unit 1 trig.notebook · unit 1 trig.notebook 5 february 04, 2013 sep 64:35 pm example 2: solve for...

Unit 1 Trig.notebook

1

February 04, 2013

Sep 79:27 AM

Area 1

Area 2

Area 3

Can you find a relationshipwith the three areas?

Bell Work

Sep 64:21 PM

Unit 1:Trigonemetry

Key Ideas:

• Pythagorean Theorem• Primary trig ratios: sine, cosine & tangent• Inverses: sin1, cos1, tan1• Obtuse angles • Solving Triangles: SOHCAHTOA, Sine Law, Cosine Law• Applications: Word Problems


2

February 04, 2013

Sep 64:19 PM

Day 1: Pythagorean Theorem:

c2 = a2 + b2

• always "c"• opposite the right angle• the longest side

WARNING: When solving for one of the other two sides use:

a2 = c2 b2 b2 = c2 a2

They mean the same thing!

hypotenuse

Sep 64:23 PM

Examples: Solve for the missing side. Round to one decimal place.

108

x

2.3

3.7

x

x2 = 102 82

a2 = c2 b2

x2 = 100 64

x2 = 36

x = √36x = 6

c2 = a2 + b2

x2 = 2.32 + 3.72

x2 = 5.29 + 13.69

x2 = 18.98

x = √19.98x ≈ 4.4


3

February 04, 2013

Sep 64:25 PM

Day 2: Primary Trigonometric Raos: Sine, Cosine & Tangent

Example 1: Evaluate to 3 decimal places.

sin 650 = cos 1240 = tan 3410 =

angles

Example 2: Solve for the angle to the nearest degree.

sin R = 0.25 cos B = 0.92 tan Q = 1.54

Recall: When solving for an angle we must use the inverse functions!

R = sin1(0.25) B = cos1(0.92) R = tan1(1.54)

0.906 0.559 0.344

R = 14.50 B = 23.10 R = 57.00

Sep 64:32 PM

When do we use these?

When we want to solve for an angle or a side in a right angle triangle.

A

BC

What is the first thing that we must do when solving a right angle triangle?

LABEL THE SIDES: OPPOSITE, ADJACENT, HYPOTENUSE

hypostenuse

adjacent

opposite


4

February 04, 2013

Sep 64:34 PM

A

BC

Recall: SOHCAHTOA

Hence, the primary trig ratios for angle A below are:

hypostenuseadjacent

opposite

sin A =

opposite

hypotenusecos A =

adjacent

hypotenusetan A =

adjacent

opposite

Sep 64:34 PM

Example 1: Solve for side b.

420

12 cm

b

opposite

adjacenthypotenuse

sin 420 = opposite

hypotenuse

sin 420 = b

12

b = 12sin420 (12 x sin420)

b = 8.0 cm


5

February 04, 2013

Sep 64:35 PM

Example 2: Solve for side p.

670

9.7 cm

p

oppositeadjacent

hypotenuse

tan 670 = opposite

adjacent

tan 670 = 9.7

p

p = 9.7tan 670

(9.7 ÷ tan670)

p = 4.1 cm

Sep 98:11 AM

Bell WorkSolve for side b in the following two triangles.

230

11 in

b

5 cm

3 cm

b

sin 230 =

opp

adj

hyp

opp

hyp

b

11

b = 11sin230

sin 230 =

b ≈ 4.3 in

b2 = c2 a2

b2 = 52 32

b2 = 25 9

b2 = 16

b = √16

b = 4 cm

1


6

February 04, 2013

Sep 98:27 AM

Day 3: Solving for a missing angle in a right angle triangle

Example 1: Evaluate to 3 decimal places.

sin 650 = cos 1240 = tan 3410 =

angles

Example 2: Solve for the angle to the nearest degree.

sin R = 0.25 cos B = 0.92 tan Q = 1.54

Recall: When solving for an angle we must use the inverse functions!

