Professor Murray

Law of Cosines

Slide Duration:

Section 1: Trigonometric Functions
Angles

39m 5s

Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
2:08
2:31
2:52
Half-Circle and Right Angle
4:00
6:24
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions

43m 16s

Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles

33m 5s

Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
9:43
Mnemonic: All Students Take Calculus
10:13
13:11
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D

52m 3s

Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions

36m 4s

Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions

27m 18s

Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions

32m 58s

Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions

31m 8s

Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Section 2: Trigonometric Identities
Pythagorean Identity

19m 11s

Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x

23m 16s

Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2

52m 52s

Intro
0:00
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas

29m 5s

Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas

43m 55s

Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
Section 3: Applications of Trigonometry
Trigonometry in Right Angles

25m 43s

Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines

56m 40s

Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines

49m 5s

Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle

27m 37s

Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry

34m 25s

Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
Vectors

46m 42s

Intro
0:00
Vector Formulas and Concepts
0:12
Vectors as Arrows
0:28
Magnitude
0:38
Direction
0:50
Drawing Vectors
1:16
Uses of Vectors: Velocity, Force
1:37
Vector Magnitude Formula
3:15
Vector Direction Formula
3:28
Vector Components
6:27
Example 1: Magnitude and Direction of Vector
8:00
Example 2: Force to a Box on a Ramp
12:25
Example 3: Plane with Wind
18:30
Extra Example 1: Components of a Vector
-1
Extra Example 2: Ship with a Current
-2
Section 4: Complex Numbers and Polar Coordinates
Polar Coordinates

1h 7m 35s

Intro
0:00
Polar Coordinates vs Rectangular/Cartesian Coordinates
0:12
Rectangular Coordinates, Cartesian Coordinates
0:23
Polar Coordinates
0:59
Converting Between Polar and Rectangular Coordinates
2:06
R
2:16
Theta
2:48
Example 1: Convert Rectangular to Polar Coordinates
6:53
Example 2: Convert Polar to Rectangular Coordinates
17:28
Example 3: Graph the Polar Equation
28:00
Extra Example 1: Convert Polar to Rectangular Coordinates
-1
Extra Example 2: Graph the Polar Equation
-2
Complex Numbers

35m 59s

Intro
0:00
Main Definition
0:07
Number i
0:23
Complex Number Form
0:33
Powers of Imaginary Number i
1:00
Repeating Pattern
1:43
Operations on Complex Numbers
3:30
3:39
Multiplying Complex Numbers
4:39
FOIL Method
5:06
Conjugation
6:29
Dividing Complex Numbers
7:34
Conjugate of Denominator
7:45
Example 1: Solve For Complex Number z
11:02
Example 2: Expand and Simplify
15:34
Example 3: Simplify the Powers of i
17:50
Extra Example 1: Simplify
-1
Extra Example 2: All Complex Numbers Satisfying Equation
-2
Polar Form of Complex Numbers

40m 43s

Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem

57m 37s

Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
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• ## Related Books

### Law of Cosines

Main formulas:

• The Law of Cosines:
 c2 = a2 + b2 − 2abcosC
This holds in any triangle - not just right triangles! This is a generalization of the Pythagorean Theorem. (If C is a right angle, then cosC = 0, so we get c2 = a2 + b2.)
• Use the Law of Cosines for triangles described in the following ways:
• SAS always has a unique solution (assuming the angle is less than 180° ).
• SSS always has a unique solution (assuming each side is less than the sum of the other two).
• (Use Law of Sines for ASA, SAA, and SSA.)
• Heron's Formula:
 Area = √ s(s−a)(s−b)(s−b) ,
where s = [1/2](a+b+c) is the semiperimeter of the triangle.

Example 1:

In triangle ABC, side a has length 3, side b has length 4, and angle C measures 60° . Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

Example 2:

The side lengths of triangle ABC are 5, 7, and 10. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

Example 3:

Find the area of a triangle whose side lengths are 5, 7, and 10.

Example 4:

The side lengths of triangle ABC are 16, 30, and 34. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

Example 5:

A triangle has two sides of length 8 and 16 with an included angle of 45° . Find the length of the third side and the area of the triangle.

