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Chapter 3
Radian Measure and Dynamic Trigonometry
222
Chapter 3 Topics
• Angle Measure in Radians
• Length, Velocity and Area of a Circular sector
• Unit Circle
• Trig and Real Numbers
333
Chapter 3.1
Angle Measure in Radians
444
Overview
• Radians for Angle Measurement
• Radian Measure of Standard Angles
• Converting
• And yes, we did a bit of this before
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The Unit Circle
• We use a “central circle”
– Circle in the x-y plane
– Center at the origin
• A central angle is an angle whose Vertex is at the center of
the circle
• An angle whose sides intersect the circle at B and C
“subtends” an arc BC
C
B
y
x
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Radians
• A radian is the measure of an angle subtended by an arc of whose
length is the radius
• Therefore, the length of an arc, s, the radius r times the angle in
radians, 2 r = s
• Now, lets make some sense of this:
What is arc length?
– The arc whose length is the entire circle is the circumference
– The circumference is 2 r
– Therefore, the angle that encompasses the whole circle
is 2 , which in degree measure is 360
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Example
If a circle has radius 8 cm, and the arc length is 18 cm, what is
the radian measure of an angle ?
= s/r = 18/8 = 9/4 = 2.25 rad
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Some More Examples
Find the radian measure of the angle with arc s and
circle radius r
• s = 24 m, r = 4 m
• s = 10 ft, r = 10 ft
• s = 5.2 mm, r = 2.6 mm
• s = 11 cm, r = 20 m
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Example (Solutions)
Find the radian measure of the angle with arc s and
circle radius r
• s = 24 m, r = 4 m; = 6
• s = 10 ft, r = 10 ft; = 1
• s = 5.2 mm, r = 2.6 mm; = 2
• s = 11 cm, r = 20 m; equivalent to 11 cm and 2000 cm:
= 11/2000
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Standard Angles
• Circumference C = 2 r, or an arc of length 2 r goes around
the circle one time
This is the same as saying that an arc whose length is the
circumference subtends an angle of 2
• Therefore, 360 = 2
• 90 = 2/4 = /2
• 180 = 2/2 =
• 45 = 2/8 = /4
• 60 = 2/6 = /3
• 30 = 2/12 = /6
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Example, Finding Angle Measure
• Find the radian measure of the following angle:
• 120=
• -225=
• 270=
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Example (Solutions)
• Find the radian measure of the following angle:
• 120= 2(60 ) = 2 (/3) = 2/3
• -225= - 5(45 ) = -5(/4) = -5/4 [= 2 - 5/4 = 3/4]
• 270= 3(90 ) = 3(/2)
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More Examples
Find the following angles in radians
• 180 =
• 30 =
• 330 =
• -225 =
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More Examples (Solutions)
Find the following angles in radians
• 180 = rad
• 30 =/6 rad
• 330 =11 /6 rad
• -225 = -5/4 rad
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Converting Degrees and Radians
360 = 2, so 180 =
• To convert from radians to degrees, multiply by 180 /
radians x deg/radians = deg
• To convert from degrees to radians, multiply by /180
deg x radians/deg = radians
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Example
Convert to radians:
-75 =
250 12’ =
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Example (Solution)
Convert to radians:
-75 = -75 ( /180 ) = 5 / 2
250 12’ = 250.2 = 259.2 ( /180 ) = 1.39
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Example (Solution)
Convert from radians to degrees
/ 24 =
-5 =
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Example
Convert from radians to degrees
/ 24 = /24 (180 / ) = 180 / 24 = 7.5
-5 = -5 (180 / ) = 900 /
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More Examples
Convert to radians:
• 230 =
• -35 =
• 200 48’ =
3.1, Even problems 38 - 64
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Example (Solutions)
Convert to radians:
•230 = 23/18
•-35 =7 /6
•200 48’ = ? For this one, you need to convert 48’ to degrees.
