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Suppose you wanted to know the height of a VERY tall building . . .

. . . maybe somebody (desperately) needs to know the height of a REALLY TALLtree!

So, that means I’m only about 15 feet from at least 32 feet of safety!

h

15 more feet!

65⁰

Pythagorean Theorem?

Family?

Special Right Δ ?

Trigonometry

helps you solve for missing measures in right triangles when you don’t have all the information you need in order to use the methods you’ve already learned!

ΔABC ~ΔADE~ΔAFG

F

All three triangles are similar by AA~.

D

B

x⁰

1

3

2

A

C

E

G

∡A is one acute corresponding angle in each of the three triangles

(reflexive property)

ΔABC ~ΔADE~ΔAFG

F

All three triangles are similar by AA~.

D

B

x⁰

1

2

3

A

C

E

G

Angles 1, 2, and 3 are right angles,

. . . and ALL right angles are congruent.

ΔABC ~ΔADE~ΔAFG

All three triangles are similar by AA~.

The third angle in each triangle has “no choice,” since the sum of the angles in any triangle is exactly 180⁰!

F

D

B

x⁰

1

3

2

A

C

E

G

(That is why AA~ is sufficient to prove the three triangles similar.)

ΔABC ~ΔADE~ΔAFG

All three triangles are similar by AA~.

F

D

B

1

2

3

x⁰

x⁰

x⁰

A

A

A

G

E

C

Since ΔABC, ΔADE, and ΔAFG are similar triangles, the following ratios are equal:

That is because they represent corresponding sides of the three similar triangles.

ΔABC ~ΔADE~ΔAFG

All three triangles are similar by AA~.

F

90 - x

D

B

90 - x

90 - x

1

2

3

x⁰

x⁰

x⁰

A

C

A

A

G

E

NOTE:

No matter what the measure of ∡A, the other acute angles equal 90 – (m ∡A).

Remember, the acute angles in each right triangle have a

complementary relationship!

ΔABC ~ΔADE~ΔAFG

F

90 - 30

D

B

90 - 30

90 - 30

30⁰

30⁰

30⁰

A

C

A

A

G

E

Before we go on, it may help to review some concepts you learned in a previous section about one of the special triangles, the 30 – 60 – 90 right triangle.

Remember in the 30-60-90 Right Triangle, there is a certain relationship that the sides always obey.

That relationship can be expressed using the following ratio,

ΔABC ~ΔADE~ΔAFG

F

90 - 30

D

g

60⁰

B

90 - 30

e

60⁰

a

c

a

90 - 30

a

60⁰

30⁰

30⁰

30⁰

A

C

d

f

b

A

A

G

E

a:b:c=a:d:e=a:f:g=1:√3:2

The short leg (usually a) is opposite the smallest (30⁰) angle,

The long leg (usually b) is opposite medium (60⁰) angle, and

The hypotenuse (usually c) is ALWAYS opposite the right (90⁰) angle,

which is the largest angle. . . so it is the LONGEST side!

ΔABC ~ΔADE~ΔAFG

Proper Notation:

F

VERTICES are labeled using capital letters

SIDES are labeled using lower case letters

D

f

60⁰

B

d

60⁰

b

60⁰

e

g

c

a

a

a

30⁰

30⁰

30⁰

A

C

A

A

G

E

a:d:e=

a:f:g=

a:b:c=

The short leg (a) is opposite the smallest (30⁰) angle,

The long leg (b – d – f ) is opposite medium (60⁰) angle, and

The hypotenuse (c – e – g) is ALWAYS opposite the right (90⁰)angle,

which is the largest angle. . . so the longest side

ΔABC ~ΔADE~ΔAFG

F

D

60⁰

B

g

e

a

60⁰

a

c

a

60⁰

30⁰

30⁰

30⁰

d

A

b

C

f

A

A

G

E

a:d:e=

a:f:g=

a:b:c=

30:60:90

small:medium:large

1:√3:2

x:x√3:2x

Using the figure from the beginning, this time we will let m∡ A = 30, BC = 1, DE = 2, and FG = x

F

D

x

B

2

1

3

30⁰

A

C

E

G

Taken apart, and according to the ratio of sides,

along with these measures we have:

F

D

60⁰

B

2x

60⁰

4

x

2

2

60⁰

1

30⁰

30⁰

30⁰

A

C

A

A

G

E

In Δ ABC: In Δ ADE: In Δ AFG

BC 1 DE 2 1 FG x 1

== = = =

AB 2AD 4 2 AF 2x 2

1

2

1

2

1

2

Taken apart, and according to the ratio of sides,

we have:

F

D

60⁰

B

2x

60⁰

4

x

2

2

60⁰

1

30⁰

30⁰

30⁰

A

C

A

A

G

E

In each of the triangles above, the comparison, or “ratio,” of the lengths of the sides of the angle across from the 30⁰ reference angle is 1: 2 or 0.5!

Opposite LEG to HYPOTENUSE = 1 : 2 or 0.5!

1

2

This ratio will be true for every 30 – 60 – 90 right triangle, no matter how small or large !

