Math 11 - 2-Trigonometry

Math 11 - 2.1-Standard Position in Cartesian Plane

When angles share the same terminal arm, they are calledco-terminal angles

The principal angle – the smallest positive angle co-terminalwith the given angle

The reference angle – the angle between the terminal arm andthe x-axis -> always positive and in between 0° - 90°

Initial Arm

Terminal Arm

Quadrant IQuadrant II

Quadrant III Quadrant IV

+θ

-θIf this is a given angle

This is the principal angleThis is the reference angle

Parallel Lines and Angles

- The corresponding angles that are formed by each parallel line and the transversal are equal

a= p , b=q , c=r , d= s ,

Interior angles→ c ,d , p ,q ,

Exterior angles →a , b , r , s ,

a=d= p=s , b=c=q=r

s

a

c

b

d

p q

r

Exp. 2.1.1 What could it be the co-terminal angle of the given one ?

Solution: θcould be=90 °+50°=140 °

So one of the co-terminal angles of 140° is 360°+140 °=500 °

Exp. 2.1.2 What could it be the co-terminal angle of the given one ?

Solution:

So one of the co-terminal angles of -50° is −50 °−360°=−410 °

50° θ

-50°

Exp. 2.1.3 The angle θ=840° , the terminal arm of this angle is located in quadrant ?

Solution: θ=840°=360+360+120 °

So the terminal arm of the angle 840° is located in Quadrant II

Exp. 2.1.4 The terminal arm of the angle θ is located in quadrant III. What should the angle θ be?

(a) 150° (b) 350° (c) 550° (d) 750°

Solution: In quadrant III, the angle should be in between 180°-270° --> not for (a) and (b)

For (c) 550° --> 550°=360 °+180 °+10° --> The angle θ should be (c)550°

For (d) 750° --> 750°=360 °+360 °+30° --> Not in quadrant III

Exp. 2.1.5 Determine the reference angles of the following angles

The givenangles

How to find them? The reference angles

(a) -50°

(b) 50°

(c) -120°

(d) 120°

(e) 650°

(f) -650°

Solution:

The givenangles

How to find them?The reference angle is the angle between the terminal arm and

the x-axis – always positive and in between 0° - 90°

The reference angles

(a) -50° -50° 50 °

(b) 50° 50° 50 °

(c) -120° 180°-120° = 60° 60°

(d) 120° 180°-120° = 60° 60°

(e) 650° 650°=360°+180°+110° --> 180° – 110° = 70° 70°

(f) -650° -650°=-360°-180°-110° --> 180° – 110° = 70° 70°

Exp. 2.1.6 Determine all unknown angles in the figure

Solution: CAD=31°= ACB , 104 °= BCD

104 °= BCD= ACD+ ACB= ACD+31° → ACD=104 °−31°=73°= BAC

ABE=76°= BCE=D

BAD= BAC+DAE=73°+31°=104 °

Exp. 2.1.7 Determine ACD ,θ

Solution:

CAD=31°=CFE= ACB , 76°+ ACD+ ACB=180 °=76°+ ACD+31° →

ACD=180°−76°−31°=73°

θ+CFE=180 °=θ+31° → θ=180 °−180 °−31°=149°

A

104°

31°

76°

B

A

C

D

E

A

31°

76°

F

A

C

D

θE

B

Math 11 - 2.2-Trigonometric Ratios

θ 0° 30° 45° 60° 90° 180° 270°

Sin θ 012

1√2

√32 1 0 -1

Cos θ 1√32

1√2

12 0 -1 0

Tan θ 01√3

1 √3 undefined 0 undefined

P(x, y)

r

θ

cos+

all+

tan+

sin+

A

B

C30°

2

1

60°

√3 A

B

C45°

1

45°

√21

C – A – S – TRULE

Q4 – cos+ ,Q1 – all+ ,Q2 – sin+ ,Q4 – tan+

sin θ= oppositehypotenuse

= yr

cosθ= adjacenthypotenuse

= xr

tanθ= oppositeadjacent

= yx

The standard positionthe point P(x, y) is on the terminal arm,

the angle θ with the radius r (hypotenuse) ,x(adjacent) and y(opposite)

Exp. 2.2.1 Given triangle as shown, P=R What is the length of PR ? ( Round the answer to the nearest

tenth of a ft.)

