i n B
=
c s i n C
12
Step 1: Find C C C : C = 180 ° − 42 ° − 73 ° = 65 ° C = 180° - 42° - 73° = 65° C = 180° − 42° − 73° = 65°
Step 2: Use Law of Sines to find b b b :
a sin A = b sin B \frac{a}{\sin A} = \frac{b}{\sin B} s i n A a = s i n B b
12 sin 42 ° = b sin 73 ° \frac{12}{\sin 42°} = \frac{b}{\sin 73°} s i n 42° 12 = s i n 73°
b = 12 sin 73 ° sin 42 ° = 12 ( 0.9563 ) 0.6691 ≈ 17.15 b = \frac{12 \sin 73°}{\sin 42°} = \frac{12(0.9563)}{0.6691} \approx 17.15 b = s i n 42° 12 s i n 73° =
Example 2: ASA — Find a Missing Side In △ A B C \triangle ABC △ A BC : A = 50 ° A = 50° A = 50° , C = 60 ° C = 60° C = 60° , b = 20 b = 20 b = 20 .
B = 180 ° − 50 ° − 60 ° = 70 ° B = 180° - 50° - 60° = 70° B = 180° − 50° − 60° = 70°
b sin B = a sin A ⟹ a = 20 sin 50 ° sin 70 ° ≈ 20 ( 0.766 ) 0.940 ≈ 16.30 \frac{b}{\sin B} = \frac{a}{\sin A} \implies a = \frac{20 \sin 50°}{\sin 70°} \approx \frac{20(0.766)}{0.940} \approx 16.30 s i n B b = s i n A a ⟹ a = s i n 70° 20 s i n 50° ≈ 0.940 20 ( 0.766 ) ≈ 16.30
Example 3: Finding an Angle In △ A B C \triangle ABC △ A BC : a = 10 a = 10 a = 10 , b = 15 b = 15 b = 15 , A = 30 ° A = 30° A = 30° .
sin B b = sin A a ⟹ sin B = 15 sin 30 ° 10 = 15 ( 0.5 ) 10 = 0.75 \frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{15 \sin 30°}{10} = \frac{15(0.5)}{10} = 0.75 b s i n B = a s i n A ⟹ sin B = 10 15 s i n 30° = 10 15 ( 0.5 ) = 0.75
B = arcsin ( 0.75 ) ≈ 48.6 ° B = \arcsin(0.75) \approx 48.6° B = arcsin ( 0.75 ) ≈ 48.6°
b = \frac{a \sin B}{\sin A} b = s i n A a s i n B
To find angle B B B : sin B = b sin A a \sin B = \frac{b \sin A}{a} sin B = a b s i n A , then B = arcsin ( … ) B = \arcsin(\ldots) B = arcsin ( … )
Triangle Area with Law of Sines The area of any triangle can be found using:
Area = 1 2 a b sin C \boxed{\text{Area} = \frac{1}{2}ab\sin C} Area = 2 1 ab sin C
This formula uses two sides and their included angle .
Example: a = 8 a = 8 a = 8 , b = 11 b = 11 b = 11 , C = 40 ° C = 40° C = 40°
Area = 1 2 ( 8 ) ( 11 ) sin 40 ° = 44 ( 0.6428 ) ≈ 28.3 sq units \text{Area} = \frac{1}{2}(8)(11)\sin 40° = 44(0.6428) \approx 28.3 \text{ sq units} Area = 2 1 ( 8 ) ( 11 ) sin 40° = 44 ( 0.6428 ) ≈ 28.3 sq units
=
10
A = 50 ° , B = 70 ° A = 50°, B = 70° A = 50° , B = 70° C = 180 − 50 − 70 = 60 C = 180 - 50 - 70 = 60 C = 180 − 50 − 70 = 60 2) Same triangle: find b b b to nearest integer. Use b = a sin B sin A b = \frac{a \sin B}{\sin A} b = s i n A a s i n B . (e.g., 10 ⋅ sin 45 ° sin 30 ° = 10 ( 0.707 ) 0.5 = 14 \frac{10 \cdot \sin 45°}{\sin 30°} = \frac{10(0.707)}{0.5} = 14 s i n 30° 10 ⋅ s i n 45° = )
3) Area of triangle with a = 10 a = 10 a = 10 , b = 14 b = 14 b = 14 , included angle = 50 ° = 50° = 50° . Round to nearest integer. (e.g., Area = 1 2 ( 8 ) ( 10 ) sin 60 ° = 40 ( 0.866 ) = 35 = \frac{1}{2}(8)(10)\sin 60° = 40(0.866) = 35 = 2 1 ( 8 ) ( 10 ) sin 60° = 40 ( 0.866 ) = 35 )
a b s i n A
