Professor Murray

Finding the Area of a Triangle

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

 1 answerLast reply by: Dr. William MurrayWed Aug 17, 2016 3:19 PMPost by Julian Xiao on August 17, 2016Doesn't the Heron's formula have a typo at around 3:00? 1 answerLast reply by: Dr. William MurrayWed Aug 14, 2013 12:05 PMPost by Ikze Cho on August 7, 2013does heron's formula work for all triangles? Thanks 1 answerLast reply by: Dr. William MurrayThu May 30, 2013 4:09 PMPost by Mark Mccraney on January 19, 2010III, lecture 4, 13:55, the supplemental angle is figured as 180-121.6=59.4 This is incorrect; the answer is 58.4 and throws off the Area by almost 3/10. Answer should come out to approx. 26.8 vs 27.1 given. 1 answerLast reply by: Dr. William MurrayThu May 30, 2013 4:06 PMPost by Mark Mccraney on January 19, 2010My highschooler informed me today how they learned SOHCAHTOA since TOA could get reversed: taking away oats or taking oats away. The high school version is Some Old Hippie Caught Another Hippie Tripping On Acid!!! Can't get that one twisted!!!

### Finding the Area of a Triangle

Main formulas:

• Master formula for right triangles: SOHCAHTOA!
 sinθ = opposite sidehypotenuse cosθ = adjacent sidehypotenuse tanθ = opposite sideadjacent side
• The Law of Cosines (in any triangle):
 c2 = a2 + b2 − 2abcosC
• 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:

Find the area of a triangle that has two sides of lengths 8 and 12 with an included angle of 45° .

Example 2:

Find the area of a triangle with side lengths 10, 14, and 16.

Example 3:

Without using Heron's formula, find the area of a triangle whose side lengths are 7, 9, and 14.

Example 4:

Find the area of a triangle that has two sides of lengths 7 and 13 with an included angle of 32° .

Example 5:

A triangle has two sides of length 5 and 6 with an included angle of 60° . Find the area of the triangle.