### Law of Cosines

In triangle ABC, side a has length 25, side c has length 15, and angle B is 40°. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.
• Start by drawing a triangle with properly labeled angles and side lengths.
• According to the triangle: SAS and the angle between the sides measures 40° which is less than 180°
• One solution
• Solve using the Law of Cosines
• b2 = a2 + c2 − 2ac  cosB ⇒ b2 = 252 + 152 − 2(25)(15)cos40° ⇒ b2 = 275.47 ⇒ b = √{275.47}
• b ≈ 16.6
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(16.62 + 152 − 252)/2(16.6)(15)] ⇒ cosA = [( − 124.44)/498] ⇒ cosA = − 0.24988 ⇒ A = arccos( − 0.24988)
• ∠A ≈ 104.5°
• ∠C = 180° − 104.5° − 40° = 35.5°
∠b ≈ 16.6, ∠A ≈ 104.5°, ∠C = 35.5°
The side lengths of triangle ABC are 6, 8, and 11. Determine how many triangles satisfy these conditions and solve the triangle(s) completetly.
• Start by drawing a triangle with properly labeled angles and side lengths.
• According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
• 6 < 8 + 11, 8 < 6 + 11, 11 < 6 + 8
• One solution
• Solve using the Law of Cosines
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(112 + 62 − 82)/2(11)(6)] ⇒ cosA = [93/132] ⇒ cosA = 0.7045 ⇒ A = arccos(0.7045)
• ∠A ≈ 45.2°
• cosB = [(a2 + c2 − b2)/2ac] ⇒ cosB = [(82 + 62 − 112)/2(8)(6)] ⇒ cosB = [( − 21)/96] ⇒ cosB = ( − 0.21875) ⇒ B = arccos( − 0.21875)
• ∠B ≈ 102.6°
• ∠C = 180° − 45.2° − 102.6° = 32.2°
∠A ≈ 45.2°, ∠B ≈ 102.6°, ∠C = 32.2°
In triangle ABC, side a has length 6, side b has length 2, and angle C is 15°. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.
• Start by drawing a triangle with properly labeled angles and side lengths.
• According to the triangle: SAS and the angle between the sides measures 15° which is less than 180°
• One solution
• Solve using the Law of Cosines
• c2 = a2 + b2 − 2ab  cosC ⇒ c2 = 62 + 22 − 2(6)(2)cos15° ⇒ c2 = 16.82 ⇒ c = √{16.82}
• c ≈ 4.1
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(22 + 4.12 − 62)/2(2)(4.1)] ⇒ cosA = [( − 15.19)/16.4] ⇒ cosA = − 0.92622 ⇒ A = arccos( − 0.92622)
• ∠A ≈ 157.9°
• ∠B = 180° − 157.9° − 15° = 7.1°
c ≈ 4.1, ∠A ≈ 157.9°, ∠B = 7.1°
The side lengths of triangle ABC are 9, 12, and 15. Determine how many triangles satisfy these conditions and solve the triangle(s) completetly.
• Start by drawing a triangle with properly labeled angles and side lengths.
• According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
• 9 < 12 + 15; 12 < 9 + 15; 15 < 9 + 12
• One solution
• Solve using the Law of Cosines
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(152 + 92 − 122)/2(15)(9)] ⇒ cosA = [162/270] ⇒ cosA = 0.6 ⇒ A = arccos(0.6)
• ∠A ≈ 53.1°
• cosB = [(a2 + c2 − b2)/2ac] ⇒ cosB = [(122 + 92 − 152)/2(12)(9)] ⇒ cosB = [0/216] ⇒ cosB = (0) ⇒ B = arccos(0)
• ∠B ≈ 90°
• ∠C = 180° − 90° − 53.1° = 36.9°
∠A ≈ 53.1°, ∠B ≈ 90°, ∠C = 36.9°
In triangle ABC, side b has length 3, side c has length 10, and angle A is 55°. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.
• Start by drawing a triangle with properly labeled angles and side lengths.
• According to the triangle: SAS and the angle between the sides measures 55° which is less than 180°
• One solution
• Solve using the Law of Cosines
• a2 = b2 + c2 − 2bc  cosA ⇒ a2 = 32 + 102 − 2(3)(10)cos55° ⇒ a2 = 74.5852 ⇒ a = √{74.5852}
• a ≈ 8.6
• cosB = [(a2 + c2 − b2)/2ac] cosB = [(8.62 + 102 − 32)/2(8.6)(10)] cosB = [164.96/172] cosB = 0.9591 B = arccos(0.9591)
• ∠B ≈ 16.4°
• ∠C = 180° − 55° − 16.4° = 108.6°
a ≈ 8.6, ∠B ≈ 16.4°, ∠C = 108.6°
Use the Law of Cosines to solve the triangle
• According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
• 6 < 12 + 8; 8 < 6 + 12; 12 < 6 + 8
• One solution
• Solve using the Law of Cosines
• cosA = [(b2 + c2 − a2)/2bc] cosA = [(82 + 122 − 62)/2(8)(12)] cosA = [172/192] cosA = 0.89583 A = arccos(0.89583)
• ∠A ≈ 26.4°
• cosB = [(a2 + c2 − b2)/2ac] cosB = [(62 + 122 − 82)/2(6)(12)] cosB = [116/144] cosB = (0.80556) B = arccos(0.80556)
• ∠B ≈ 36.3°
• ∠C = 180° − 26.4° − 36.3° = 36.9°
∠A ≈ 26.4°, ∠B ≈ 36.3°, ∠C = 36.9°
Use the Law of Cosines to solve the triangle
• According to the triangle: SAS and the angle between the sides measures 55° which is less than 180°
• One solution
• Solve using the Law of Cosines
• a2 = b2 + c2 − 2bc  cosA ⇒ a2 = 152 + 302 − 2(15)(30)cos30o ⇒ a2 = 345.573 ⇒ a = √{345.573}
• a ≈ 18.6
• cosB = [(a2 + c2 − b2)/2ac] ⇒ cosB = [(18.62 + 302 − 152)/2(18.6)(30)] ⇒ cosB = [1020.96/116] ⇒ cosB = 0.91484 ⇒ B = arccos(0.91484)
• ∠B ≈ 23.8°
• ∠C = 180° − 30° − 23.8° = 126.2°
a ≈ 18.6, ∠B ≈ 23.8°, ∠C = 126.2°
Use the Law of Cosines to solve the triangle in which a = 8 b = 3 c = 9
• 43 < 9 + 8; 8 < 3 + 9; 9 < 3 + 8
• One solution
• Solve using the Law of Cosines
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(32 + 92 − 82)/2(3)(9)] ⇒ cosA = [26/54] ⇒ cosA = 0.48148 ⇒ A = arccos(0.48148)
• ∠A ≈ 61.2°
• cosB = [(a2 + c2 − b2)/2ac] ⇒ cosB = [(82 + 92 − 32)/2(8)(9)] ⇒ cosB = [136/144] ⇒ cosB = (0.9444) ⇒ B = arccos(0.9444)
• ∠B ≈ 19.2°
• ∠C = 180° − 61.2° − 19.2° = 99.6°
∠A ≈ 61.2°, ∠B ≈ 19.2°, ∠C = 99.6°
Use the Law of Cosines to solve the triangle with C = 105 a = 10 b = 5
• Solve using the Law of Cosines
• c2 = a2 + b2 − 2ab  cosC ⇒ c2 = 102 + 52 − 2(10)(5)cos105° ⇒ c2 = 150.882 ⇒ c = √{150.882}
• c ≈ 12.3
• cosA = [(b2 + c2 − a2)/2bc] ⇒ cosA = [(52 + 12.32 − 102)/2(5)(12.3)] ⇒ cosA = [76.29/123] ⇒ cosA = 0.62024 ⇒ A = arccos(0.62024)
• ∠A ≈ 51.7°
• ∠B = 180° − 51.7° − 105° = 23.3°
c ≈ 12.3, ∠A ≈ 51.7°, ∠B = 23.3°
Use the Law of Cosines to solve the triangle with a = 5 b = 7 c = 2
• First check the following 2 < 7 + 5; 5 < 7 + 2; 7 < 5 + 2
No solution