R = sin1(0.25) B = cos1(0.92) R = tan1(1.54)

0.906 0.559 0.344

R = 14.50 B = 23.10 R = 57.00

Recall from yesterday our primary trig ratios: sine, cosine and tangent

Sep 98:30 AM

When do we use these?

When we want to solve for an angle or a side in a right angle triangle.

A

BC

What is the first thing that we must do when solving a right angle triangle?

LABEL THE SIDES: OPPOSITE, ADJACENT, HYPOTENUSE

hypostenuse

adjacent

opposite


7

February 04, 2013

Sep 98:32 AM

A

BC

Recall: SOHCAHTOA

Hence, the primary trig ratios for angle A below are:

hypostenuseadjacent

opposite

sin A = opposite

hypotenuse

cos A = adjacent

hypotenuse

tan A = adjacent

opposite

Sep 98:32 AM

Example 1: Solve for angle Q.

11 m

opposite

adjacent

hypotenuse

cos Q = hypotenuse

adjacent

cos Q = 11

17

Q = (cos1 (11÷17) )

Q = 49.70

Q

RS

17 m

11

17cos1 [ [


8

February 04, 2013

Sep 98:33 AM

Example 2: Solve for angle D.

8.7 inopposite

adjacenthypotenuse

tan D = opposite

adjacent

tan D = 8.7

12.1

D = (tan1 (8.7÷12.1) )

D ≈ 35.70

E F

D

12.1 in

tan1 [ [8.712.1

Sep 910:23 AM

Bell WorkSolve for the indicated variable.

15 cm

8 cmx

5.1 in

9.7 in

X

Tool:Pythagorean Theorem

Solution:

c2 = a2 + b2

x2 = 82 + 152

x2 = 64 + 225

x2 = 289x = √289x = 17 cm

Pythagorean TheoremLongest Side

Tool: SOHCAHTOA

Inverse (sin1, cos1, tan1)

Solution:

tan X = opp

adj

tan X = 9.75.1

opp

adj

hyp

X = tan1( )9.75.1X ≈ 62.30


9

February 04, 2013

Sep 112:57 PM

Day 4: Sine, Cosine, and Tangent of Obtuse Angles

Recall:

• An ACUTE ANGLE is less than 900.

• A RIGHT ANGLE is equal to 900.

• An OBTUSE ANGLE is bigger than 900 and less than 1800.

Sep 113:05 PM

THE CARTESIAN GRID: Quadrant I ACUTE ANGLES (Less than 900)

Quadrant IQuadrant II

X

Quadrant III Quadrant IV

Summary: In Quadrant I (ACUTE ANGLES) sine, cosine and tangent are ALL POSITIVE!

opp

adj

hyp


10

February 04, 2013

Sep 113:18 PM

THE CARTESIAN GRID: Quadrant II OBTUSE ANGLES (900 1800)

Quadrant IQuadrant II

X

Quadrant III Quadrant IV

Summary: In Quadrant II (OBTUSE ANGLES) sine is POSITIVE

cosine and tangent are NEGATIVE!

opp

adj

hyp

Sep 113:03 PM

R = 570In Summary,

Sin A Cos A Tan AFor Acute angles (0°


11

February 04, 2013

Sep 113:24 PM

Example 1: State whether the following angle is obtuse, acute or either.

a) sin A = 0.2545 b) cos A = 0.9684 c) tan A = 1.5684

Either Acute Obtuse

Example 2: Solve for the following OBTUSE angle.

a) sin A = 0.2545 b) cos B = 0.5468 c) tan C = 0.2547

A = 1800 sin1(0.2545)

A ≈ 1800 14.70

A ≈ 165.30

B = cos1(0.5468)

B ≈ 123.10 C = 1800 [tan1(0.2547)]

C ≈ 1800 14.30

C ≈ 165.70

Sep 113:39 PM

Question 1: State whether the following angle is obtuse, acute or either.

a) sin A = 0.9657 b) cos A = 0.9562 c) tan A = 0.8564

Either AcuteObtuse

Question 2: Solve for the following OBTUSE angle.

a) sin A = 0.5812 b) cos B = 0.2568 c) tan C = 1.7854

A = 1800 sin1(0.5812)

A ≈ 1800 15.50

A ≈ 144.50

B = cos1(0.2568)

B ≈ 104.90 C = 1800 [tan1(1.7854)]

C ≈ 1800 60.70

C ≈ 119.30

Bell Work


12

February 04, 2013

Sep 113:47 PM

Can we use the Pythagorean Theorem? NO

Can we use SOHCAHTOA? NO

Then ... what can we use?

