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Page 1: Areas and Approximations - math.berkeley.edu

Areas and Approximations

Jeff Hicks

Apr. 17, 2019

UC Berkeley

Jeff Hicks (UC Berkeley) Areas and Approximations Apr. 17, 2019 1 / 28

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Page 2: Areas and Approximations - math.berkeley.edu

Outline

• Computing Areas of Triangles

• Computing Areas of other polygons

• Approximations of sin(θ) and cos(θ) for small θ.

• Calculating distances

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Page 3: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Page 4: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Computing the Area of a Triangle

Formula

A triangle has area b·h2 where b is the length of one of the sides (the

“base”), and h is the height of the triangle as measured by an altitude.

b

h

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Area of a Triangle

How can we use trig to help

Let’s take a closer look at this triangle:

b

hc

θ

We see that the height h = c · sin(θ). This makes the area

1

2b · h =

1

2b · c · sin(θ)

Notice that the area formula does not contain the height of the triangle.

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Page 6: Areas and Approximations - math.berkeley.edu

Area of a Triangle

How can we use trig to help

Let’s take a closer look at this triangle:

b

hc

θ

We see that the height h = c · sin(θ). This makes the area

1

2b · h =

1

2b · c · sin(θ)

Notice that the area formula does not contain the height of the triangle.

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Page 7: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Formula

The area of a triangle with sides b and c meeting at an angle θ is:

A =1

2b · c · sin(θ)

Example

Find the area of an equalateral triangle which has a side length of 5.

5

5

π3

1

2b · c sin(θ) =

1

2· 5 · 5 · sin(π/3)

=1

2c 2̇5 ·

√3

2

=25√

3

4.

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Page 8: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Formula

The area of a triangle with sides b and c meeting at an angle θ is:

A =1

2b · c · sin(θ)

Example

Find the area of an equalateral triangle which has a side length of 5.

5

5

π3

1

2b · c sin(θ) =

1

2· 5 · 5 · sin(π/3)

=1

2c 2̇5 ·

√3

2

=25√

3

4.

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Page 9: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Some other observations about the rule

When θ is close to 0, our triangle looks like this:

b

c

θ

Notice that even if b and c are large, the area of this triangle will be small.

This is because when θ ∼ 0, then sin(θ) ∼ 0.

1

2b · c · sin(θ) ∼ 1

2b · c · 0 ∼ 0

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Page 10: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Some other observations about the rule

When θ is close to 0, our triangle looks like this:

b

c

θ

Notice that even if b and c are large, the area of this triangle will be small.

This is because when θ ∼ 0, then sin(θ) ∼ 0.

1

2b · c · sin(θ) ∼ 1

2b · c · 0 ∼ 0

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Page 11: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Some other observations about the rule

When θ is close to π/2, our triangle looks like this:

b

c

θ

When θ ∼ 0, then sin(θ) ∼ 1.

1

2b · c · sin(θ) ∼ 1

2b · c

which looks a lot like the formula for the area of a right triangle.

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Page 12: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Some other observations about the rule

When θ is close to π/2, our triangle looks like this:

b

c

θ

When θ ∼ 0, then sin(θ) ∼ 1.

1

2b · c · sin(θ) ∼ 1

2b · c

which looks a lot like the formula for the area of a right triangle.

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Area of a Triangle

Some Other observations about the rule

The formula1

2b · c sin(θ)

treats the variables b and c on equal footing.

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Page 14: Areas and Approximations - math.berkeley.edu

Area of a Triangle

What about other shapes?

b

hc

θ

The area of a parallelogram is

A = b · h

Formula

The area of a parallelogram with sides b and c meeting at an angle θ is:

A = b · c · sin(θ)

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Page 15: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Example

What is the area of the following parallelogram

5

3√

2

π/4

A =b · c · sin(θ)

=5 · 3√

2 · sin(π/4)

=5 · 3√

2 ·√

2

2

=15.

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Page 16: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Example

What is the area of the following parallelogram

5

3√

2

π/4

A =b · c · sin(θ)

=5 · 3√

2 · sin(π/4)

=5 · 3√

2 ·√

2

2

=15.

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Page 17: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Relation between Parallelograms and Triangles

We could have also could have gotten this formula from cutting the

parallelogram into two .

5π/4

π/4

This is a trick we can use to find the area of many different shapes.

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Page 18: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Area of a regular Pentagon

What is the area of a regular pentagon when the distance from the center

to the vertex is 3?

The area of each triangle is

1

23 · 3 sin(θ)

The angle θ = 2π5 . This gives us that

the area of the whole pentagon is

5·(

1

23 · 3 sin(2π/5)

)=

75

3((√

10 + (2√

5))/4)

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Page 19: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Area of a regular Pentagon

What is the area of a regular pentagon when the distance from the center

to the vertex is 3?

The area of each triangle is

1

23 · 3 sin(θ)

The angle θ = 2π5 . This gives us that

the area of the whole pentagon is

5·(

1

23 · 3 sin(2π/5)

)=

75

3((√

10 + (2√

5))/4)

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Page 20: Areas and Approximations - math.berkeley.edu

Area of a Triangle

Area of a n-gon

Repeating on a shape with n equal sides and radius r ,

r

Area is

n·(

1

2r · r · sin

(2π

n

))=

n

2

(sin

(2π

n

))r2

This looks suspiciously like the formula

for the area of the circle. Notice that

as n becomes very large, this better and

better approximates the area of a circle.

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Page 21: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Page 22: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Comparing circles

Our estimate from the last slide says that when n is very large

n

2

(sin

(2π

n

))r2 ≈ π · r2

where the right hand side is the area of a circle.