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Law of Cosines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Law of Cosines Formula 0:23
• Graphical Representation
• Relates Sides to Angles
• Any Triangle
• Generalization of Pythagorean Theorem
• When to Use Law of Cosines 2:26
• SAS, SSS
• Heron's Formula 4:49
• Semiperimeter S
• Example 1: How Many Triangles Satisfy Conditions, Solve Completely 5:53
• Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:19
• Example 3: Find Area of a Triangle Given All Side Lengths 26:33
• Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
• Extra Example 2: Length of Third Side and Area of Triangle

### Transcription: Law of Cosines

Hi! We are doing some more examples on using the law of cos and Heron’s formula.0000

We are given one now where the side lengths triangle (a, b and c) are (16, 30, 34).0005

We are asked to determine how many triangles satisfy these conditions and to solve the triangles completely.0013

What we are given here are three side lengths that is a (side, side, side) situation.0020

Because it is (side, side, side), it has a unique solution if it satisfies that check where each side is less than the sum of the other two.0026

Let us check that out for these three sides, let us check if (16 < 30+34) that is certainly true.0059

Is (30 < 16+34)? Clearly true, Is (34 < 16+30)? That is certainly true.0068

There is a unique triangle satisfying these three lengths of sides.0083

Let us try to find out what the angles would be in that triangle because we know what the side lengths are, let me draw the triangle.0099

(16, 30, 34) I am going to label an angle here, I will label this angle C and then we are going to use the law of cos to find the angles in the triangle.0112

That is why I started out by labeling angle C because I know the way I remember cos is with an angle C in it.0126

Let me write that down, c2=(a2) +(( b2 - (2ab)cos(C)).0133

The way I have labeled angle C, that makes the opposite side (c) and then these sides must be (a) and (b) here.0145

I can fill everything into the law of cos and then I can solve for angle (C), I will fill that in 162= (302+342) – 2(30)(34) x cos(C).0154