Solution: Q2

=42 °2

=21° --> sin PQS=sin 42°2

=sin 21°= PS15

=0.3584 →

PS=(0.3584)(15)=5.3755 --> PR=2RS=2(5.3755)=10.75≈10.8 ft

Exp. 2.2.2 In standard position, the terminal arm of the angle β is in quadrant IV

(a) cos β is negative, sin β is positive (b) cos β is positive, sin β is negative

(a) tan β is negative, sin β is negative (b) cos β is negative, tan β is negative

Solution: (b) cos β is positive, sin β is negative is correct

42°

RP R

Q

15 ft

S

P(x, y)

r

θ

cos+

all+

tan+

sin+

Exp. 2.2.3 The point P( 5, -4) is on the terminal arm of the angle α in standard position, find cos α , sin α ?

Solution: The point P (5, -4) --> Pythagoreantheorem : r2=x2+ y2=52+(−4)2=25+16=41→

r=√41 --> cosα= xr= 5

√41 , sin α= y

r= −4

√41

Exp. 2.2.4 0°≤α≤360 ° , find the angle α which cos α = cos 415°

Solution: 415 °=360°+55 ° --> the terminal arm is in the quadrant I, the reference angle is 55°

cos 415 °=0.5735=cos 55°=cosα→ α=55°

Exp. 2.2.5 cos(P+50 °)=−√22

, (P+50°) is the angle in quadrant II. What is the value of P ?

Solution: cos(P+50 °)=−√22

=−1

√2=sin 45° where the reference angle in quadrant II is 45° -->

the given angle in quadrant II must be 180°-45° --> cos(P+50 °)− 1√2

=sin (180 °−45°)→

(P+50°)=(180 °−45 °)→ P=135°−50 ° → B=85°

Exp. 2.2.6 The point M(0,1) is on the terminal arm of the angle α as shown, find cos α , sin α

Solution: cos α= xr=0

1=0 , sin α= y

r=1

1=1

Exp. 2.2.7 Find the value of cos−1( √32

) , sin−1( 1

√2) , tan−1(√3)

Solution: cos−1( √32

)=30°

sin−1( 1√2

)=45°

tan−1(√3)=60 °

Exp. 2.2.8 Find the value of α to the nearest tenth of a degree, if cosα=59

Solution: α=cos−1(59)=56.251° → α≈56.3 °

Exp. 2.2.9 The 2-right triangles are given as shown, find out (a) PRQ=? (b) PT = ?(Round the answer to

the nearest tenth of a centimeter)

Solution: (a) PQR+ PRQ+QPR=180 ° →

46 °+ PRQ+90 °=180 ° →

PRQ=180 °−90 °−46 °=44 °

(b) cos PRQ= PRQR

=cos 44 °= PR6.22cm

PR=(6.22cm)(cos 44 °)

sin PRT = PTPR

=sin 69 °= PT(cos 44 °)(6.22cm )

PT=(sin 69 °)(cos 44°)(6.22cm )=(0.9335)(0.7193)(6.22cm )=4.1767cm→ PT≈4.2 cm

6.22

cm

69°

46°

P

Q

R

Q

T

Exp. 2.2.10 The point M(4, y) PRQ=? is the terminal arm of the angle α in the standard position, r = 5

units – round all the answers to the nearest tenth. Find out (a) the value of y = ? (b) The value of the angle α = ? (c) tan α = ?

Solution: (a) PythagoreanTheorem :r 2=x2+ y2=52=42+ y2→ y2=25−16=9→

y=3 units

(b) α=cos−1( xr)=cos−1(4

5)=36.8698° → α≈36.9°

(c) tan α= yx=3

4→ tan α=0.25

Exp. 2.2.11 The angle 0≤α≤180° and tan α = -1.4281, what is the value of α ? Round the answer to

the nearest tenth.

Solution: tan α=−1.4281=tan (−55° )→ The reference angle of the angle α is (55 °)

Because the angle 0≤α≤180° --> so α=180 °−55°=125 °

Exp. 2.2.12 The point Q(-3, -10) is on the terminal arm of the angle α in the standard position. (Round the answer to the nearest tenth.)

(a) What is the value of α ? (b) What are the values of cos α , sin α and tan α ?

Solution: (a) PythagoreanTheorem :r 2=x2+ y2=(−3)2+(−10)2=9+100=109 →

r=√109=10.44

(b)

cos α= xr=

(−3)(√109)

=−0.2873

sin α= yr=

(−10)(√109)

=−0.9578

tan α= yx=

(−10)(−3)

=3.333 ..