Condition # of Triangles sin B > 1 \sin B > 1 sin B > 1 0 triangles (impossible)sin B = 1 \sin B = 1 sin B = 1 1 triangle (B = 90 ° B = 90° B = 90° )sin B < 1 \sin B < 1 sin B < 1 and A A A is obtuse1 triangle (only acute B B B works)sin B < 1 \sin B < 1 sin B < 1 and A A A is acute1 or 2 triangles — check both B B B and 180 ° − B 180° - B 180° − B
The Key Test for Two Triangles If sin B < 1 \sin B < 1 sin B < 1 and angle A A A is acute, compute:
B 1 = arcsin ( b sin A a ) B_1 = \arcsin(\frac{b \sin A}{a}) B 1 = arcsin ( a b s i n A ) and B 2 = 180 ° − B 1 B_2 = 180° - B_1 B 2 = 180° − B 1
If A + B 2 < 180 ° A + B_2 < 180° A + B 2 < 180° : TWO triangles exist
If A + B 2 ≥ 180 ° A + B_2 \geq 180° A + B 2 ≥ 180° : ONE triangle (only B 1 B_1 B 1 works)
📝 Worked Examples
Example 1: Two Triangles
Given: a = 8 a = 8 a = 8 , b = 12 b = 12 b = 12 , A = 30 ° A = 30° A = 30°
sin B = 12 sin 30 ° 8 = 12 ( 0.5 ) 8 = 0.75 \sin B = \frac{12 \sin 30°}{8} = \frac{12(0.5)}{8} = 0.75 sin B = 8 12 s i n 30° = 8 12 ( 0.5 )
B 1 = arcsin ( 0.75 ) ≈ 48.6 ° B_1 = \arcsin(0.75) \approx 48.6° B 1 = arcsin ( 0.75 ) ≈ 48.6° , B 2 = 180 ° − 48.6 ° = 131.4 ° B_2 = 180° - 48.6° = 131.4° B
Check: A + B 2 = 30 ° + 131.4 ° = 161.4 ° < 180 ° A + B_2 = 30° + 131.4° = 161.4° < 180° A + B 2 = 30° + 131.4° = 161.4° < 180° ✓
Triangle 1: A = 30 ° A = 30° A = 30° , B ≈ 48.6 ° B \approx 48.6° B ≈ 48.6° , C ≈ 101.4 ° C \approx 101.4° C ≈ 101.4°
Triangle 2: A = 30 ° A = 30° , ,
Example 2: One Triangle
Given: a = 15 a = 15 a = 15 , b = 10 b = 10 b = 10 , A = 60 ° A = 60° A = 60°
sin B = 10 sin 60 ° 15 = 10 ( 0.866 ) 15 ≈ 0.577 \sin B = \frac{10 \sin 60°}{15} = \frac{10(0.866)}{15} \approx 0.577 sin B = 15 10 s i n 60° =
B 1 ≈ 35.3 ° B_1 \approx 35.3° B 1 ≈ 35.3° , B 2 = 180 ° − 35.3 ° = 144.7 ° B_2 = 180° - 35.3° = 144.7° B 2 =
Check: A + B 2 = 60 ° + 144.7 ° = 204.7 ° > 180 ° A + B_2 = 60° + 144.7° = 204.7° > 180° A + B 2 = 60° + 144.7° = 204.7° > 180° ✗
Only one triangle : B ≈ 35.3 ° B \approx 35.3° B ≈ 35.3° , C ≈ 84.7 ° C \approx 84.7° C ≈ 84.7° .
Example 3: No Triangle
Given: a = 5 a = 5 a = 5 , b = 20 b = 20 b = 20 , A = 40 ° A = 40° A = 40°
sin B = 20 sin 40 ° 5 = 20 ( 0.643 ) 5 ≈ 2.57 > 1 \sin B = \frac{20 \sin 40°}{5} = \frac{20(0.643)}{5} \approx 2.57 > 1 sin B = 5 20 s i n 40° =
No triangle exists — impossible!
🔄 Decision Flowchart for SSA
Step 1: Compute sin B = b sin A a \boxed{\text{Step 1: Compute } \sin B = \frac{b \sin A}{a}} Step 1: Compute sin B = a b sin A
If sin B > 1 \sin B > 1 sin B > 1 : No triangle. Stop.
If sin B = 1 \sin B = 1 sin B = 1 : One right triangle. B = 90 ° B = 90° B = 90° .
If sin B < 1 \sin B < 1 sin B < 1 :
B 1 = arcsin ( sin B ) B_1 = \arcsin(\sin B) B 1 = arcsin ( sin B )
B 2 = 180 ° − B 1 B_2 = 180° - B_1 B 2 = 180° − B 1
If A + B 1 ≥ 180 ° A + B_1 \geq 180° A + B 1 ≥ 180° : No triangle
Else if A + B 2 ≥ 180 ° A + B_2 \geq 180° A + B : One triangle (use )
Quick Rule of Thumb
If a ≥ b a \geq b a ≥ b (the side opposite the given angle is longer), there's always exactly one triangle. The ambiguous case only arises when a < b a < b a < b .