We conclude that when n is very large,

n

2

(sin

(2π

n

))≈π

sin

(2π

n

)≈2π

n

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Page 23: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Comparing circles

Our estimate from the last slide says that when n is very large

n

2

(sin

(2π

n

))r2 ≈ π · r2

where the right hand side is the area of a circle.

We conclude that when n is very large,

n

2

(sin

(2π

n

))≈π

sin

(2π

n

)≈2π

n

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Page 24: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Letting θ = 2πn when n is very big...

Identity

When θ is very small, sin(θ) ≈ θ.

Remark

This only works if θ is measured in radians! Fun Fact: This is the real

reason why we use radians!

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Page 25: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Identity

When θ is very small, sin(θ) ≈ θ.

sin(θ) θ

A large difference be-

tween sin(θ) and θ

sin(θ) θ

A large difference be-

tween sin(θ) and θ

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Page 26: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

In red, we have the graph of y = sin(θ), while in blue we have the graph of

y = θ.

−1 1

−1

0

1

θ

y

y

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Page 27: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Example

Apprxomate the value of 5 · sin(.01).

5 · sin(0.1) ≈ 5 · (0.1) = .5

Example

Given that sin(θ) = .02, approximate θ.

θ is approximately .02.

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Page 28: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Example

Apprxomate the value of 5 · sin(.01).

5 · sin(0.1) ≈ 5 · (0.1) = .5

Example

Given that sin(θ) = .02, approximate θ.

θ is approximately .02.

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Page 29: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Example

Apprxomate the value of 5 · sin(.01).

5 · sin(0.1) ≈ 5 · (0.1) = .5

Example

Given that sin(θ) = .02, approximate θ.

θ is approximately .02.

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Page 30: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Example

Apprxomate the value of 5 · sin(.01).

5 · sin(0.1) ≈ 5 · (0.1) = .5

Example

Given that sin(θ) = .02, approximate θ.

θ is approximately .02.

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Page 31: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

In fact, there are better and better approximations.

Fact

sin(θ) = θ − θ3

3! + θ5

5! −θ7

7! + · · ·

−3−2−1 1 2 3

−2

−1

0

1

2

θ

y

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Page 32: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

A similar statement is true of cos(θ).

Identity

When θ is very small, cos(θ) ≈ 1− θ2

2 .

In red, we have the graph of y = cos(θ), while in blue we have the graph

of y = 1− θ2/2.

−1 10

1

θ

y

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Page 33: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Fact

cos(θ) = 1− θ2

2! + θ4

4! −θ6

6! + · · ·

−3−2−1 1 2 3

−2

−1

0

1

2

θ

y

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Page 34: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

For Beyond this Course

In almost every field of quantitative study, one desires to study some

function in terms of simpler functions. In this class so far we’ve seen:

ex = 1 + x +x2

2!+

x3

3!+

x4

4!+ · · ·

cos(θ) = 1− θ2

2!+θ4

4!− θ6

6!+ · · ·

sin(θ) = θ − θ3

3!+θ5

5!− θ7

7!+ · · ·

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Page 35: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

The approximations which have been important to us in this course come

from considering just the first few terms of these sums:

ex ≈ 1 + x

cos(θ) ≈ 1− θ2

2‘

sin(θ) ≈ θ

We call this a “leading order approximation”. One portion of calculus is

understanding why we can make these approximations in general

situations.

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Page 36: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Measing distance to a star

r

θ

H

The length H is given by H = rsin(θ) .

Since θ is very small, we can approximate sin(θ) ≈ θ

H ≈ r

θ

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Page 37: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Measing distance to a star

r

θ

H

The length H is given by H = rsin(θ) .

Since θ is very small, we can approximate sin(θ) ≈ θ

H ≈ r

θ

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Page 38: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Measing distance to a star

H

The length H is given by H = rsin(θ) .

Since θ is very small, we can approximate sin(θ) ≈ θ

H ≈ r

θ

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Page 39: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Rule of Thumb for far away things

The distance to something far away is about

H ≈ r

θ

Where θ is the number of radians it moves in your view, and r is the

distance between the two points that you make an observation from.

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Page 40: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Where else does this occur? In humans, when judging the distance to

far away objects! Our depth position is controlled by our abilities to

resolve angles.

• Without additional information, humans can tell the distance of a

point to about 10 meters.

• Human eyes are about 66mm apart.

The angle which a point moves between the left and right eye when at 10

meters is roughly

10 meters =33

sin(θ)≈ 33mm

θ

θ ≈ 0.0033 Radians.

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Page 41: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Where else does this occur? In humans, when judging the distance to

far away objects! Our depth position is controlled by our abilities to

resolve angles.

• Without additional information, humans can tell the distance of a

point to about 10 meters.

• Human eyes are about 66mm apart.

The angle which a point moves between the left and right eye when at 10

meters is roughly

10 meters =33

sin(θ)≈ 33mm

θ

θ ≈ 0.0033 Radians.

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Page 42: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

θ ≈ 0.0033 Radians

That angle roughly corresponds to the size of a small beachball at 1

football field away. In fact, humans can resolve angles as small as

0.0003radians.

That’s about a small beachball at a kilometer away. This means that there

are other difficulties involved in measuring distance (for instance, the

relative angle of your eyes to eachother.)

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Page 43: Areas and Approximations - math.berkeley.edu

Estimates for Trig Functions

Short Survey

https://forms.gle/3jQ1h2HAJVab1W5H7

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