(162)=256, (302=900)+(342=1156), ((2 x 30 x 34=2040 cos(C)).0178

The angles are a little bit messy here, I’m going to move cos(C) over the other and I will get 2040 cos(C)=900+1156-256 that is 1800 because we moved the 256 over on the other side.0196

Cos(C) = 1800/2040, (C)=arcos of that horrible fraction (180/204) but I will go ahead and plug that straight into my calculator.0220

(180/204) I’m using degree mode for this, you have to be careful because otherwise it would give you an answer in radians, but that gives me answer of about 28.1 degrees.0253

That fills in one of my angles for my triangle there 28.1 degrees.0271

We got one of the angles, we will find out the other two exactly the same way, let me go ahead and work them out for you, I will redraw my triangle.0281

(16, 30, 34) this was 28.1, we already figured that out.0293

In order to find the other two angles, what I’m going to do is draw angle (C) in a different place now.0302

I am relabeling which angle is which, the point is that is I do not have to rewrite my law of cos switching around the (a, b, c) there.0306

This time I’m going to draw my angle (C) there and that makes side (c) the one opposite and (a), (b) other one next to it.0317

I will write down my law of cos, (c2)=(a2+b2)-((2ab cos (C)).0328

I will work through and I will solve for (C), that is (302)=(162+342)-((2(16)(34)cos(C)).0338

Now, it is just a little algebra to simplify this, 900=256+1156, now (2×16×34)=1088, and we still have cos(C).0355

I want to simplify this, I want to move the cos(C) on the other side, 1088 cos(C)=(256+1156-900), that simplifies down to 512.0374

Cos(C) =512/1088, I will take the inverse cos of that in my calculator.0399

I get 61.9 degrees approximately, that tells me one more of my angles, I only have one to go.0418

I could find the third angle by adding up the two angles and subtracting it 180, but I think I would like to practice the law of cos again.0440

At the end we can use that adding up to 180 as to check if we did the law of cos right.0448

We are going to find the third angle by using the law of cos again and in order to use my law of cos with the same (a,b, c) formula, I’m going to rotate the (a, b, c) on the triangle.0453

I’m going to cross out my old (a, b, c) and I will re-label the angle that I do not know as (C) which means its opposite side is 34 (c).0468

(a)(b) are now the two sides next to it, I will plug into the law of cos to solve for (C).0481

c2 is (342)=a2is (162)+b2 is now (302)- 2(16)(30) cos(C).0491

342 is 1156, 162 is 256, 302 is 900, 2×16×30 is 960, cos (C).0510

Now an interesting thing happens because I am going to subtract 256+900 for both sides, that is equal to exactly 1156.0527

What do I get is 0 is -960 cos (C) and if I divide it by -960, I will get cos(C)=0.0537

What angle has cos(0)? Well that is exactly a right triangle. (C) is exactly a 90 degree angle, let me fill that one in.0547

We could have noticed that this was in fact a right triangle because 162 + 302 = 342.0558

It does not matter because the law of cos works in all triangles.0568

It works just as well in right triangles as in other triangles but of course the rules for right triangles SOHCAHTOA do not work in other triangles.0571

Now we solved the triangle completely, we got all three sides and we got all three angles but I want to check and see if those three angles actually do add up to 180 degrees.0581

To check here, I look at (28.1+61.9+90) if you add those together you will indeed get 180 degrees which means we must have done the problem right.0594

Let us just recap what we did there, we are given a (side, side, side) presentation of a triangle.0612

The first thing that we did was check that each one of those sides was less than the sum of the other two.0621

If that check have not work out, we would have stopped right there and said there is no such triangle, but that check did worked out.0626

Then we go on to using the law of cos to find the each of the angles of the triangle.0632

We take the law of cos here and we fill in the lengths of the three sides, and then we solve down to find the cos of a missing angle.0641

Once we know the cos of the missing angle, we can use our cos to find the angle itself.0652

We did that three times just applying it to each angle and term, we got each of the three angles and then we check in the end if they added up to 180.0656

A triangle has two sides of length 8 with an included angle of 45 degrees, we want to find the length of the third side and the area of the triangle.0000

Let me try drawing that, there is 8, there is 16, that is about a 45 degree angle but it is not really intended to be drawn on the scale.0010

Notice here that we are given here a (side, angle, side) presentation of a triangle, (side, angle, side) and the angle is less than 180 degrees.0026

We know that there is a unique triangle satisfying this data.0036

We want to find the length of the third side, now this is tailor made for the law of cos.0041

Let me write down the law of cos to get started, the law of cos says that (c2)=(a2+b2) – ((2ab cos(C)).0046

Let me call the missing side (c), which means that its opposite angle is (C) and (a) and (b) will be the sides that we know.0061

Now, we can put all that information into the law of cos and we can solve for the missing side (c).0069