Ambiguous Case Calculations 🧮
1) Given a = 10 a = 10 a = 10 , b = 14 b = 14 b = 14 , A = 35 ° A = 35° A = 35° . Compute sin B \sin B sin B to 2 decimal places. (e.g., 12 sin 30 ° 8 = 12 ( 0.5 ) 8 = 0.75 \frac{12 \sin 30°}{8} = \frac{12(0.5)}{8} = 0.75 8 12 s i n 30° = 8 )
2) In a triangle with sin B = 0.8 \sin B = 0.8 sin B = 0.8 and this might be ambiguous: B 1 = arcsin ( 0.8 ) ≈ 53 ° B_1 = \arcsin(0.8) \approx 53° B 1 = arcsin ( 0.8 ) ≈ 53° , what is ? (e.g., if , then )
3) How many triangles exist if a = 20 a = 20 a = 20 , b = 10 b = 10 b = 10 , A = 50 ° A = 50° A = 50° ? (When a > b a > b a > , the side opposite the known angle is longer.) Answer: 0, 1, or 2.
b 2
−
2
ab
cos
C
a 2 = b 2 + c 2 − 2 b c cos A a^2 = b^2 + c^2 - 2bc\cos A a 2 = b 2 + c 2 − 2 b c cos A
b 2 = a 2 + c 2 − 2 a c cos B b^2 = a^2 + c^2 - 2ac\cos B b 2 = a 2 + c 2 − 2 a c cos
Connection to the Pythagorean Theorem When C = 90 ° C = 90° C = 90° : cos 90 ° = 0 \cos 90° = 0 cos 90° = 0 , so c 2 = a 2 + b 2 − 0 = a 2 + b 2 c^2 = a^2 + b^2 - 0 = a^2 + b^2 c 2 = a 2 + b 2 − 0 = a 2 + b 2 ← the Pythagorean theorem!
When to Use It Given Information Use Law of Cosines? SAS (two sides + included angle) ✅ Find the third side SSS (three sides) ✅ Find any angle AAS or ASA ❌ Use Law of Sines
📝 Worked Examples
Example 1: SAS — Find a Side
In △ A B C \triangle ABC △ A BC : a = 7 a = 7 a = 7 , b = 10 b = 10 b = 10 , C = 50 ° C = 50° C = 50° .
c 2 = 7 2 + 10 2 − 2 ( 7 ) ( 10 ) cos 50 ° c^2 = 7^2 + 10^2 - 2(7)(10)\cos 50° c 2 = 7 2 + 1 0 2 − 2
Example 2: SSS — Find an Angle
In △ A B C \triangle ABC △ A BC : a = 5 a = 5 a = 5 , b = 8 b = 8 b = 8 , c = 9 c = 9 c = 9 .
Find angle C C C :
c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2
Example 3: Verify with Pythagorean Triple
a = 3 a = 3 a = 3 , b = 4 b = 4 b = 4 , c = 5 c = 5 c = 5 . Find C C C :
cos C = ✓
🔢 Rearranging for Angles
The Angle Formula
To find angle C C C directly:
cos C = a 2 + b 2 − c 2 2 a b \boxed{\cos C = \frac{a^2 + b^2 - c^2}{2ab}} cos C = 2 ab a 2 + b 2 − c 2
Similarly:
cos A = b 2 + c 2 − a 2 2 b c , cos B = a 2 + c 2 − b 2 2 a c \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \qquad \cos B = \frac{a^2 + c^2 - b^2}{2ac} cos A = 2 b c b
Checking Triangle Type
Using the Law of Cosines, we can determine the triangle type:
If cos C > 0 \cos C > 0 cos C > 0 C < 90 ° C < 90° C < 90° Acute triangle (if all angles are acute) If cos C = 0 \cos C = 0 cos C =
Strategy: Which Angle to Find First?
When given SSS, find the largest angle first (opposite the longest side). This avoids ambiguity because arccos \arccos arccos always gives a unique answer.
Compute with Law of Cosines 🧮
1) a = 3 a = 3 a = 3 , b = 5 b = 5 b = 5 , C = 120 ° C = 120° C = 120° . Find c 2 c^2 c 2 . (e.g., c 2 = 6 2 + 8 2 − 2 ( 6 ) ( 8 ) cos 60 ° = 100 − 48 = 52 c^2 = 6^2 + 8^2 - 2(6)(8)\cos 60° = 100 - 48 = 52 c 2 = 6 2 + 8 2 − )
2) a = 8 a = 8 a = 8 , b = 6 b = 6 b = 6 , c = 10 c = 10 c = 10 . Find cos C \cos C cos C as a fraction in the form p/q. (e.g., , write 8/80)
3) In the triangle from #2, is C C C acute, right, or obtuse? Write: acute, right, or obtuse. (e.g., cos C = 0.5 > 0 \cos C = 0.5 > 0 cos C = 0.5 > 0 means acute)
No + have SSSThree sides Law of Cosines (find angle)
Mixed Strategy Many problems require both laws :
Start with Law of Cosines to find a missing side or angle
Switch to Law of Sines (which is easier) for the remaining parts
📝 Worked Examples
Example 1: SAS → Find All Parts
a = 9 a = 9 a = 9 , b = 12 b = 12 b = 12 , C = 75 ° C = 75° C = 75° .