Math 11 - 2.3-The Law of Sine - Cosine

AB

C

c

ab

α

The Law of SINE

sin Aa

=sin Bb

=sin Cc

The Law of COSINE

a2=b2+c2−2bc(cos A)

b2=a2+c2−2ac(cos B)

c2=a2+b2−2ab(cosC )

The Ambiguous Case: 2 sides and an angle are given to determine the length of an unknown side.

There may be no solution, one solution or two solutions.

Things to know

A+ B+C=180 °

exterior angle α= A+C

Exp. 2.3.1 The triangle ABC is given as shown , round the answers to the nearest tenth.(a) What is the value of AB? (b) What are the values of the angle A ?

Solution: (a)

c2=a2+b2−2ab (cos C )=(5)2+(15)2−2(5)(15)(cos 36 °)=25+225−150 (0.809)=128.64745 →

c=11.3423 --> AB=c≈11.3

(b) a2=b2+c2−2bc (cos A)=(5)2=(15)2+(11.3)2−2(15)(11.3)(cos A)→

−328.6475=−339 cos A→ cos A=0.9695=cos(14.196 °)→ A=14.196 °

Exp. 2.3.2 The triangle ABC is given as shown , round the answers to the nearest tenth.What is the value of AB?

Solution: A+B+C=180 ° → A+119°+36 °=180 ° → A=180 °−155°=25°

sin Aa

= sin Bb

= sin Cc

→ sin 25°5

= sin 36 °c

→c=5sin 36 °sin 25°

=5(.5877)(0.4226)

=6.954 → AB=c≈7.0

AB

C

c

a=5b=15

36°

AB

C

c

a=5b

36°

119°

Exp. 2.3.3 The triangle ABC is given as shown , round the answers to the nearest tenth.What is the value of AB?

Solution:

sin A

a= sin B

b= sin C

c= sin A

6= sin 19

3→sin A= 6sin 19 °

3=2(0.32556)=0.6511=sin 40.63 ° →

sin A=0.6511=sin 40.63°=sine139.37 ° → A=40.63° ,139.37 °

If A=40.63° → A+B+C=180 ° →(40.63°)+(19° )+C=180 ° →C=180°−(40.63 °)−(19°)→

C=120.37 ° → sin Aa

= sin Bb

= sin Cc

=→c= bsinCsin B

=2.8sin 120.37°sin 19°

=2.8 (0.86277)(0.32556)

=7.42→

c=AB≈7.4

If

A=139.37 ° → A+B+C=180 ° →(139.37 °)+(19 °)+C=180° →C=180 °−(139.37 °)−(19°)→

C=21.63 ° → sin Aa

= sin Bb

= sin Cc

=→c= bsinCsin B

=2.8sin 21.63 °sin 19°

=2.8(0.3686)(0.32556)

=3.17

c=AB≈3.2

AB

C

c

a =5.4

b=2.8

19°

Exp. 2.3.4 A wildlife photographer takes pictures of a bear and a deer on a hill side as shown, the bear is at A, the deer is at B. From his camera at the point C, he notices that the positions of the 2 animals is 40° How far is the deer from the bear? Round the answer to the nearest tenth.

Solution:

AB2=AC2+BC2−2(AC )(BC )(cos C)=(91m )2+(109.2m )2−2(91m )(109.2m15)(cos 40 °)=→

AB2=4980.3263 → AB=70.571 m→ The deer is 70.6 m from the bear.

Exp. 2.3.5 What is the length of AB? (Round the answer to the nearest tenth.)

Solution: A+B+C=180 °=40 °+B+72° → B=180 °−40 °−72 °=68°

sin Aa

= sin Bb

= sin Cc

=→c= bsinCsin B

=78sin 72°sin 68°

=78(0.95105)(0.92718)

=80.0086 →

AB=c=80.0 ft

91 m

109.2 m

B

A C40°

B

A C40° 72°

78 ft

Exp. 2.3.6 What is the measure of the angle O? (Round the answer to the nearest tenth.)

Solution:

(mo)2=(mn)2+(no)2−2(mn)(no)(cos N )=(4.5m)2+(3m )2−2(4.5m)(3m)(cos 25°)=→

(mo)2=4.77968 → mo=2.186 m

sin Mm

= sin Nn

= sin Oo

=→sin O= osin N(n=MO)

=(4.5m)sin 25°

(2.186m )=

(4.5m )(0.4226)(2.186m )

=0.86998 →

sin O=0.86998=sin 60.4566°=sin(180 °−60.4566°)=sin 119.54 ° -->

For sin θ in quadrant I and II --> they are the same value, in this solution --> we need the bigger

angle due to the the specific value of the angle N = 25° --> So O=119.5°

Exp. 2.3.7 Which equation is not used to solve the length of r ?