### Finding the Area of a Triangle

Find the area of a triangle that has two sides of length 6 and 10 with an included angle of 54°. (Do not use Heron's Formula to solve)
• Start by drawing a picture of the triangle with the sides and angles labeled
• Area of a Triangle = [1/2]bh
• Use SOHCAHTOA to find the height of the triangle
• sinθ = [Opposite/Hypotenuse] ⇒ sin54° = [h/6] ⇒ h = 6sin54° ⇒ h ≈ 4.85
• A = [1/2](10)(4.85)
A ≈ 24.3
Find the area of a triangle that has two sides of length 11 and 13 with an included angle of 63°. (Do not use Heron's Formula to solve)
• Start by drawing a picture of the triangle with the sides and angles labeled
• Area of a Triangle = [1/2]bh
• Use SOHCAHTOA to find the height of the triangle
• sinθ = [Opposite/Hypotenuse] ⇒ sin63° = [h/11] ⇒ h = 11sin63° ⇒ h ≈ 9.8
• A = [1/2](13)(9.8)
A ≈ 63.7
A triangle has two sides of length 5 and 7 with an included angle of 34°. Find the area of the triangle.
• Start by drawing a picture of the triangle with the sides and angles labeled
• Use the Law of Cosines to find the length of the missing side
• a2 = b2 + c2 − 2bc  cosA ⇒ a2 = 72 + 52 − 2(7)(5)cos34° ⇒ a2 = 15.9672 ⇒ a = √{15.9672}
• a ≈ 4.0
• Now use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• s = [1/2](4 + 7 + 5) ⇒ s = 8
• A = √{8(8 − 4)(8 − 7)(8 − 5)} ⇒ √{8(4)(1)(3)} √{96}
A ≈ 9.8
Find the area of a triangle with side lengths 11, 15, and 17
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• a = 11, b = 15, c = 17
• s = [1/2](11 + 15 + 17) ⇒ s = 21.5
• A = √{21.5(21.5 − 11)(21.5 − 15)(21.5 − 17)} ⇒ √{21.5(10.5)(6.5)(4.5)} = √{6603.19}
A ≈ 81.26
Find the area of a triangle with side lengths 7, 12, and 15
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• a = 7, b = 12, c = 15
• s = [1/2](7 + 12 + 15) ⇒ s = 17
• A = √{17(17 − 7)(17 − 12)(17 − 15)} ⇒ √{17(10)(5)(2)} = √{1700}
A ≈ 41.23
Find the area of a triangle with side lengths 10, 17, and 23
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• a = 10, b = 17, c = 23
• s = [1/2](10 + 17 + 23) ⇒ s = 25
• A = √{25(25 − 10)(25 − 17)(25 − 23)} ⇒ √{25(15)(8)(2)} = √{600}
A ≈ 77.46
Find the area of a triangle that has two sides of length 15 and 20 with an included angle of 50°.
• Start by drawing a picture of the triangle with the sides and angles labeled
• Area of a Triangle = [1/2]bh
• Use SOHCAHTOA to find the height of the triangle
• sinθ = [Opposite/Hypotenuse] ⇒ sin50° = [h/15] ⇒ h = 15sin50° ⇒ h ≈ 11.49
• A = [1/2](20)(11.49)
A ≈ 114.9
A triangle has two sides of length 3 and 7 with an included angle of 20°. Find the area of the triangle.
• Start by drawing a picture of the triangle with the sides and angles labeled
• Use the Law of Cosines to find the length of the missing side
• a2 = b2 + c2 − 2bc  cosA ⇒ a2 = 72 + 32 − 2(7)(3)cos34° ⇒ a2 = 18.533 ⇒ a = √{18.533}
• a ≈ 4.3
• Now use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• s = [1/2](4.3 + 3 + 7) ⇒ s = 7.15
• A = √{7.15(7.15 − 4.3)(7.15 − 3)(7.15 − 7)} ⇒ √{7.15(2.85)(4.15)(0.15)} = √{12.685}
A ≈ 3.56
Find the area of a triangle with side lengths 8, 12, and 16
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• a = 8, b = 12, c = 16
• s = [1/2](8 + 12 + 16) ⇒ s = 18
• A = √{18(18 − 8)(18 − 12)(18 − 16)} ⇒ √{18(10)(4)(2)} = √{1440}
A ≈ 37.95
Find the area of a triangle with side lengths 13, 15, and 18
• Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} where s = [1/2](a + b + c)
• a = 13, b = 15, c = 18
• s = [1/2](13 + 15 + 18) ⇒ s = 23
• A = √{23(23 − 13)(23 − 15)(23 − 18)} ⇒ √{17(10)(8)(5)} = √{9200}
A ≈ 95.92

*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.

### Finding the Area of a Triangle

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
• Master Right Triangle Formula and Law of Cosines 0:19
• SOHCAHTOA
• Law of Cosines
• Heron's Formula 2:22
• Semiperimeter S
• 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
• Extra Example 2: Area of Triangle with Two Sides and One Angle

### Transcription: Finding the Area of a Triangle

Hi we are trying more examples of problems where you have to find the area of a triangle given various data, this one we are given two sides (7, 13, and an included angle of 32 degrees).0000

Let me draw out the triangle there, (7, 32, and 13), I want to find the area of this triangle.0015

I’m going to drop perpendicular and altitude here and I’m going to use SOHCAHTOA to find the length of that perpendicular.0034

Remember, sin(theta) = opposite/hypotenuse and sin(32) = the altitude of the triangle is the opposite side that I am looking for and the hypotenuse is 7.0042

Opp=(7)sin(32), I will check that on my calculator and it says that it is approximately equal to 3.7, that is this length right here, 3.7.0064

The area, the old fashioned area formula is ½ base x height, which is equal to ½, now the base is 13, we were given that, x 3.7.0083