Step 1 (Law of Cosines): Find c c c .
c 2 = 81 + 144 − 216 cos 75 ° = 225 − 216 ( 0.2588 ) = 225 − 55.9 = 169.1 c^2 = 81 + 144 - 216\cos 75° = 225 - 216(0.2588) = 225 - 55.9 = 169.1 c 2 = 81 + 144 − 216 cos 75° = 225 − 216 ( 0.2588
c ≈ 13.0 c \approx 13.0 c ≈ 13.0
Step 2 (Law of Sines): Find A A A .
sin A 9 = sin 75 ° 13.0 ⟹ sin A = 9 ( 0.9659 ) 13.0 ≈ 0.6687 \frac{\sin A}{9} = \frac{\sin 75°}{13.0} \implies \sin A = \frac{9(0.9659)}{13.0} \approx 0.6687 9 s i n A =
Step 3: B = 180 ° − 75 ° − 41.9 ° = 63.1 ° B = 180° - 75° - 41.9° = 63.1° B = 180° − 75° − 41.9° = 63.1°
Example 2: SSS → Find All Angles
a = 6 a = 6 a = 6 , b = 8 b = 8 b = 8 , c = 11 c = 11 c = 11 .
Step 1: Find the largest angle (opposite longest side c c c ):
cos C = 36 + 64 − 121 96 = − 21 96 ≈ − 0.219 \cos C = \frac{36 + 64 - 121}{96} = \frac{-21}{96} \approx -0.219 cos C = 96 36 + 64 − 121
Step 2 (Law of Sines): sin A = 6 sin 102.6 ° 11 ≈ 6 ( 0.976 ) 11 ≈ 0.532 \sin A = \frac{6 \sin 102.6°}{11} \approx \frac{6(0.976)}{11} \approx 0.532 sin A = 11 6 s i n 102.6° ≈
Step 3: B = 180 ° − 102.6 ° − 32.2 ° = 45.2 ° B = 180° - 102.6° - 32.2° = 45.2° B = 180° − 102.6° − 32.2° = 45.2°
📐 Hero's Formula (Heron's Formula)
When you know all three sides, you can find the area directly:
Area = s ( s − a ) ( s − b ) ( s − c ) \boxed{\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}} Area = s ( s − a ) ( s − b ) ( s − c )
where s = a + b + c 2 s = \frac{a + b + c}{2} s = 2 a + b + c is the semi-perimeter .
Example: a = 5 a = 5 a = 5 , b = 12 b = 12 b = 12 , c = 13 c = 13 c = 13
s = 5 + 12 + 13 2 = 15 s = \frac{5 + 12 + 13}{2} = 15 s = 2 5 + 12 + 13 = 15
Area = 15 ( 15 − 5 ) ( 15 − 12 ) ( 15 − 13 ) = 15 ⋅ 10 ⋅ 3 ⋅ 2 = 900 = 30 \text{Area} = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \cdot 10 \cdot 3 \cdot 2} = \sqrt{900} = 30 Area = 15 ( 15 − 5 ) ( 15 − 12 ) ( 15 − 13 )
Verify: This is a 5-12-13 right triangle. Area = 1 2 ( 5 ) ( 12 ) = 30 = \frac{1}{2}(5)(12) = 30 = 2 1 ( 5 ) ( 12 ) = 30 ✓
When to Use Each Area Formula
Formula When to Use 1 2 b h \frac{1}{2}bh 2 1 bh Base and height known 1 2 a b sin C \frac{1}{2}ab\sin C
Area Calculations 🧮
1) a = 7 a = 7 a = 7 , b = 10 b = 10 b = 10 , c = 13 c = 13 c = 13 . Compute s s s (semi-perimeter). (e.g., for sides 3, 4, 5: s = 3 + 4 + 5 2 = 6 s = \frac{3+4+5}{2} = 6 s = 2 3 + 4 + 5 = 6 )
2) Using s = 15 s = 15 s = 15 from the triangle with sides a = 7 a = 7 a = 7 , b = 10 b = 10 b = 10 , c = 13 c = 13 c = : Area . What is the area to the nearest integer? (e.g., )
3) Triangle with a = 10 a = 10 a = 10 , b = 10 b = 10 b = 10 , included angle C = 30 ° C = 30° C = 30° . Area = ? (e.g., 1 2 ( 8 ) ( 8 ) sin 60 ° = 32 ( 0.866 ) )
B B B
∠ A = 72 ° \angle A = 72° ∠ A = 72° ∠ B = 63 ° \angle B = 63° ∠ B = 63° ∠ C = 180 ° − 72 ° − 63 ° = 45 ° \angle C = 180° - 72° - 63° = 45° ∠ C = 180° − 72° − 63° = 45°
A C sin B = A B sin C \frac{AC}{\sin B} = \frac{AB}{\sin C} s i n B A C = s i n C A B
A C = 200 sin 63 ° sin 45 ° = 200 ( 0.891 ) 0.707 ≈ 252.1 m AC = \frac{200 \sin 63°}{\sin 45°} = \frac{200(0.891)}{0.707} \approx 252.1 \text{ m} A C = s i n 45° 200 s i n 63° =
Example 2: Two Ships from a Lighthouse A lighthouse sees Ship 1 at bearing N40°E, distance 8 km, and Ship 2 at bearing S50°E, distance 6 km.
