UNIT 5 – TRIGONOMETRIC RATIOS

Date Lesson Text TOPIC Homework

Apr.

24

5.1

(48)

Trigonometry Review WS 5.1 # 3 – 5, 9 – 11,

(12, 13)doso

Apr.

26

5.2

(49)

Related Angles Complete lesson shell

& WS 5.2

Apr.

29

5.3

(50)

5.3

5.4

Solving for Angles Pg. 299 # 2, 6ace, 8abc, 9, 10a,

12

Apr.

30

5.4

(51) 5.2

Special Angles Pg. 287 # 4, 7, 9, 11

May 1 5.5

(52)

5.6

5.7

Solving for Obtuse Angles

QUIZ (5.1 – 5.3)

Pg. 318 # 1, 4

Pg. 326 # 3, 4

May 2 5.6

(53)

The Ambiguous Case WS 5.6

Pg. 318 # 3, 5

May 3 5.7

(54)

5.6

5.7

Trig Word Problems WS 5.7 # 1 - 10

May 6 5.8

(55)

5.6

5.7

More Trig Word Problems

QUIZ (5.4 - 5.6)

WS 5.7 # 12 – 16

Pg. 319 # 7, 8

Pg. 326 # 5, 6, 7, 10

May 7 5.9

(56) 5.8

3D Trig Word Problems/Inaccessible

Distances

WS 5.9

Q # 5 – “DIRECTLY ABOVE”

Pg. 332 # 3a, 4, 5, 9

May 8 5.10

(59)

Review for Unit 5 Test Pg. 338 # 1 – 5, 8 – 12,

Pg. 340 # 1b, 2, 3, 6, 7, 8

May 10 5.11

(60)

UNIT 5 TEST

15

8.4 cm

x

x

8.2 km

42

12.1

cm

8.2

cm

C

BA

MCR3U Lesson 5.1 Trigonometry Review

1. Determine length x in each triangle. Round your answer to one decimal place.

a)

b)

2. Determine the measure of A , to the nearest degree.

3. In ABC , 2.8b cm, 75C , 1.10c cm. Determine the measure of B .

3

B

C

3

4

5

A

A

A

A

tan

cos

sin

C

C

C

tan

cos

sin

8 cm

6 cm

65C

B

A

3.8 m 5.4 m

6.5 mCB

A

41

60

15 cm

p

R

P

Q

4. Solve the following triangles.

6.0 cm

8.0 cm

HW: WS 5.1

hypotenuse

adjacent

opposi te

cb

aC B

A

cb

aCB

A

cb

aCB

A

Trig Formula Sheet

Right triangle

hypotenuse

oppositesin

opposite

hypotenusecsc

hypotenuse

adjacentcos

adjacent

hypotenusesec

adjacent

oppositetan

opposite

adjacentcot

222 bac

Acute and Obtuse Triangles:

The Sine Law:

In any ABC , C

c

B

b

A

a

sinsinsin (when solving for sides)

or c

C

b

B

a

A sinsinsin (when solving for angles)

The Cosine Law

bc

acbA

2cos

222

ac

bcaB

2cos

222

ab

cbaC

2cos

222

Abccba cos2222

Baccab cos2222

Cabbac cos2222

Use this version when given 2 sides and a contained angle.

When 3 sides are known the Cosine Law can be written as follows:

MCR3U Lesson 5.2 Related Angles

Draw each of the following angles in standard position, clearly showing the RAA (Related Acute Angle). The RAA

is the angle that the terminal arm makes with the x axis. You must draw the rotation direction of the angle and

label it for it to be considered the desired angle.

130 210 250 310

140 230 340 80

For ease of calculation, angles can be placed on the Cartesian plane within a unit circle (circle with a radius

of 1 unit). Sketch a 50 angle in the unit circle below

(1,0)

(0,1)

(-1,0)

(0,-1)

x

y

Note that in a

circle with radius r,

sin

cos

tan

In the unit circle

(r = 1),

sin

cos

tan

Now we will move this triangle to each of the four quadrants:

Sketch each of the following angles in standard position. Clearly show the RAA (related acute angle) and

label the point on the terminal arm which intersects the unit circle.

130 230 310

(1,0)

(0,1)

(-1,0)

(0,-1)

x

y

(1,0)

(0,1)

(-1,0)

(0,-1)

x

y

(1,0)

(0,1)

(-1,0)

(0,-1)

x

y

50 130 230 310

sin

cos

tan

CAST Rule:

We say that 50 , 130 , 230 , and 310 are related angles, and that 50 is the related acute angle

(RAA).