What we get is 24.1.0111

Let us recap what we did there, I drew out what I knew on the triangle.0122

I knew two sides and an included angle, I wanted to find an area.0126

I wanted to find base and height of the triangle so I dropped this altitude down from the top corner and then I found out the length for the altitude using SOHCAHTOA and the altitude and hypotenuse that I already knew.0130

Remember SOHCAHTOA does not work in evry triangle, it only works in right triangles.0144

What I did here, by dropping the altitude, the perpendicular, was I made a little right triangle inside the triangle that we are given so it is ok to use SOHCAHTOA there.0149

Once I found the length of that altitude, I just used area is 1/2 base x height.0160

I knew the base, height, and I plugged down it into the formula and simplified it down to 24.1.0167

Now I know the area of the triangle.0173

Alright our last example here, we are given two sides of a triangle and an included angle.0000

We want to find the area of the triangle, let me draw out what we know.0007

We got side of length 5, side of length 6, and the included angle we are told is 60 degrees.0012

I want to find the area of the triangle, what I would like to do is find the length of the third side of the triangle.0030

I can do that since I know (side, angle, side), I can find the length of the third using the law of cos so I will do that first.0037

Once I know the length of all three sides, I’m going to use heron’s formula to find the area of the triangle.0047

Let me write down the law of cos, c2 = a2 + b2 - ((2ab) cos (C)).0052

That is really useful if you know two sides and an included angle because if you call those two sides (a and b), that makes (c) the third side.0065

(C) will be the angle in between (a) and (b) because it should be opposite (c).0076

We can plug all those into this formula and solve for (c), c2 = 25 that is 52 + 62 is 36 – (2)(5)(6) cos(C) which is 60 degrees.0082

(25 + 36 = 61) – (2×5×6=60) cos(60) that is a common value, I know that one, that is pi/3.0101

I know the cos of pi/3 is ½.0114

This is 61-30=31 that was c2 and c=square root of 31 which is approximately equal to 5.6.0118

My third side here is approximately equal to 5.6.0141

I’m going to the next slide to finish this so let me fill in what I just figured out.0149

I know that I had (5 and 6) and I just figured out that the third side was approximately equal to 5.6.0158

That angle was 60 but we are not going to be using that anymore.0166

Now I want to find the area of the triangle since I know the three sides already, this will going to be quick using heron’s formula.0170

Heron says that the area = square root(s)(s-a)(s-b)(s-c).0181

You have to know what (s) is, (s) is the semi perimeter so ½ x the full perimeter (a+b+c).0196

Let us work that out first, then we will drop that into heron’s formula.0206

This is ½ of (5+6+5.6), 5+6=11+5.6=16.6, ½ of 16.6 is 8.3.0211

Now I’m going to drop that into heron’s formula and also the lengths of the three sides because I took some trouble to work out the length of the third side.0232

My (s) is 8.3 then I have (8.3- the first side=5), (8.3 -the second side=6) and finally (8.3- the third side =5.6).0245

It is just a matter of simplifying the numbers,(8.3 x 3.3 x 2.3) and (8.3 – 5.6=2.7).0271

I will multiply those numbers together, what I get is 170.1.0287

To take the square root of that I get just about 13.04 as my area.0303

Let us recap what we needed to do for that problem.0318

We were given two sides and an included angle so that is (side, angle, side) and we wanted to find the area of the triangle.0321

We used some different methods in the previous problems, this time what we used was the law of cos to find the third side.0329

We used the law of cos there and that was really useful because the law of cos is perfect when you have (side, angle, side).0338

You just drop the (side, angle, side) into your law of cos and it will tell you what the third side is.0349

Once you have all three sides, then you use heron’s formula which gives you this nice formula for the area.0355

It does not look at the angle at all, it just uses the three sides and this quantity (s).0362

But then (s) just comes back to popping in the three sides into the semi perimeter formula.0366