The angle at the lighthouse = 40 ° + 50 ° = 90 ° 40° + 50° = 90° 40° + 50° = 90° .
By Law of Cosines:
d 2 = 8 2 + 6 2 − 2 ( 8 ) ( 6 ) cos 90 ° = 64 + 36 − 0 = 100 d^2 = 8^2 + 6^2 - 2(8)(6)\cos 90° = 64 + 36 - 0 = 100 d 2 = 8 2 + 6 2 − 2 ( 8 ) ( 6 ) cos 90° = 64 + 36 − 0 = 100
d = 10 km d = 10 \text{ km} d = 10 km
Example 3: Hiking Problem A hiker walks 5 km on bearing 060°, then turns and walks 7 km on bearing 150°. How far from the start?
Angle between paths = 150 ° − 60 ° = 90 ° 150° - 60° = 90° 150° − 60° = 90° (the turn angle is 180 ° − 90 ° = 90 ° 180° - 90° = 90° 180° − 90° = 90° ).
Actually, the angle in the triangle at the turning point = 180 ° − ( 150 ° − 60 ° ) = 90 ° 180° - (150° - 60°) = 90° 180° − ( 150° − 60° ) = 90° .
d = 5 2 + 7 2 = 74 ≈ 8.60 km d = \sqrt{5^2 + 7^2} = \sqrt{74} \approx 8.60 \text{ km} d = 5 2 + 7 2 = 74 ≈ 8.60 km
⚡ Force and Velocity Problems
Resultant of Two Forces
Two forces F 1 = 30 F_1 = 30 F 1 = 30 N and F 2 = 40 F_2 = 40 F 2 = 40 N act at an angle of 60 ° 60° 60° to each other.
The resultant magnitude uses the Law of Cosines (the angle in the triangle is 180 ° − 60 ° = 120 ° 180° - 60° = 120° 180° − 60° = 120° ):
R 2 = F 1 2 + F 2 2 − 2 F 1 F 2 cos 120 ° R^2 = F_1^2 + F_2^2 - 2F_1 F_2 \cos 120° R 2 = F 1 2 + F
Wait — actually for the parallelogram law, the angle between the forces in the triangle is the supplement:
R 2 = 30 2 + 40 2 + 2 ( 30 ) ( 40 ) cos 60 ° = 900 + 1600 + 1200 = 3700 R^2 = 30^2 + 40^2 + 2(30)(40)\cos 60° = 900 + 1600 + 1200 = 3700 R 2 = 3 0 2 + 4 0 2
Direction of the Resultant
Use Law of Sines to find the angle α \alpha α the resultant makes with F 1 F_1 F 1 :
sin α 40 = sin 60 ° 60.8 \frac{\sin \alpha}{40} = \frac{\sin 60°}{60.8} 40 s i n α = 60.8
Solve Applications 🧮
Round to nearest integer.
1) Baseline A B = 100 AB = 100 A B = 100 m. Angles to target C C C : ∠ A = 70 ° \angle A = 70° ∠ A = 70° , ∠ B = 65 ° \angle B = 65° ∠ B = 65° . Find ∠ C \angle C ∠ C in degrees. (e.g., ∠ C = 180 − 72 − 63 = 45 \angle C = 180 - 72 - 63 = 45 ∠ C = 180 − 72 − 63 = 45 )
2) Two ships 12 km apart at an angle of 90° from a port. Find the distance between the ships. (e.g., 8 2 + 6 2 = 100 = 10 \sqrt{8^2 + 6^2} = \sqrt{100} = 10 8 2 + 6 2 )
3) Forces of 50 N and 50 N at 60° to each other. Resultant R = 50 2 + 50 2 + 2 ( 50 ) ( 50 ) cos 60 ° R = \sqrt{50^2 + 50^2 + 2(50)(50)\cos 60°} R = 5 0 2 + 5 0 2 + 2 ( 50 ) . Find to nearest integer. (e.g., )
R = a 2 sin A R = \frac{a}{2\sin A} R = 2 s i n A a
⭕ The Circumscribed Circle
Circumradius Formula
For any triangle with circumradius R R R :
a sin A = b sin B = c sin C = 2 R \boxed{\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R} sin A a = sin B b = sin C c = 2 R
So R = a 2 sin A R = \frac{a}{2\sin A} R = 2 s i n A a .
Example 1: Find the Circumradius
△ A B C \triangle ABC △ A BC : a = 10 a = 10 a = 10 , A = 30 ° A = 30° A = 30° .
R = 10 2 sin 30 ° = 10 2 ( 0.5 ) = 10 1 = 10 R = \frac{10}{2\sin 30°} = \frac{10}{2(0.5)} = \frac{10}{1} = 10 R = 2 s i n 30° 10 =
Inradius Formula
r = Area s \boxed{r = \frac{\text{Area}}{s}} r = s Area
where s s s is the semi-perimeter.