I plug that in, (c2=(82 is 64), (162 is 256) – ((2(8)(16) cos(C)).0077

(64+256=320) – (2x8x16=256), (C) is the given angle, that was given as 45 a degree angle and I know what is the cos of 45 degree is.0094

That is one of the common values that we learned earlier on the trigonometry lectures, the cos of that is the same as pi/4, the cos of the is square root 2/2.0114

That simplifies down to 320-128 square root of 2, (C) is equal to the square root of 320-128 square root of 2.0127

That is probably something worth checking out on the calculator, I work out the square root of 320-128 square root of 2.0144

It tells me that it is approximately 11.8, we have that third side is approximately 11.8 units long.0157

That is the first problem of the example here, now we are asked to find the area of the triangle.0173

I want to do a little more trigonometry to find that area, I am going to drop altitude from this top angle here and I want to try to find the length of that altitude.0182

The reason I’m trying to find is that is I remember the area formula, area=1/2 base x height.0192

I know the base is 16, that is even labeled with side length (b), do I have it labeled on my triangle.0201

The height is the length of that altitude, I have got to solve for that length right there.0211

I’m going to use SOHCAHTOA here because I have the hypotenuse of the triangle and I have the angle here.0217

I know that by SOHCAHTOA, sin(theta)=opposite/hypotenuse, sin(45)=opposite/8 and so the opposite=(8)sin(45) is equal to 8.0224

sin(45) is something I know because it is a common value, it is pi/4.0257

That is square root 2/2, that is 4 square root 2=length of the opposite there, that is 4 square root 2.0261

My area, which is ½ base x height which is ½(16)(4 square root of 2) is what I figured out the height was.0275

That is (8 x 4), that is 32 square root of 2, is my area.0290

If you want that to be a decimal, we can approximate that in the calculator as about 45.3.0297

That gives us the area of the triangle based on the ½ base x height calculation, we really done with this one.0311

I like to check it using another formula we learned from trigonometry which is heron’s formula.0319

Let me remind you what heron’s formula is.0326

Heron’s formula says that the area of the triangle, if you know all three sides, which we did figure out, that is the square root of s(s-a)(s-b)(s-c), that was heron’s formula.0331

(a, b, c) are the lengths of the sides of the triangle but this mysterious quantity (s) is the semi perimeter, that means ½ of the perimeter which is the sum of the sides.0350

Let us work that out first, we know that two of the sides were 8 and 16, and we worked out on the previous side that the third side is approximately equal to 11.8.0363

We put that together, 8 + 16 = 24 +11.8 =35.8 and half of that is 17.9, that is the semi perimeter.0381

Let me drop that in to heron’s formula now, the area is equal to (17.9 x 17.9 – a(8)) ((17.9-b(16))((17.9-c(11.8)).0395

Now it is just a matter of simplifying that, that is 17.9–8 = 9.9, 17.9-16=1.9, 17.9-11.8=6.1.0430

I’m going to multiply those together on my calculator.0452

I get 2053.9, take the square root of that and I get approximately 45.3, which is what we figured out on the previous side.0466

That is a very useful check that we are doing everything right on the previous side.0483

Let us recap what we did there, we are given a triangle with two sides and an included angle (8, 16, and the included angle which is 45).0488

We use the law of cos to find the length of the third side of the triangle.0498

The law of cos is tailor made if you have (side, angle, side), you use the law of cos to find the length of the third side.0504

We then have to find the area of the triangle, I did that the first time by dropping an altitude of the triangle.0512

Find the length of the altitude and use the old geometry formula ½ base x height to find the area.0519

The other way we could possibly find the area was to use heron’s formula, which is useful when you know all three sides of the triangle.0526

You find the semi perimeter which is what we did here and then you take that and you drop it into heron’s formula for the area and you drop all three sides in there.0534

It is just a matter of working through some arithmetic to find the area.0544

That is the end of our lecture on the law of cosines and solving triangles and using heron’s formula.0550

These are the trigonometry lectures on www.educator.com.0555

This is Will Murray for educator.com and we're here today to talk about the law of cosines, which is the second of the two big trigonometric rules.0000

Remember, last time we talked about the law of sines.0007

You kind of put those together, and together those enable you to find the length of any side and the measure of any angle in a triangle, if you're given enough information to start with.0011

The law of cosines is c2=a2+b2-2abcos(C).0024

Let me draw you a a triangle so we can see how that applies.0031

Remember, the convention is that you use lowercase a, b, and c for the sides of the triangle, and uppercase A, B, and C for the angles.0037

You use the same letter for the angle and the side opposite it.0048

My uppercase A goes here, and my B goes here because it's opposite of side b, here's angle C.0053

The point of the law of cosines is it relates the lengths of the three sides a, b, and c, little a, little b and little c, to the measure of one of the angles which is capital C here.0061