(a)sin 65°

r=sin 25 °

3.3cm

(b) (7.8cm)2=(3.3cm)2+(r)2−2(3.3cm)(r )sin RPQ

(c) sin 90°7.8cm

= sin 65 °r

(d) (7.8cm)2=(3.3cm)2+(r)2

Solution: (b) is not the cosine law

M

N O

25°3 m

4.5 m

P

Q25°

R

65°

7.8 cm

3.3

cm

Exp. 2.3.8 Find the angle SRT from the figure.( Round the answer to the nearest tenth.)

Solution:

sin Tt

= sin Rr

= sin Ss

=→sin R= rsinT(t=RS )

=(8.1m)sin 18°

(3.1m)=

(8.1m)(0.309)(3.1m )

=0.8074 →

sin R=0.8074=sin 53.8457°=sin (180 °−53.8457°)=sin 126.1542 ° --> 53.8457° is the

reference angle of 126.1542 ° --> the solution is sin R=sin 126.1542 ° → SRT =R=126.2°

Exp. 2.3.9 which equation could be used to calculate the length of b?

(a) sin 35124

= sin 75b

(b) b=√(124)2+(209)2+2(124)(209)cos 70 °

(c) b=√(124)2+(209)2−2(124)(209)cos 70 °

(d) b=209sin 69sin 75

Solution: The equation in (c) b=√(124)2+(209)2−2(124)(209)cos 70 ° could be used, it is the

COSINE LAW.

R

S T

3.1 m

18°8.1 m

B

A C

75° 35°

b

124 m

209 m

Exp. 2.3.10 Find the angle α and AC from the figure.( Round the answer to the nearest tenth.)

Solution:

sin CAB

= sin ABC

= sin BAC

=→sin A= BCsinC(AB)

=(6.84cm)sin 15°

(3cm)=

(6.84cm )(0.2588)(3cm)

=0.5901→

sin A=sin α=0.5901=sin 36.1646° → A=α≈36.2 °

A+B+C=180 ° → B=180°−A−C=180 °−36.2 °−15 °=128.835°≈128.8°

sin CAB

= sin ABC

= sin BAC

=→ AC= ABsin Bsin C

=(3cm )sin 128.8 °

sin 15°=

(3cm)(0.77895)0.2588

=9.02cm →

AC≈9.0 cm

B

C15°

3 cm6.84 cm

A

α

Exp. 2.3.11 A triangle ABC is given as shown. What is the possible value of the angle A ?

(a) 45° (b) 40°

(c) 35° (d) 30°

Solution: To find out the angle A = α , draw the height line - h

sin α= hAB

→h=AB sin α→ h=(72.4m )sin α→

If α=45° → h=(72.4m )sin 45 °=(72.4m)(0.7071)=51.19 m

If α=40 ° →h=(72.4m)sin 40°=(72.4m)(0.6428)=46.54 m

If α=35 ° →h=(72.4m)sin 35 °=(72.4m)(0.5736)=41.53 m

If α=30 ° → h=(72.4m )sin 30°=(72.4m)(0.5)=36.2 m

To maintain the reality --> BC must be greater than h --> BC>h

So α=45° is not possible --> α could be 40 ° , 35° , 30 °

h72.4 m

47.8

m

CA

B

α

Exp. 2.3.12 A triangle MNO is given as shown, which equation is correct?

(a) sin α= MN sin MNO

(b)sin OMO

= sin MNO

(c) MO2=MN 2+NO2−2(MN )(NO)(cosα)

(d) MO2=MN 2+NO2−2(MN )(NO)(cos M )

Solution: (c) MO2=MN 2+NO2−2(MN )(NO)(cosα) is correct , it is the Law of COSINE.

Exp. 2.3.13 A triangle XYZ is given as shown, find the perimeter of this triangle?

Solution: The angles X +Y+Z=180°=X +38°+18 ° → X =180°−56 °=124°

sin XYZ

= sin YXZ

= sin ZXY

→ sin 124 °YZ

= sin 38°48.6

= sin 18 °XY

→

YZ=48.6 (sin 124 °)

sin 38 °=

48.6(0.829)0.6157

=65.44

XY=48.6(sin 18° )

sin 38°=

48.6 (0.309)0.6157

=24.39 -->

The perimeter of the triangle XYZ is XY+YZ+ XZ=24.39+65.44+48.6=138.43 m

ON

M

α

ZY

X

38° 18°

48.6 cm