48’ = 48/60= 0.8 degrees. Then do the multiplication:
(200.8/180)
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Convert to Degrees:
• /4 =
• /2 =
• 5 /6 =
• 6 =
22
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Solutions
Convert to Degrees:
•/4 = 45
•/2 =90
•5 /6 =150
•6 = 1080
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More Examples
Find the terminal side quadrant:
• = 3.9
• = 5.4
• = -4.3
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Solutions
Find the terminal side quadrant:
• = 3.9, Q 3
• = 5.4, Q 4
• = - 4.3, Q 2
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Summary
We can convert from radians to degrees and vice versa: The key
is that 360 degrees is 2 radians
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Chapter 3.2
Arc Length, Velocity, Area of Circular Sector
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Overview
• Use radians to compute the length of a subtended arc
• Solve problems involving angular velocity and linear velocity
• Calculate the area of a circular sector
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Length of an arc
• For , a subtended arc of a circle of radius r, the arc length is
s = r , for in radians
• For example: radius is10 cm, angle is 3.5 radians,
subtended arc is s = 10(3.5) = 35 cm
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Angular and Linear Velocity
Angular velocity equals the amount of rotation per unit time
• Often designated as (omega)
• = / t
For example, a Ferris wheel rotating at 10 revolutions per minute
has angular velocity of
= / t
= 10 revolutions/min
each revolution is 2 rad
= 10 (2 ) / 1 min = 20 rad/min
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Linear Velocity
Linear velocity is distance traveled (or change in position)
per unit time
D = r t, r = D / t
For angular motion, the distance D is the length of the arc, s
rate = v = s / t
Since s = r , v = r / t = r ( / t ) = r
Linear velocity v = r
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Example
A point P rotates around the circumference of a circle with radius
r = 2 ft at a constant rate. If it takes 5 sec to rotate through an
angle of 510,
a. What is the angular velocity of P
b. What is the linear velocity of P
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Solution
a. 510 = 510 ( / 180 )
= (510/180) = 17/6
= / t = (17/6) / 5 = (17/30) rad/sec
b. V = r = 2ft (17/30 ) = (17/15) ft/ sec
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More Examples
Point P passes through angle in time t as it travels around the
circle. Find its angular velocity in radians
• = 540 , t = 9 yr
• = 270 , t = 12 min
• = 690 , t = 5 sec
• = 300 , t = 5 hr
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Solutions
Covert to radians, = angle / time
• = 540 , t = 9 yr; 540 = 540 / 180 = 3 , = /3 rad/yr
• = 270 , t = 12 min; = /8 rad/min
• = 690 , t = 5 sec; = 23 / 30 rad/sec
• = 300 , t = 5 hr; = /3 rad/hr
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Examples
Point P travels around a circle of radius r, find its linear velocity
1. = 12 rad/min, r = 15 ft
2. = 2312 rad/sec, r = 0.01 km
3. = 282, t = 4.1 min, r = 1.2 yd
4. = 45, t = 3 hr, r = 2 mi
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Solutions
V = r
1. 180 ft/min
2. 72.6 mph
3. 1.44 yd/min
4. 30 mph
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Example
Using Angular Velocity to Determine Linear Velocity
The wheels on a bicycle have a radius of 13 in. How fast, mph, is
the cyclist going if the wheels turn at 300 rpm?
= 300 rev/min = 300 (2 ) / min = 600 /min
V = r = 13 in (600 /min)
1 mile = 5280 ft x 12 in /ft, 1 hour = 60 min
V = 13 in (600 /min) (60 min/hr) ( 1 mi / (5280 x 12 in)
V = 23.2 mph
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The Area of a Circular Sector
• The area of a circle is r2
• The area of one half the circle is r2/2
• The area subtended by an arbitrary angle = r2/2,
for in radians
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Examples
Find the area of a central angle of 3 /4 if the radius is 72 ft.
A = (3 /4 ) (72ft)2 /2 = 1944 ft2
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Another Example
The second hand on a wristwatch is 12 mm long. Find the area
of the watch face the second hand passes over in 20 sec
= / t = 2 / 60 rad/sec
in 20 sec, = 20 sec (2 / 60 rad/sec) = 2 / 3 rad
Area A = r2/2 = (2 / 3 rad) (12mm) 2 /2
= 48 mm 2
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Summary
• Arc Length: s = r , in radians
• Angular Velocity: = / t, in radians
• Linear Velocity: v = r
• Area of a circular sector subtended by angle : r2/2,
in radians
All assume is in radians!!!!