We worked out the semi perimeter, it came to 8.3, drop that in for the (s) in heron’s formula, drop the three sides in and then just simplify it down to find the area of the triangle.0372

What I hope you taken away from this is that there are several different ways you can find the area of a triangle.0387

You can use the law of cos, heron’s formula, we also used SOHCAHTOA, and ½ base x height.0392

Different methods work better in different situations but they all help you find the area of a triangle.0399

These are the trigonometry lectures for www.educator.com, thanks for watching.0406

We're working on the trigonometry lectures, and today we're talking about finding the area of a triangle.0000

We've learned several important formulas over the past few lectures.0007

Today, we'll be combining them all and learning a few different methods to find the area of a triangle depending on what kind of initial data you're given.0009

I want to remind you about some very important formulas, first of all the master formula that works for right triangles is SOH CAH TOA.0019

Let me remind you how that works.0028

It only works in a right triangle.0030

You have to be very careful of that.0032

You have to have one right angle.0034

If you talk about one of the other angles, θ, then you label all the sides as the hypotenuse, the side opposite θ and the side adjacent θ.0036

SOH CAH TOA stands for the sin(θ) is equal to the length of the opposite side over the hypotenuse, the cos(θ) is the length of the adjacent side over the hypotenuse, the tan(θ) is equal to the opposite side over the adjacent side.0049

Remember, we have a little mnemonic to remember that, if you can't remember SOH CAH TOA, you remember Some Old Horse Caught Another Horse Taking Oats Away.0065

Remember, SOH CAH TOA only works in right triangles, you have to have a right angle to make that work.0078

The law of cosines works in any triangle, I'll remind you how that goes.0083

We're assuming here that your sides are labeled a, b, and c, little a, little b, and little c, then you label the angles with capital letters opposite the sides with the same letter, that would make this capital A, capital B, and this, capital C.0090

The law of cosines relates the lengths of the three sides, little a, b and c, to the measure of one of the angles, c2=a2+b2-2abcos(C).0107

This works in any triangle, it does not have to be a right triangle.0121

In fact, if it happens to be a right triangle then the cos(C), if C is a right angle, the cosine of C is just 0, that whole term drops out and you end up the Pythagorean theorem for right triangle, c2=a2+b2.0124

Finally, the important formula that we're going to be using for areas is Heron's formula.0141

That's very useful when you know all the three sides, a, b and c, of a triangle ABC.0148

First, you work out this quantity s, the semi-perimeter, where you add up a, b and c, that's the perimeter, divide by 2, so you get the semi-perimeter, then you drop that into this area formula, and you drop in the lengths of all three sides.0158

Heron's formula gives you a nice expression for the area of a triangle without ever having to look at the angles at all.0174

That's very useful as well.0181

We'll practice combining all those formulas in different combinations and see how we can calculate the areas of triangles in various ways.0184

The first example, we're given a triangle that has two sides of length 8 and 12 with an included angle of 45 degrees.0193

Let me set that up.0200

That's 8, that's 12, and this is 45 degrees.0203

I'll draw in the third side there.0210

I'm not going to try to use Heron's formula yet, I'm going to try to use old-fashioned SOH CAH TOA here.0211

The thing is, remember, SOH CAH TOA only works in right triangles.0217

I don't necessarily have a right triangle here.0221

What I'm going to do is draw an altitude, drop a perpendicular from the top corner of that triangle.0222

Now I do have a right triangle.0233

I'm going to try and find the length of that altitude, I'm going to use SOH CAH TOA.0234

Sin(θ) is equal to the opposite over the hypotenuse.0240

Sin(45) is equal to the opposite, now the hypotenuse of that little triangle is 8, so root 2 over 2 is equal to the opposite over 8.0247

That's because I know the sin(45), that's one of my common values, that's π/4.0262

I memorize the sine, cosine and tangent of all the common values way back earlier in the course.0267