If 30⁰, then opposite side to hypotenuse =

What do you think would happen if we changed ∡ A

to another acute measure?

Do you think the ratio of the length of the side located opposite of the angle to the length of the hypotenuse will remain “constant” ?

The answer is YES!

opposite

Question: Why?

hypotenuse

Well . . . what have you learned that proves figures “similar”?

57⁰

74⁰

63⁰

Answer:

A

The ANGLES are CONGRUENT, and

CORRESPONDING SIDES of SIMILAR figures are PROPORTIONAL!

In fact, in right triangles, given an ACUTE angle of “x degrees,” the length of the side opposite the angle always compares to the length of the hypotenuse by the same ratio !

Just in case you are not convinced, let’s try it with the other special right triangle, the 45 – 45- 90.

x

2

1

45⁰

Taken apart, and according to the ratio of sides,

we have:

F

45⁰

D

45⁰

x√2

x

B

2√2

2

√2

45⁰

1

45⁰

45⁰

45⁰

A

C

A

G

A

E

In Δ ABC: In Δ ADE: In Δ AFG

BC 1 DE 2 1 FG x 1

== = = =

AB √2AD 2√2 √2 AF √2x √2

1

√2

1

√2

1

√2

Sure enough, in each similar triangle from before, the ratio of the leg opposite the reference angle does compare to the hypotenuse length in the same way:

In the 30 – 60 – 90 triangle,

the ratio of the leg opposite the 30⁰ angle to the hypotenuse = 1 : 2.

And,

in the 45 – 45 – 90 triangle, the ratio of the leg opposite the 45⁰ angle to the hypotenuse = 1 : √2 (In simplest radical form, √2 )

2

Notice that in these examples, we have been comparing the side located opposite of the 30⁰ or 45⁰ degree angle to the hypotenuse of these triangles.

For any given acute angle, the ratio of the sides opposite of the angle will be the SAME, regardless of the size of the RIGHT triangle.

From now on we will refer to this type of ratio by its special name:

THE SINE RATIO.

The SINE RATIO of an acute angle of a right triangle is

the comparison of the length of the side opposite the reference angle to the length of the hypotenuse.

hypotenuse

Opposite to Hypotenuse

Opposite

side

Reference Angle

Opposite : Hypotenuse

Opposite

Hypotenuse

= SINE Ratio

SIN

Calculator = SIN

Either ACUTE ANGLE can be the “reference angle,” but NEVER the

RIGHT angle!

Okay, so now we will tie up some loose ends!

Reference Angle: The acute angle in the right triangle from which you view the other parts of the given triangle.

Opposite Side

(LEG)

Adjacent Side

(LEG)

Hypotenuse

Opposite Side

Hypotenuse

Reference Angle

Adjacent Side

Opposite: across from

Adjacent: next to

So, working from the other acute angle, we have:

A reversed perspective!

Reference Angle: The acute angle in the right triangle from which you view the other parts of the given triangle.

Reference Angle

Opposite Side

(LEG)

Adjacent Side

(LEG)

Hypotenuse

Adjacent Side

Hypotenuse

Opposite Side

Opposite: across from

Adjacent: next to

In the earlier examples, we were exploring one of the three trigonometric ratios introduced in this section:

The SINE ratio

So again, the sineratioof an acute angle in any right triangle is the value you get when the length of the side opposite the reference angle

is divided by the length of the hypotenuse.

Opposite Leg

SINE ratio =

Hypotenuse

In a right triangle, other ratios may also be formed, and each has a special name:

SINE,COSINE,andTANGENT

(SohCahToa)

SINE

COSINE

TANGENT

Opposite Leg

Adjacent Leg

Opposite Leg

Hypotenuse

Hypotenuse

Adjacent Leg

Soh

Cah

Cah

Toa

Toa

Soh

SohCahToais a memory device that helps you remember which sides of the right triangle you need for each trigonometric ratio, and in what order they should be entered into the ratio!

(SohCahToa)

tan ∡A =

sin ∡A =

cos ∡A =

a

h

o

a

o

h

SINE

COSINE

TANGENT

Opposite Leg

Adjacent Leg

Opposite Leg

Hypotenuse

Hypotenuse

Adjacent Leg

Soh

Cah

Cah

Toa

Toa

Soh

Think!

Think!

Think!

Definition: A trigonometric ratio is the comparison of the lengths of any two sides of a RIGHT triangle when taken from the perspective of one of the two acute angles.

Given: Right Δ ABC, with legs 3 and 4, and hypotenuse 5, find:

A

0.6

Soh

1) sin ∡A =

4

5

a

h

o

a

3

4

o

h

3

5

0.8

Cah

5

4

2) cos ∡A =

Hypotenuse

leg

0.75

Toa

3) tan ∡A =

B

leg

C

3

Remember, every ratio can be written as a decimal!

See the table on PAGE 424 in your book for ratios related to acute angles whose measures are between 0 and 90 degrees!

What is the result if we use ∡B as our reference angle?