The point is that, first of all, you can use this in any triangle.0077

It's not just valid in right triangles.0080

Remember the big rule we had, SOH CAH TOA, is only valid in right triangles.0084

The law of cosines is valid in any triangle.0088

It's a generalization of the Pythagorean theorem, in a sense that, remember the old Pythagorean theorem was just c2=a2+b2, that only works in a right triangle.0093

If you look at the law of cosines, if angle C is a right angle, then the cos(π/2) or the cos(90) is zero.0105

If angle C is a right angle, then this term 2ab-cos(c), drops out,the law of cosines just reduces down to the Pythagorean theorem, c2=a2+b2.0115

You can kind of think of the Pythagorean theorem as just being a consequence of the law of cosines.0128

The law of cosines is the more general one that applies to any triangle.0134

The Pythagorean theorem is the more specific one that just applied when angle C happens to be a right angle.0137

Let's see how it's used.0142

The law of cosines is really used in two situations.0146

First of all, it's used in a side angle side situation.0151

That means where you know two sides of a triangle and the included angle.0155

The reason it's useful ...0160

Let me write the law of cosines again c2=a2+b2-2abcos(C).0162

The point here is that if you label this sides as little a and little b here, that makes this angle capital C and little c is down there.0173

If you know the side angle side, in other words, if you know, little a, little b, and capital C, then you know all of the right-hand side of the law of cosines.0185

You can then solve for little c.0196

That's why the law of cosines is useful for side angle side situations, it's because you can fill in everything you know on one side of the law of cosines, then you can solve for little c.0199

It's also useful for side side side situations.0211

Let me draw that.0218

Side side side means you know all three sides of a triangle, but you don't necessary know any of the angles yet.0219

The point is, if you know little, little b, and little c, then you know all of these parts of the law of cosines, so you can solve for the cosine of capital C, and you can figure out what that angle capital C is.0227

Then you can figure out what one of the angles is.0245

You can just kind of rotate the triangle, and relabel what a, b, and c are to find the other two angles.0248

If you know all three sides of a triangle, the law of cosines is very useful for finding the angles, one at a time.0254

Remember, there's a couple other ways that you can be given information for triangles.0261

You can be given, angle side angle, or side angle angle, or side side angle.0266

Those two don't really lend themselves very well to solution by the law of cosines.0273

If you're given one of those situations then you really want to use the law of sines which we learned about in the previous lecture.0280

There's one more formula we're going to be using in this lecture which is Heron's formula.0289

The point here is that, if you know all three lengths of sides of a triangle, I'll call them a, b, and c, as usual, then you have a nice formula for the area.0294

It's got one more variable in it, this s.0309

S is 1/2 (a+b+c), that's the semi perimeter.0312

Remember, the perimeter is the distance around the edge of a triangle, that's a+b+c.0317

The semi perimeter is just 1/2 of a+b+c.0323

We worked that out ahead of time and we call it s.0327

We plugged that into this formula, that's a fairly simple formula just involving s and then a, b, and c, and it spits out the are of the triangle for us.0331

That's very useful if you know the lengths of the sides.0339

You never really have to look at any angles, and you don't have to get into any sines or cosines, no messy numbers there, hopefully.0342

Let's try out some examples here for law of cosines.0350

First example, we're given a triangle ABC.0353

Let me go ahead and draw that.0355

I'm going to put a here, and b here, and c here, which forces the angles, remember the angles go opposite the sides of the same letter.0362

We're given the a=3,b=4, and angle c measures 60 degrees.0374

We want to first of all determine how many triangles satisfy these conditions and then we want to solve the triangles completely.0383

To answer the first question, we have a side angle side situation.0392

What we know is that side angle side always has a unique solution assuming the angle is less than 180.0400

In this case, the angle is 60 which is less than 180, so there's a unique solution.0408

There's exactly one triangle satisfying these conditions.0419

That answers the first question, how many triangles satisfy those conditions, exactly one.0431

Now we have to solve the triangle completely, this is where the law of cosines is going to be useful.0438

Let me copy down the law of cosines c2=a2+b2-2abcos(C).0444

This is very useful because we know a and b, and capital C.0453

I can just plug all those in and solve for little c.0460

Let me do that, a2 would be 9, b2 would be 16 because 42 is 16, minus 2×3×4=24, cos(60) ...0464

This is 25-24, now cos(60), that's one of my common values, that's π/3, I remember that the cos(60) is 1/2.0483

This is 25-24×1/2, 25-12, c2 is 13, c is equal to the square root of 13.0493

I can get an approximation for that on my calculator, that's about 3.61.0513

That's approximately equal to 3.61.0521

I'll fill that in on my triangle.0530

Now, I've got the third side of the triangle.0536

The only thing that's left is it says to solve the triangles completely.0540

I need to find the other two angles A and B.0544

To do that, I'm going to use the law of sines.0547

Let me write down the law of sines to refresh your memory, sin(A)/a=sin(C)/c.0551