• Sine Law • Cosine Law

Day 5: Solving triangle that are NOT right angles ‐ THE SINE LAW

Sep 113:49 PM

Before we begin let's recall a few things about triangles:

#1 Sum of the angle in a triangle:

A

B

C

A + B + C = 1800

#2 Labelling Conventions:

P

Q

R

pr

• Angles are labelled with

• Sides are labelled with

upper case letters

lower case letters

• sides and angles correspond.Opposite


13

February 04, 2013

Sep 113:49 PM

Sine Law:

A

B

C

a

b

c a = b = c sin A sin B sin C

OR

sin A = sin B = sin C a b c

HINT: Look for the OPPOSITE SIDEANGLE PAIR!

If you are given two angles, always start by finding the third angle (i.e. 1800 minus the other two)

i.e. Are you given A & a, B & b or C & c?

Sep 113:51 PM

A

B

C

a

b

12

320

650

Example 1:

A = 1800 320 650 = 830

a = c sin A Sin C

Find the value of side a in the triangle below.

a = 12 sin 83 0 sin 65 0

12sin83 sin65 0

a = 0

a = 13.14

The size of opposite sideangle pairs should correspond (i.e. the biggest side should be across from the biggest angle).


14

February 04, 2013

Sep 113:54 PM

A

B

C

2.3

b

3.1

470

Example 2:

sin C = sin A c a

Find the value of angle C in the triangle below.

3.1sin47 2.3

sin C = 0

sin C = sin 47 3.1 2.3

0

3.1sin47 2.3

0C = sin-1 [ [

C = 80.30

Sep 113:55 PM

Bell Work

Question 1: How do we know to use the Sine Law?

Question 2: Solve for the missing side in the following triangles.

Not a right angle triangle given a side/angle pair.

350

620

x

15.1 cm

x570 15.1 cm

1800 350 620 = 830

830

x

sin83015.1

sin350=

x = 15.1sin830

sin350

x ≈ 26.1 cmo

ha

SOHCAHTOA

cos570 = x

15.1

x = 15.1cos570

x ≈ 8.2 cm


15

February 04, 2013

Sep 113:54 PM

Can we use the Pythagorean Theorem? NO

Can we use SOHCAHTOA? NO

Then ... what can we use?

• Sine Law • Cosine Law

Day 6: Solving triangle that are NOT right angles ‐ THE COSINE LAW

Sep 113:57 PM

Cosine Law:

A

B

C

a

b

c

a2 = b2 + c2 - 2bc cosA .OR

cosA = a2 + b2 - c2

2bc


16

February 04, 2013

Sep 113:58 PM

Example 1: Find the value of side b in the triangle below.

A

B

C

1411

230

b

b2 = 142 + 112 - 2(14)(11)cos230

b2 = 142 + 112 - 2(14)(11)cos230

b2 = a2 + c2 - 2ac cosB .

b = √(142 + 112 - 2(14)(11)cos230)

b ≈ 5.8

Sep 113:58 PM

Example 2: Find the value of angle A in the triangle below.

A

B

C

26

15

19

cosA = a2 + b2 - c2

2bc

cosA = 192 + 152 - 262

2(19)(15)

A = cos-1 192 + 152 - 262

2(19)(15)[ [

A = cos-1 -90

570[ [


17

February 04, 2013

Sep 115:18 PM

Bell Work

Question 1: How do we know to use the Cosine Law?

Question 2: Solve for the missing angle in the following triangles.