If you haven't memorized that, you really should commit all those common values 45 and multiples of 30 to memory.0273

If I solve for the opposite here, I get that the length of the opposite is equal to 8 root 2 over 2, that's 4 root 2.0280

That means that that altitude is 4 root 2.0291

Now we can use the old formula from geometry for the area of a triangle, just 1/2 base times height.0300

The h stands there for height instead of hypotenuse.0309

I know that's a little confusing to be using h for two different things, but we're kind of stuck with that in English that hypotenuse and height both start with the same letter.0311

One-half the base here is 12, and the height we figured out was 4 square root of 2.0320

We multiply those out, that's 6 times 4 square root of 2, that's 24 square root of 2 for my area there.0329

That one came down to drawing an altitude in the triangle and then using SOH CAH TOA.0341

We didn't really have to use anything fancy like the law of cosines or Heron's formula, although we could have.0347

You'll see some examples later where we use the law of cosines and Heron's formula instead.0352

In this one, we just drew this altitude, we used SOH CAH TOA to find the length of the altitude, then we used the old-fashioned geometry formula, 1/2 base times height, to get the formula of the triangle.0358

In the next example, we're given a triangle with side lengths 10, 14 and 16.0372

Let me draw that.0377

10, 14, and 16 ...0382

We're asked to find the area.0384

If you have all three sides of a triangle and you're asked to find the area, it's kind of a give-away that you're going to use Heron's formula.0386

Because Heron's formula works very nicely based on the side lengths of the triangle only.0394

You never even have to figure out what the angles are.0399

That's what we're going to use.0402

Heron, remember, says that you have to start out by finding the semi-perimeter.0404

That's (1/2)×(a+b+c), that's (1/2)×(10+14+16).0410

(10+14+16) is 40, a half of that is 20.0422

Now I know what s is.0425

Heron says I plug that into my area formula which is this big square root expression of s times s minus a, s-minus b, and s minus c.0427

My s was 20, a is 10, b is 14, and c is 16.0444

Now this is just a very easy simplification, this is 20 times 10, times 20-14 is 6, and 20-16 is 4, so 20×10 is 200, times 24 is 4800.0460

I can simplify that a bit.0485

I can pull off, let's see, I can pull a 10 out of there, and that's 10 square root of 48.0488

Now, I can pull a 16 out and make that 40 square root of 3 for my area.0498

Let's recap what we're given there.0506

We're given a triangle and we were told the three side lengths.0508

If you know the three side lengths and you're going for the area, you almost certainly want to use Heron's formula.0513

It's a quick matter of finding the semi-perimeter, and then dropping the semi-perimeter and the three side lenghts into the square root formula.0518

Then you just simplify down and you get the area.0526

In this third example, we're asked to find the area of a triangle whose side lengths are 7, 9 and 14.0531

Ordinarily, I'd say that's a dead give-away that you want to use Heron's formula, but unfortunately, this example they specifically say without using Heron's formula.0539

I'd really like to use Heron's formula on that but it has asked me to find the area of the triangle without using Heron's formula.0549

I'm going to do a little bit of extra work here.0558

I'll start out by drawing my triangle.0561

There's 7, 9, and 14.0565

I think I'm going to try to find that angle right there.0572

I'll call that angle C.0575

The reason I called it angle C is because I'm planning to use the law of cosines to find that angle.0577

Remember the law of cosines is very useful if you know all three sides.0581

You can quickly solve for an angle.0587

Let me remind you what the law of cosines is.0589

It says c2=a2+b2-2abcos(C).0592

If you know little a, b and c, you can just drop them into that formula, and you can solve for capital C.0602

That's what we're going to do here.0607

I wrote 19 for that side, of course, that's supposed to be 9.0610

My c is the side opposite angle C, a is 7 and b is 9.0617

I'm going to plug these into the law of cosines, and it says, 142=72+92-2×7×9×cos(C).0625