I know what side a, side c, and capital C are.0569

I'm going to cross multiply this and I get csin(A)=asin(C).0574

I'll fill in the values that I know.0583

I know little a is 3, sin(C)=sin(60), sin(A), I don't know that yet, and little c, I figured out, is 3.61.0585

I'm solving for sin(A)=3sin(60)/3.61.0604

I'll just work that out on my calculator.0616

What I get is approximately 0.72.0629

By the way, it's very important that your calculator be in degree mode if you're using degrees here.0637

I gave angle C as 60 degrees.0642

It's very important that you set your calculator to degree mode.0646

If your calculator's in radian mode, then it will interpret that 60 as a radian measure, so your answers will be way off, so set your calculator to be in degrees before you try this calculation.0650

A is arcsin(0.72), I'll work that out on my calculator.0661

That tells me that a is just about 46.0 degrees.0673

Now, I've got a measure for angle A.0683

I'm going to use the law of sines to find the measure for angle B, but I need a little more space.0688

Let me redraw my triangle.0694

We've got a, b, and c.0704

I figured out the c was 3.61, a was given as 3, b was given as 4, C was given as 60.0711

I figured out that A was 46 degrees.0721

I'm just trying to find the measure of angle B now.0727

I'm going to use again the law of sines.0729

Sin(B)/b=sin(C)/c, I'll fill in what I know here, I know that little b is 4, sin(B), I don't know, sin(C)=sin(60), little c is 3.61.0734

I'll cross multiply that, 3.61sin(B)=4sin(60), sin(B), that's what we're solving for is equal to 4sin(60)/3.61.0760

Let me work that out on my calculator, 4sin(60)/3.61=0.96.0784

B is arcsin(0.96) and I'll work that out, that's 73, just about 74 degrees, rounds to 74 degrees.0801

Now I figured out angle B, 74 degrees.0828

Now, I've solved for all three sides of the triangle, and all three angles of the triangle.0837

It's nice at this point, even though we're done with the problem to get some kind of check because we've done lots of calculations here, we could have made a mistake.0841

What I'm going to do is add up all three angles in the triangle, and make sure that they come up to be 180 degrees.0848

To check on my work here, I'll add up 60+46+74, that does indeed come out to be 180 degrees.0856

That suggests that we probably didn't make a mistake in solving all those angles.0877

Just to recap this problem here, we're given a side angle side situation, that's a definite tip-off that you're going to be using the law of cosines.0882

I filled in my side, the included angle, and a side.0891

The first thing I did was I used the law of cosines to find the missing side.0895

To solve the triangle completely, I still had two angles that I didn't know, I used the law of sines after that to find the two angles that I didn't know based on knowing the other sides and the other side, and angle.0903

Let's try another one now.0918

In this one, we're given the side lengths of the triangle, a, b and c are 5, 7 and 10.0922

Let me draw a possible triangle like that, 5, 7 and 10.0926

We want to find out how many triangles satisfy this conditions and solve the triangles completely.0935

The first thing to do with this problem is to identify what we're given.0941

We're given a side side side configuration.0945

That usually gives you a unique triangle.0950

What you have to do is check that each side is less than the sum of the other two.0952

Let's check that out.0960

Unique if each side is less than the sum of the other two.0963

That'll be a quick real check.0980

We're comparing 5 with 7+10, 5 is certainly less than 17, so that works.0983

7 should be less than 5+10, that certainly works, 7 is less than 15.0989

10 should be less than 5+7, that certainly works.0995

If any of those checks had failed, for example 10 if it had been 13, 5, and 7, instead of 10, 5, and 7, then the last check would have failed because 13 is not less than 5+7.1001

At that point, we would have stopped and said, "This is invalid. There is no triangle that satisfies those conditions."1015

Since all those three conditions checked, it does mean that there is a unique solution.1022

There's exactly one triangle with those three lengths of sides.1036

We've found all the side lengths.1043

We need to find the angles, this is where the law of cosines is really handy.1046

Let me write that down, the law of cosines says c2=a2+b2-2abcos(C).1051

You can label the sides and angles whichever way you want.1061

I'm going to label the top one c.1067

Let me write that outside of the triangle.1075

That makes the bottom side c and then the two other sides a and b.1077

What I can do here is I can plug in little a, little b and little c, and then I can solve for the cosine of capital C, then in turn solve for what angle capital C is.1085