List the related angles for:

a) 20 b) 220

Given the value of one trig ratio, provide 3 related evaluated trig ratios.

a) 1736.010sin b) 6820.047cos

b) 7475.270tan d) 6428.0220sin

For each of the following trig expressions write an equivalent trig expression. Use angles between 0˚ and

360˚.

a) 50sin b) 110cos c) 345tan

d) 280cos e) )160sin( f) )95tan(

g) 280cos = h) 70tan = i) 305sin =

j) 210sin = k) 170tan = l) 125cos =

m) 20cos = n) 306tan = o) 123sin =

p) )100tan( = q) )125cos( = r) )305sin( =

OR = OR = OR =

MCR3U Lesson 5.3 Solving For Angles

Solve each of the following to the nearest degree, where 3600 .

a) 6.0sin b) 2.0cos

c) 9626.1tan d) 7547.0sin

Outside The Unit Circle

1. 4,3 lies on the terminal arm of angle in standard position.

a) Draw a sketch of angle .

b) Determine the primary trig ratios for angle

.

c) Calculate the value of to the nearest

degree.

2. is a third quadrant angle such that 7.0sin

a) Draw a sketch of angle .

b) Determine the primary trig ratios for angle

.

c) Calculate the value of (to the nearest

degree) using both a counterclockwise and

clockwise rotation.

HW: p. 299 #2, 6ace,

8abc, 9, 10a, 12

1 2 3 4 5–1–2–3–4–5 x

1

2

3

4

5

–1

–2

–3

–4

–5

y

x

y

MCR3U Lesson 5.4 Special Angles

The Unit Circle

Ex. Evaluate each of the following: (exact answers required)

a) 45sin45tan b) 30cos30tan c) 45cos60sin 22

(cos , sin )

Ex. Evaluate each of the following. Answers must be exact. No calculators permitted.

120sin 180cos 30tan 315sin

210cos 300tan 270sin 60cos

240sin 225tan 0tan 135cos

150sin 300sin 330cos 150tan

Ex. Solve each of the following, where 3600 .

a) 2

3sin __________ or __________ .

b) 2

2cos __________ or __________ .

c) 2

1sin

__________ or __________ .

d) 3tan __________ or __________ .

e) 0cos __________ or __________ .

f) 2

3cos

__________ or __________ .

g) 1tan __________ or __________ .

h) 1sin __________ or __________ .

TEXT HW: Pg. 287 #4,7,9,11

3.5 cm

5.5 cm

2.5 cm

MCR3U Lesson 5.5 Solving Obtuse Triangles

1. Find the value of , correct to the nearest degree.

2. Solve the following triangle: (round angle measures to 1 decimal place)

For each of the following, draw diagrams (somewhat to scale would be nice)

3. Solve ABC , where 48A , 0.5b cm, and 3.6c cm.

4. Solve ABC , where 30A , 0.2a cm, and 0.5b cm.

5. Solve ABC , where 30A , 0.7a cm, and 0.5b cm.

6. Solve ABC , where 30A , 0.4a cm, and 0.5b cm. Pg. 318 # 1, 4

Pg. 326 # 3, 4

MCR 3U Lesson 5.6 The Ambiguous Case

Any triangle without a right angle is called an oblique triangle. The cosine law and the sine law can

be used to determine angles or sides in all triangles (acute, right or obtuse).

In ∆ABC, with sides a, b, and c,

The Sine Law The Cosine Law

sin sin sin

a b c

A B C

2 2 2 2 cosa b c bc A

2 2 2

cos2

b c aA

bc

sin sin sinA B C

a b c

To solve an oblique triangle, you need to know the

measure of at least one side and two other parts

of the triangle. There are four cases in which this

can happen.

Given Information What can be found Law Required

1. Two angles and any side (AAS or ASA) side Sine law

2. Two sides and the contained angle (SAS) side Cosine law

3. Three sides (SSS) angle Cosine law

4. Two sides and the angle opposite one of them (ASS) angle Sine law

Case 4 is called the ambiguous case because sometimes it is possible to draw more than one triangle

for the given information. In this case, there are four possible outcomes if A is acute and two

possible outcomes if A is obtuse. These possibilities are shown for the given A and the given

sides a and b, in ∆ABC. The side opposite the given angle is always a and bsinA represents the

possible height of the triangle.

Determining whether a given angle (A) is acute or obtuse, and then comparing the size of a, b,

and bsinA allows you to see which situation you are dealing with and, in turn, the number of possible

solutions.

C B

A

a

b c

Is it really a triangle?

A triangle ABC with angle A of 63.9°, side AB = 8.3 cm, and side BC = 6.9 cm cannot exist! So how

can you tell?