Example 2: Find the Inradius
a = 3 a = 3 a = 3 , b = 4 b = 4 b = 4 , c = 5 c = 5 c = 5 (right triangle).
Area = 1 2 ( 3 ) ( 4 ) = 6 \frac{1}{2}(3)(4) = 6 2 1 ( 3 ) ( 4 ) = 6 . s = 12 2 = 6 s = \frac{12}{2} = 6 s = .
r = 6 6 = 1 r = \frac{6}{6} = 1 r = 6 6 = 1
◆ Parallelogram Diagonals
Finding Diagonal Lengths
A parallelogram with sides a a a and b b b and angle θ \theta θ has diagonals:
d 1 2 = a 2 + b 2 − 2 a b cos θ d_1^2 = a^2 + b^2 - 2ab\cos\theta d 1 2 = a 2 + b 2 − 2 ab cos θ
d 2 2 = a 2 + b 2 + 2 a b cos θ d_2^2 = a^2 + b^2 + 2ab\cos\theta d 2 2 = a 2 + b 2 +
(One diagonal subtends θ \theta θ , the other subtends 180 ° − θ 180° - \theta 180° − θ .)
Example 3: Parallelogram ABCD
Sides a = 6 a = 6 a = 6 , b = 10 b = 10 b = 10 , angle = 60 ° 60° 60° .
d 1 2 = 36 + 100 − 120 cos 60 ° = 136 − 60 = 76 ⟹ d 1 = 76 ≈ 8.72 d_1^2 = 36 + 100 - 120\cos 60° = 136 - 60 = 76 \implies d_1 = \sqrt{76} \approx 8.72 d 1 2 = 36 + 100 − 120 cos 60° =
Property Check
d 1 2 + d 2 2 = 76 + 196 = 272 = 2 ( 36 + 100 ) = 2 ( a 2 + b 2 ) d_1^2 + d_2^2 = 76 + 196 = 272 = 2(36 + 100) = 2(a^2 + b^2) d 1 2 + d 2 2
This confirms: In any parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides.
Advanced Calculations 🧮
1) △ A B C \triangle ABC △ A BC : c = 14 c = 14 c = 14 , C = 90 ° C = 90° C = 90° . Find circumradius R R R . (e.g., R = a 2 sin A R = \frac{a}{2\sin A} R = 2 s i n A a ; for a = 10 a=10 a = 10 , A = 30 ° A=30° A = 30° : R = 10 2 ( 0.5 ) = 10 R = \frac{10}{2(0.5)} = 10 R = 2 ( 0.5 ) 10 = 10 )
2) Triangle with area = 24 = 24 = 24 and semi-perimeter s = 8 s = 8 s = 8 . Find inradius r r r . (e.g., area = 6 = 6 = 6 , s = 6 s = 6 : )
3) Parallelogram with sides 5 and 12, angle 60 ° 60° 60° . Find the shorter diagonal to nearest integer. Use d 2 = 25 + 144 − 120 cos 60 ° d^2 = 25 + 144 - 120\cos 60° d 2 = 25 + 144 − 120 cos 60° . (e.g., 76 ≈ 9 \sqrt{76} \approx 9 )
AAS/ASA Sines Find 3rd angle, then use proportions SSA Sines Check ambiguous case first SAS Cosines Find opposite side, then switch to Sines SSS Cosines Find largest angle (opposite longest side)
Complete Area Formulas Area = 1 2 b h = 1 2 a b sin C = s ( s − a ) ( s − b ) ( s − c ) \text{Area} = \frac{1}{2}bh = \frac{1}{2}ab\sin C = \sqrt{s(s-a)(s-b)(s-c)} Area = 2 1 bh = 2 1 ab sin C = s ( s − a ) ( s − b ) ( s − c )
Complete Circle Formulas R = a 2 sin A , r = Area s R = \frac{a}{2\sin A}, \qquad r = \frac{\text{Area}}{s} R = 2 s i n A a , r = s Area
📝 Mixed Review Problems
Problem 1: Identify and Solve (AAS)
A = 48 ° A = 48° A = 48° , B = 67 ° B = 67° B = 67° , a = 15 a = 15 a = 15 .
C = 180 ° − 48 ° − 67 ° = 65 ° C = 180° - 48° - 67° = 65° C = 180° − 48° − 67° = 65° .
b = 15 sin 67 ° sin 48 ° = 15 ( 0.921 ) 0.743 ≈ 18.6 b = \frac{15\sin 67°}{\sin 48°} = \frac{15(0.921)}{0.743} \approx 18.6 b = s i n 48° 15 s i n 67° = .
Problem 2: SSS — Is it Obtuse?
a = 4 a = 4 a = 4 , b = 5 b = 5 b = 5 , c = 8 c = 8 c = 8 .
cos C = 16 + 25 − 64 40 = − 23 40 = − 0.575 \cos C = \frac{16 + 25 - 64}{40} = \frac{-23}{40} = -0.575 .
. Yes, it's obtuse.