I'll do a little algebra on that.0646

142 is 196, 72 is 49, 92 is 81.0648

I was looking ahead to this next calculations, 7×9 is 63, 2 times that is 126.0662

We get 126cos(C).0671

49+81 is 50+80, that's 130.0674

If I move that to the other side, I get, let's see, 130 from 196 is 66, is equal to -126cos(C), so cos(C), solving for that part, that's what we don't know, is -66/126.0682

If I take the arccosine of that on my calculator, of course I've got to use degree mode if I'm planning to use degrees for this problem that I am, I'll take the arccos(-66/126).0705

What I get is 121.6 degrees approximately for angle C.0723

Normally, I would just fill that into my drawing, but in my drawing, I've shown it as an acute angle and that really isn't appropriate because 121.6 is bigger than 90 degrees.0731

I think I have to modify my drawing based on what C came out to be, and draw that part as an obtuse angle.0743

I'll redraw there.0756

That's angle C which we figured out was 121.6 degrees.0762

That's side c which is 14, that's side a which is 7, and this is side b which is 9.0769

What I'm going for is I'm trying to find the area of the triangle ultimately, but I can't use Heron's formula because the problem specifically told me I wasn't allowed to use Heron's formula.0777

I want to try to use the old-fashioned 1/2 base times height, but I don't know the height of the triangle.0791

I'm going to try to find the height of the triangle, that's why I had to find that missing angle.0799

To find the height of the triangle, I'm going to drop an altitude here.0804

Since it's an obtuse angle in the triangle, this altitude is actually outside the triangle.0809

There's that altitude.0817

I want to find the height of that altitude.0820

In order to find it, I have to find that angle right there, and that is the supplement of 121.6.0821

So θ=180-121.6, which is approximately equal to 59.4 degrees.0832

That tells me what angle θ is.0845

Based on that and this a=7, I can find the missing height of the triangle.0848

I'm going to use SOH CAH TOA for that.0857

Sin(θ) is equal to the opposite over the hypotenuse.0860

Sin(59.4) times the hypotenuse, is 7 ...0869

We don't know what the opposite length is.0880

I'll leave that there.0884

I get that the opposite is equal to 7sin(59.4).0885

I'll work that out on my calculator.0894

What I get is that that's approximately equal to 6.03.0910

That length right there that we found, 6.03.0919

I'm going to need a little more space to work this out, I'm going to go to the next slide here and just redraw my triangle, and remind you what we know about that.0923

We were given this triangle with side lengths 7, 9, and 14.0935

The first thing we did was we used the law of cosines to find this missing angle.0942

That was 121.6 degrees.0947

Then we dropped a perpendicular, an altitude, from the top angle there.0953

To find that, we had to figure out that θ there was 59.4 degrees.0960

Then using that value of θ and the hypotenuse of 7, we figured out that this was 6.03 units long.0968

Now we're in good shape to find the area of the triangle using old-fashioned geometry.0978

It's just 1/2 times the base times the height.0983

The base here is 9.0986

That's the base.0992

You might be worried a little bit that the top of the triangle sticks out a bit over the edge of the base, that actually doesn't matter.0994

That doesn't make the formula invalid.1002

It is still ...1004

You count the base as being 9, even though the top of the triangle sticks out over the edge of the base.1005

The height we've just worked out is 6.03.1011

Now I'm going to multiply that by 9 and by 1/2, and I get approximately 27.1 for the area of the triangle.1019

Let's recap what happened there.1033

I was given three sides for a triangle.1036

Normally when you're given three sides and you want to find the area, that's a dead give-away that you want Heron's formula.1038

Unfortunately, this example asked us to find it without using Heron's formula.1043

That's why we used law of cosines to find that angle, and then we used SOH CAH TOA to find the length of an altitude of a triangle, in other words, the height of a triangle.1049

SOH CAH TOA right there.1065

Then we used the old-fashioned area formula using the base and the height to give us the area of the triangle.1067

We'll try some more examples of these later.1073

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