math 11 - 2-trigonometry€¦ · exp. 2.1.2 what could it be the co-terminal angle of the ... find...
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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