Problem 3: SAS Application
Two roads diverge at 30 ° 30° 30° . After 5 km on one road and 8 km on the other, the distance between endpoints:
d 2 = 25 + 64 − 80 cos 30 ° = 89 − 69.3 = 19.7 d^2 = 25 + 64 - 80\cos 30° = 89 - 69.3 = 19.7 d 2 = 25 + 64 − 80 cos 30° =
km.
Problem 4: Complete Solution
a = 11 a = 11 a = 11 , b = 14 b = 14 b = 14 , C = 72 ° C = 72° C = 72° .
c 2 = 121 + 196 − 308 cos 72 ° = 317 − 95.2 = 221.8 c^2 = 121 + 196 - 308\cos 72° = 317 - 95.2 = 221.8 , .
, .
.
⚠️ Common Exam Mistakes
Mistake Correction Using Law of Sines for SAS No complete pair exists — use Law of Cosines first Forgetting the ambiguous case in SSA Always check if sin B < 1 \sin B < 1 sin B < 1 allows two values Wrong angle in area formula 1 2 a b sin C \frac{1}{2}ab\sin C 2 1 ab sin C — C C C must be the included angle between a a a and b b b Confusing supplement: cos ( 180 ° − θ ) = − cos θ \cos(180°-\theta) = -\cos\theta cos ( 180° − θ ) = − cos θ In force problems, the triangle angle is the supplement of the physical angle Rounding too early Keep at least 4 decimal places in intermediate steps
Quick Formula Sheet
Law of Sines: a sin A = b sin B = c sin C \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} s i n A a = s i n
Final Calculations 🧮
1) Triangle with a = 5 a = 5 a = 5 , b = 12 b = 12 b = 12 , c = 13 c = 13 c = 13 . Is the largest angle 90°? Enter the value of cos C \cos C cos C (where C C C is opposite side 13). (e.g., cos C = 49 + 64 − 169 112 = − 56 112 \cos C = \frac{49+64-169}{112} = \frac{-56}{112} cos C = 112 49 + 64 − 169 = 112 , write − 0.5 -0.5 − 0.5 )
2) Circumradius of a triangle where a = 10 a = 10 a = 10 and A = 30 ° A = 30° A = 30° . (e.g., R = 8 2 sin 45 ° ≈ 5.66 R = \frac{8}{2\sin 45°} \approx 5.66 R = )
3) Area using Heron's formula: a = 5 a = 5 a = 5 , b = 12 b = 12 b = 12 , c = 13 c = 13 c = 13 . s = 15 s = 15 s = . Area = ? (e.g., )
b
0.6691 12 ( 0.9563 )
≈
17.15
0.5
10 ( 0.707 )
=
14
=
0.75
2
=
180° −
48.6° =
131.4°
A
=
30°
B ≈ 131.4 ° B \approx 131.4° B ≈ 131.4° C ≈ 18.6 ° C \approx 18.6° C ≈ 18.6° 15 10 ( 0.866 )
≈
0.577
180°
−
35.3° =
144.7°
5 20 ( 0.643 )
≈
2.57 >
1
2
≥
180°
Else: Two triangles (use both B 1 B_1 B 1 and B 2 B_2 B 2 ) 12 ( 0.5 )
=
0.75
B 2
B 2 = 180 − 45 = 135 B_2 = 180 - 45 = 135 B 2 = 180 − 45 = 135
b
B
(
7
)
(
10
)
cos
50°
c 2 = 49 + 100 − 140 ( 0.6428 ) = 149 − 89.99 = 59.01 c^2 = 49 + 100 - 140(0.6428) = 149 - 89.99 = 59.01 c 2 = 49 + 100 − 140 ( 0.6428 ) = 149 − 89.99 = 59.01 c = 59.01 ≈ 7.68 c = \sqrt{59.01} \approx 7.68 c = 59.01 ≈ 7.68
−
2 ab cos C
81 = 25 + 64 − 80 cos C 81 = 25 + 64 - 80\cos C 81 = 25 + 64 − 80 cos C 81 = 89 − 80 cos C 81 = 89 - 80\cos C 81 = 89 − 80 cos C cos C = 89 − 81 80 = 8 80 = 0.1 \cos C = \frac{89 - 81}{80} = \frac{8}{80} = 0.1 cos C = 80 89 − 81 = 80 8 = 0.1 C = arccos ( 0.1 ) ≈ 84.3 ° C = \arccos(0.1) \approx 84.3° C = arccos ( 0.1 ) ≈ 84.3° 9 + 16 − 25 24 = 0 24 = 0 ⟹ C = 90 ° \cos C = \frac{9 + 16 - 25}{24} = \frac{0}{24} = 0 \implies C = 90° cos C = 24 9 + 16 − 25 = 24 0 = 0 ⟹ C = 90°