If we were to draw the

"triangle" we could

immediately tell that the

sides don't meet! However

there is a more convenient

way. height of = bsinA

If a < b sin A then there

is no possible triangle.

Lets check that relationship

here.

sin(63.9) x 8.3 = 7.4

> 6.8

Since the calculated value is

greater than the length of

side a, there is no possible

triangle. Now we have an

easy way to check.

if ASS or SSA ambiguous

case

h = bsinA

A 90 (acute) Conditions

a bsinA

a bsinA

bsinA a b

a b

NoTriangle

One Right Triangle

Two Triangles one acute

one obtuse

onetriangle

Number and Type of

Triangles Possible

bsinA

bsinA

b

b

a C

C

A

A

B

B

A B

bsinA

C

b

A

a a

B1 B2

bsinA b

C

a

A 90 (obtuse) Conditions

Number and Type of

Triangles Possible

A

a

b a b

A 90 (acute) Conditions

a bsinA

a bsinA

bsinA a b

a b

NoTriangleie: side a is shorter than the

altitude of the triangle.

One Right Triangle

Two Triangles one acute

one obtuse

onetriangle

Number and Type of

Triangles Possible

bsinA

bsinA

b

b

a C

C

A

A

B

B

A B

bsinA

C

b

A

a a

B1 B2

bsinA b

C

a

A

b

B

a

C

a b Oneobtuse triangle

NoTriangleie: side a is shorter than the

altitude of the triangle.

NONE

ONE OBTUSE

TRIANGLE a > b

a

a

b

Ex. 1 For ∆ABC, a = 3.0 cm, c = 5.0 cm and A = 30°, solve the triangle(s).

Two sides and an angle opposite one side are given (SSA)

This is the ambiguous case of the sine law, where the given angle is acute.

check the relation ship between a, c, and csinA.

Ex. 2 ∆DEF is given with D = 130°, d = 50.0 cm and e = 20.0 cm. Solve ∆DEF.

Two sides and an angle opposite one side are given (SSA)

This is the ambiguous case of the sine law, where the given angle is obtuse.

WS 5.6 and Pg. 318 #3,5

MCR 3U Lesson 5.7 Solving Problems using Trigonometry

1. From a window, a ladder extends down to the ground with an angle of depression of 65 . The base

of the ladder is 4.8 m from the building.

a) How high is the window?

b) How long is the ladder?

2. From an observation tower the angle of elevation of a weather balloon is 68 . In the same plane, on

the other side of the balloon 35.0 km away, the balloon is sighted from another location with an

angle of elevation of 47 . Calculate the distance from the weather balloon to the observation

tower.

WS 5.7 # 1 - 10

WS 5.7 # 12 – 16

Pg. 319 # 7, 8

Pg. 326 # 5, 6, 7, 10

MCR3U Lesson 5.9 Inaccessible Distances / 3 Dimensions

Ex.

1. To find the height of a cliff that is inaccessible, a surveyor measures a baseline AC of 400 m. In

the horizontal plane ABC, 27A and 35C . In the vertical place BTC, 18BCT .

Determine the cliff height to the nearest metre. (67 m)

2. From the top of a 120 m fire tower, a fire ranger observes 2 fires on the ground below (not in the

same line of sight). One fire has an angle of depression of 6 and the other has an angle of

depression of 3 . The angle between the lines of sights is 105 . Calculate the distance between

the two fires (to the nearest metre). (2811 m)

3. To determine the height (from base X to top Y) of the Peace Tower in Ottawa, measurements were

taken from a baseline AB of length 50 metres. It was found that 6.42XAY , 60XAB ,

and 65.81ABX . Calculate the height of the Peace Tower to the nearest metre. (73 m)

4. The crows-nest of a yacht is 50.0 m above the water level. The angle of depression from the

crows-nest to a buoy due west of the boat is 40 . The angle of depression to another buoy S 70

W of the yacht is 34 . How far apart are the buoys? (27.3 m)

5. Two roads intersect at 34 . Two cars leave the intersection on different roads at speeds of 80

km/h and 100 km/h. After 2 h, a traffic helicopter which is above and between the two cars takes

readings on them. The angle of depression to the slower car is 20 and the distance to it is 100

km. How far is the helicopter from the faster car? (38.7 km)

6. Jennifer and Alex were flying a hot air balloon when they decided to calculate the straight line

distance from Beaverton to Tandy. From a height of 340 m they recorded the angles of depression

of Beaverton and Tandy as 2 and 3 respectively. The angle between the line of sights to the two

towns was 80 . Find the distance from Beaverton to Tandy. (10.7 km)

Pg. 332 # 3a, 4, 5, 9