Let me do that.1098

If c is 10, that's 102 equals, a and b is 52+72-2×5×7×cos(C).1100

Now it's just a little bit of arithmetic, 100=25+49-70cos(C).1115

25+49=74, if we pull that over to the other side, we get 26, -70cos(C).1128

Cos(C)=-26/70.1145

I'm going to figure out that C is arccosine or inverse cosine of -26/70, I'll do that part on my calculator.1153

Remember, you have to be in degree mode for this.1180

What I get is that C is approximately equal to 111.8 degrees.1186

That's 111.8 in that corner.1196

Now I'm going to go over in the next page.1201

I'm going to keep going to find out the other two angles.1204

We'll find them exactly the same way.1206

Let me go ahead and redraw my triangle, 5, 7, 10.1207

We've already figured out that that angle is 111.8.1216

What I'm going to do is relabel the sides and the angles because I still want to use that law of cosines c2=a2+b2-2abcos(C).1220

I'm going to relabel everything here so that I can label C as a new angle, and I can solve for a new angle.1233

Relabel that angle as capital C, then my a and b will be 7 and 10.1242

Sorry, small C will be 7, and a and b will be 5 and 10.1256

We'll go through and we'll work that one out.1263

Little c2, that's 49 is equal to a2, that's 100, plus b2 is 25, minus 2ab, that's 2×10×5×cos(C).1267

Let me work this out, this is 125, 49 equals 125 minus 2×10×5, that's 100, cos(C).1285

If I subtract 125 from both sides, I get -76 is -100cos(C), divide both sides by -100, we get 76/100=cos(C).1300

C=arccos(76/100), I'll plug that into my calculator.1324

It tells me that that's approximately 40.5 degrees for that angle right there.1342

Now, there's one angle left to find.1362

Again, I'm going to relabel the sides and the angles so that I can continue to use the law of cosines in its standard form.1364

I don't have to change around what a, b and c are in the law of cosines.1373

I'll do this in red.1375

In red, I'm going to call this angle capital C, and that means I have to relabel my sides, that means little c is equal to 5, a and b are 7 and 10.1380

Now I'm going to plug those values into the law of cosines.1399

c2 is 25, equals 49, 72, plus b2 is 100, minus 2×a×b, 7×10, times cos(C).1403

Now it's a matter of solving that for capital C again.1422

I get 25 equals 149, minus 2×7×10, that's 140, cos(C).1424

If I subtract 149 from both sides, I get -124, is equal to -140cos(C).1440

Cos(C)=124/140, C=arccos(124/140), I'll go to the calculator on that.1455

I get 27.7 degrees.1481

Let me fill that in, 27.7 degrees.1490

Now we've solved the triangle completely.1494

We started out with all three side lengths and we found all three angles in the triangle.1497

We're really done but it's always good to find a way to check your work.1503

Let me check my work in blue here.1506

Again, I found all three angles.1508

I'm going to add them together and see if I get 180 to check that out.1511

I'm going to add up 111.8+40.5+27.7, what I get is exactly 180 degrees.1514

That tells me that I must have been right in getting those three angles for the triangle.1535

Just to recap here, what we were given was three sides of a three angle.1540

We were given three side lengths, that was a side side side situation.1546

We had to check that the three lengths, that we didn't have a situation where two sides added up to be less than the third side.1554

We had to check that each side was less than the sum of the other two.1560

Once we did that, we knew we have exactly one solution.1564

Then we filled in the side lengths of the triangle, and we used the law of cosines.1568

The law of cosines lets you fill in three side lengths and then solve for the cosine of one of the angles.1574

You work it through, solve for the cosine, then you get each one of the angles by taking the arccosine, then you just go through a separate procedure like that for each one of the angles.1580

In our third problem, we're trying to find the area of a triangle whose side lengths are 5, 7 and 10.1595

Now, this one is really a set up for Heron's formula, because Heron's formula works perfectly when you know the three sides of a triangle.1603

We're going to use Heron here.1612

Heron says that the area is equal to the square root of s times s minus a, s minus b, and s minus c.1618

You've got to figure out what s is.1632

S is the semi-perimeter of a triangle which means 1/2 of the perimeter, 1/2 of the sum of the three side lengths.1635

That's 1/2 of 5+7+10, which is 1/2 of 22, which is 11.1646

I'm going to plug that in to Heron's formula, wherever I see an s, that's 11×(11-5)×(11-7)×(11-10).1658

Then I'll just work on simplifying that, that's 11×6×4×1, that's the square root of 264, 11×24.1677

That problem was really pretty quick if you remember Heron's formula.1697

Heron's formula is very useful.1700

If you know three side lengths of a triangle, then what you do is you work out the semi-perimeter, you just drop the side lengths into this formula for the semi-perimeter, then you drop the semi-perimeter and the three side lengths for the Heron's formula for the area.1703

It simplifies down pretty quickly to give you the area of the triangle.1720

We'll try some more examples later on.1722

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