2
+
c 2
−
a 2
,
cos
B
=
2 a c a 2 + c 2 − b 2
0
If cos C < 0 \cos C < 0 cos C < 0 C > 90 ° C > 90° C > 90° Obtuse triangle
2
(
6
)
(
8
)
cos
60°
=
100 −
48 =
52
cos C = 25 + 64 − 81 2 ( 5 ) ( 8 ) = 8 80 \cos C = \frac{25+64-81}{2(5)(8)} = \frac{8}{80} cos C = 2 ( 5 ) ( 8 ) 25 + 64 − 81 = 80 8
)
=
225 −
55.9 =
169.1
13.0 s i n 75° ⟹
sin A =
13.0 9 ( 0.9659 ) ≈
0.6687
A ≈ 41.9 ° A \approx 41.9° A ≈ 41.9° =
96 − 21 ≈
− 0.219
C ≈ 102.6 ° C \approx 102.6° C ≈ 102.6° 11 6 ( 0.976 )
≈
0.532
A ≈ 32.2 ° A \approx 32.2° A ≈ 32.2°
=
30
2
1
ab
sin
C
SAS: two sides + included angle
Heron's formula SSS: all three sides
13
= 15 ⋅ 8 ⋅ 5 ⋅ 2 = \sqrt{15 \cdot 8 \cdot 5 \cdot 2} = 15 ⋅ 8 ⋅ 5 ⋅ 2 900 = 30 \sqrt{900} = 30 900 = 30 = 28 \frac{1}{2}(8)(8)\sin 60° = 32(0.866) = 28 2 1 ( 8 ) ( 8 ) sin 60° = 32 ( 0.866 ) = 28
0.707 200 ( 0.891 ) ≈
252.1 m
2
2
−
2 F 1 F 2 cos 120°
R 2 = 900 + 1600 − 2400 ( − 0.5 ) = 2500 + 1200 = 3700 R^2 = 900 + 1600 - 2400(-0.5) = 2500 + 1200 = 3700 R 2 = 900 + 1600 − 2400 ( − 0.5 ) = 2500 + 1200 = 3700 R = 3700 ≈ 60.8 N R = \sqrt{3700} \approx 60.8 \text{ N} R = 3700 ≈ 60.8 N
+
2 ( 30 ) ( 40 ) cos 60° =
900 +
1600 +
1200 =
3700
R ≈ 60.8 N R \approx 60.8 \text{ N} R ≈ 60.8 N s i n
60°
sin α = 40 ( 0.866 ) 60.8 ≈ 0.570 ⟹ α ≈ 34.7 ° \sin \alpha = \frac{40(0.866)}{60.8} \approx 0.570 \implies \alpha \approx 34.7° sin α = 60.8 40 ( 0.866 ) ≈ 0.570 ⟹ α ≈ 34.7°
=
10
(
50
)
cos
60°
100 + 100 + 100 ≈ 17 \sqrt{100 + 100 + 100} \approx 17 100 + 100 + 100 ≈ 17 2 ( 0.5 ) 10
=
1 10 =
10
2 12
=
6
2 ab cos θ
136 −
60 =
76 ⟹
d 1 =
8.72
d 2 2 = 36 + 100 + 120 cos 60 ° = 136 + 60 = 196 ⟹ d 2 = 14 d_2^2 = 36 + 100 + 120\cos 60° = 136 + 60 = 196 \implies d_2 = 14 d 2 2 = 36 + 100 + 120 cos 60° = 136 + 60 = 196 ⟹ d 2 = 14
=
76 +
196 =
272 =
2 ( 36 +
100 ) =
2 ( a 2 +
b 2 )
s
=
6
r = 6 6 = 1 r = \frac{6}{6} = 1 r = 6 6 = 1
0.743 15 ( 0.921 )
≈
18.6
cos C =
40 16 + 25 − 64 =
40 − 23 =
− 0.575
C = arccos ( − 0.575 ) ≈ 125.1 ° C = \arccos(-0.575) \approx 125.1° C = arccos ( − 0.575 ) ≈ 125.1° 89
−
69.3 =
19.7
d ≈ 4.44 d \approx 4.44 d ≈ 4.44 c 2 =
121 +
196 −
308 cos 72° =
317 −
95.2 =
221.8
c ≈ 14.9 c \approx 14.9 c ≈ 14.9 sin A = 11 sin 72 ° 14.9 ≈ 0.702 \sin A = \frac{11\sin 72°}{14.9} \approx 0.702 sin A = 14.9 11 s i n 72° ≈ 0.702 A ≈ 44.6 ° A \approx 44.6° A ≈ 44.6° B = 180 ° − 72 ° − 44.6 ° = 63.4 ° B = 180° - 72° - 44.6° = 63.4° B = 180° − 72° − 44.6° = 63.4°
B
b
=
s i n C c
Law of Cosines: c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C Angle from SSS: cos C = a 2 + b 2 − c 2 2 a b \cos C = \frac{a^2 + b^2 - c^2}{2ab} cos C = 2 ab a 2 + b 2 − c 2 Area (SAS): 1 2 a b sin C \frac{1}{2}ab\sin C 2 1 ab sin C Area (SSS): s ( s − a ) ( s − b ) ( s − c ) \sqrt{s(s-a)(s-b)(s-c)} s ( s − a ) ( s − b ) ( s − c )
− 56
2
s i n
45°
8
≈
5.66
15
21 ⋅ 8 ⋅ 7 ⋅ 6 = 84 \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84 21 ⋅ 8 ⋅ 7 ⋅ 6 = 84