Given: Right Δ ABC, with legs 3 and 4, and hypotenuse 5, find:

A

0.8

Soh

1) sin ∡B =

3

5

a

h

4

3

o

a

o

h

4

5

0.6

Cah

5

4

2) cos ∡B =

Hypotenuse

leg

1.333…

Toa

3) tan ∡B =

B

leg

C

3

Next let’s compare the ratios written from the two different reference angles,

∡A and ∡B:

Given: Right Δ ABC, with legs 3 and 4, and hypotenuse 5, find:

A

Soh

sin ∡A =

sin ∡B =

3

5

4

5

4

3

3

4

4

5

3

5

Cah

5

4

cos ∡A =

cos ∡B =

Hypotenuse

leg

Toa

tan ∡A =

tan ∡B =

B

leg

C

3

1.333…

0.6

0.6

0.8

0.8

0.75

The measure of the angles don’t change!

It is simply the opposite perspective!

Is it possible to create an equilateral triangle using two 3-4-5 right triangles?

Nope!

5

5

4

3

6

So, it is impossible for the acute angles in a 3 – 4 – 5 right triangle, and those in the same family, to have measures of 30 and 60 degrees!

RIGHT?

So, to find the measure of the acute angles we can use the ratios from before:

(= 0.75)

Toa

A

tan ∡CAB =

3

4

0.75

37⁰

5

tan ∡CAB =

4

∡CAB = (tan-1) (0.75)

∡CAB ≈ 37⁰

B

C

3

Negative Exponent NOTE: “ tan-1 “ means you are dividing 0.75 by “tan.”

Now, if m∡CAB = 37⁰ , then the m∡B = 90 – 37 = 53⁰ . . . RIGHT?

Check it by using the tangent ratio written from ∡B!

(= 1.333. . .)

Toa

A

tan ∡B =

4

3

1.333. . .

5

tan ∡B =

4

∡B = (tan-1) (1.333. . .)

53⁰

∡B ≈ 53⁰

B

C

3

Negative Exponent NOTE: “ tan-1 “ means you are dividing 1.333. . . by “tan.”

Now, if m∡B = 53⁰ , then the m∡CAB = 90 – 53 = 37⁰ . . . Check!

We will now verify the results achieved from the tangent ratio by calculating the angles with the sine and cosine ratios.

VERY IMPORTANT!

This means we could have calculated the angle measures using any of the three trigonometric ratios!

A

√

√

sin ∡A =

0.6

∡A = (sin-1)(0.6)

∡A ≈ 37⁰

√

37⁰

sin

cos

cos

sin

cos

cos

sin

sin

5

4

√

0.8

sin ∡B =

∡B = (sin-1)(0.8)

∡B ≈ 53⁰

53⁰

0.8

cos ∡A =

B

C

√

∡A ≈ 37⁰

3

∡A = (cos-1) (0.8)

√

√

Sine, Cosine or Tangent? The one you use depends on the measures you know!

cos ∡B =

0.6

√

∡B = (cos-1) (0.6)

∡B ≈ 53⁰

Example:ΔABC is isosceles, with legs AB and AC.

Find sin ∡C.

A

Step 1: First we must have a RIGHT Δ !

If isosceles, then the altitude is also a MEDIAN!

Draw an altitude from ∡A

Step 2: Use “families of right triangles” or the Pythagorean Theorem to find the length of the altitude.

opp

hyp

24

25

25

24

WHY?Remember”Soh”CahToa!

Thesine ratio is the oppositeleg

divided by the hypotenuse

Step 3: Write the ratio in fraction form,

opposite leg : hypotenuse

B

C

7

7

14

sin ∡C =

Step 1: Use the ratio you wrote forsin ∡C.

A

Step 2: Write the ratio as a decimal

0.96

sin ∡C =

32⁰

24

25

sin

sin

Step 3: Solve for ∡C: sin ∡C = 0.96

25

24

sin ∡C = 0.96

∡C = sin-1 (0.96)

74⁰

74⁰

B

C

m∡C ≈ 74⁰

7

7

Question:What is the measure of ∡A?

m∡A = 180 – 2( 74) =

32⁰

180 – 148 =

EXAMPLE: What if all you know is the measure of one of the acute angles

and only one side length of the RIGHT triangle?

Opposite and Adjacent. . . Toa!

Use the TANGENT Ratio!

Problem: Find the height of the flag pole.

Step 1: From the angle you know, determine

which trigonometric ratio to write.

o

a

h

27

We know: ReferenceAngle: 50⁰

Adjacent Side:27 ft

We want to know: Opposite Side

≈ 32 ft

h ft

Step 2: Write equation from the reference angle

Tan 50⁰ =

1.1918 =

Tan 50⁰ = 1.1918

27(1.1918) = h

50⁰

32 ft ≈ h

27 ft

When solving for a missing side,

You use these keys:

These keys convert the acute angle measure

to its corresponding side ratio value, in decimal form.

SIN

COS

TAN

When solving for an unknown angle measure:

The inverse keys divide the decimal value of the side ratio and render the corresponding angle measure.

You use these keys:

TAN-1

SIN-1

COS-1

It’s almost as easy as A, B, C!

Student response to exam question #3:

But it will require a bit more effort than this student thought!

9.9 Assignment:

Pp 420 (2 – 4; 9 – 14; 18)

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