Not a right angle triangle can't use sine law last resort :)

X

9.7 cm

15.1 cm

12.3 cm 9.7 cm

X15.1 cm

9.72 + 15.12 12.32

2(9.7)(15.1)

o

ha

SOHCAHTOAcosX =

9.7

15.1

cosX ≈ 0.642

X ≈ cos-1(0.642)

Not a right angle triangle SAS sandwich OR SSS

X ≈ 500

cosX =

cosX = 170.81292.94

cosX ≈ 0.583

X ≈ cos-1(0.583)

X ≈ 540

Sep 153:11 PM

Example:

DRAW A PICTURE!

Two planes took off from Pearson International Airport at the same time. The first plane heads due west at an angle of elevation of 25o and a speed of 168 Km/hr. The second plane heads due east at an angle of elevation of 32o and a speed of 156 Km/hr. How far apart are the planes after 3 hours?

Plane 1 Plane 2

Airport

d

504 4681230

1800 250 320 = 1230

SIMPLIFY YOUR PICTURE!

Plane 1 Plane 2

Airport

250 320

d

DECIDE WHETHER TO USE SOHCAHTOA, SINE LAW OR COSINE LAW!

COSINE LAW

∴ the planes are 282.8 km apart after 3 hours.

168 km/hr x 3 hr = 504 km 156 km/hr x 3 hr = 468 km

d2 = 5042 + 4682 2(504)(468)cos1230

d2 = 5042 + 4682 2(504)(468)cos1230

d2 ≈ 729,970.2d ≈ √729,970.2d ≈ 282.8 km

Applications


18

February 04, 2013

Sep 153:32 PM

A

B

C230

21

14

Bell Work/Test PreparationTIPS: Sovle for angle B to the nearest degree.

x

x2 = 212 + 142 2(21)(14)cos230

x2 ≈ 95.7x ≈ √95.7x ≈ 9.89.8

sinB14

sin230

9.8=

sinB14sin230

9.8=

sinB ≈ 0.558

B ≈ sin1(0.558)

B ≈ 400

Sep 153:47 PM

Application Review Test Preparation

Jamie's house is 15 Km away from Becky's house and 9 Km away from Kevin's house. From his posion, Becky's and Kevin's house are separated by an angle of 70 0. How far apart, to the nearest kilometer are Becky's and Kevin's houses? (Draw a diagram)

Example 1:

Remember: Keep the picture simple!

J

B K

15 Km 9 Km700

x

x2 = 152 + 92 2(15)(9)cos700

x2 ≈ 213.65x ≈ √213.65

x ≈ 15

∴ Becky's and Kevin's houses are approx. 15 Km apart.


19

February 04, 2013

Sep 208:21 AM

Application Review Test Preparation

Two cabins A and B are separated by a large pond. To determine the distance between the cabins, Beth walks 2 km from A to a point K. She measured angle A equal to 230 and angle K equal to 560. Determine the distance between the cabin, correct to 2 decimal places. (Draw a diagram)

Example 2:

Remember: Keep the picture simple!

pond

A B

K

x230

560

1010

1800 230 560 = 10102 Km

x

sin5602

sin1010=

x = 2sin560

sin1010

x ≈ 1.69 cm

∴ The two cabins are approx. 1.69 km apart.

Sep 208:51 AM

Trigonometry Decision Tree

Is it a rightangle triangle?

YES NO

Are you solving for a side?

YES Solving for an angle!

Use: SOHCAHTOA Inverses (sin1, cos1, tan1)

NO

Do you HAVE two sides?

Pythagorean Theorem

YES

c2 = a2 + b2 a2 = c2 b2

hypotenuse other leg

ADD SUBTRACT

NO

SOHCAHTOALabel Carefully!

Variable on the...

top

bottom

MULTIPLY DIVIDE

x = 14sin230 x = 14÷sin230

Do you have a SIDEANGLE PAIR?

Sine Law

Solving for a ...

side angle

xsinX

ysinY

=

xsinX

ysinY

=

INVERSE!

Cosine Law

x2 = a2 + b2 2ab cosX.

side

Solving for a ...

cosX = a2+b2x2

2ab

INVERSE!

angle

Attachments

Cartesian Grid

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