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Chapter 3.3
The Unit Circle
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Overview
• Locate points on a unit circle
• Use special triangles to find points on a unit circle
• Define the 6 trig functions in in terms of points on the
unit circle
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Unit Circle
• Circle of radius 1
• Center at the origin
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x
y
(0,0)
(1,0)
(0,1)
(-1,0)
(0,1)
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Exercise
• Find a point of the unit circle if y = 1/2
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Solution
x2 + y2 = 1
x = ± sqrt (1 – ¼) = ± sqrt(3)/2
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Symmetry
• Find the quadrant containing (-3/5, -4/5) and verify it is on the
unit circle
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Solution
• Quadrant : x and y < 0, so is in quadrant 3
• Check: x2 + y2 = 1
(3/5) 2 + (4/5) 2 = (9 + 16)/25 = 1
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More on Symmetry
• If the point (a,b) is on the unit circle, so are
(-a, b)
(a, -b)
(-a, -b)
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Special Triangles
Find points on the unit circle associated with /4: /4: /2 triangle
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Solution
• In quadrant 1, the triangle has x and y values sqrt(2)/2
• In quadrant 2 it is (- sqrt(2)/2, sqrt(2)/2)
• In quadrant 3, both signs are –
• In quadrant 4, x is positive, y is negative
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More Examples
• Do the same for a /6: /3: /4 triangle
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Solutions
(± sqrt(3)/2, ± 1/2)
(± 1/2, ± sqrt(3)/2)
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Points Associated with Rotation
Find the points on the unit circle associated with
a. 5/6
b. 4 /3
c. 7 /4
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Solution
a. (-sqrt(3)/2, 1/2)
b. (-1/2, - sqrt(3)/2)
c. (sqrt(2)/2, -sqrt(2)/2)
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Trig Functions and Rotations
Find the six trig functions for = 5 /4
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Solution
Quadrant 3
• cos = -1/sqrt(2)
• sin = -1/sqrt(2)
• tan = 1
• sec = -sqrt(2)
• csc = -sqrt(2)
• cot = 1
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Summary
• Can find points on the unit circle for angles in radians
• Can calculate trig functions for points on the unit circle
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Chapter 3.4
Trigonometry of Real Numbers
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Overview
• Define trig functions in terms of a real number t
• Find the number associated with special values of the
trig functions
• Find the real number t associated with any trig value
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Real Numbers
• Integers … -2, -1, 0, 1, 2, …
• Rationals: any number that can be expressed as a fraction
Rationals include integers
• Reals: the non-imaginary numbers. Includes 2 and all
rational numbers
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Trigonometry of Real Numbers
• Work with the unit circle: r=1.
if r = 1, then arc length s = r =
• That way, any function of is a function of arc length, s
• We can also treat reference arc as reference angle
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Examples
Give the six trig functions of:
a. 11 / 6
b. 3 / 2
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Solution
a. Quadrant 4:
cos = sqrt(3)/2
sin = -1/2
tan = -1/sqrt(3)
b. Along y axis
cos = 0
sin = -1
tan = undefined (-)
Given these, can find csc, sec, cot
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More Examples
• csc (/6) =
• csc (5/6) =
• csc (11/6) =
• csc (- /6) =
• csc (-17/6) =
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Solutions
• csc (/6) = 2
• csc (5/6) = 2 ; /6 away from 180
• csc (11/6) = - 2 ; /6 away from 360
• csc (- /6) = - 2 ; /6 away from 0
• csc (-17/6) = -2 ; /6 away from -18 /6 = -3 = -180
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Examples
• cot =
• cot 0 =
• cot /2 =
• cot 3 /2
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Solutions
• cot = undefined
• cot 0 =undefined
• cot /2 = 0
• cot 3 /2 = 0
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Special Angles
Find t such that
a. cos t = -1
2
in quadrant 2
b. tan t = 3
in quadrant 3
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Solutions
Find t such that
a. cos t = -1/sqrt(2); Q2 (like a 45 deg angle)
t = 3 /4
b. tan t = sqrt (3); Q3(like a 60 deg angle)
t = 4 /3
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Summary
• Know how to find special angles on the unit circle
• Can calculate the trig functions of these angles
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Chapter 3 Review
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• Give the reference angle for:
7 /6
24 /3
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Solution
• Give the reference angle for:
7 /6 : / 6
24 /3: 8 or 0
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Find
• Sin /3 =
• Cos 2 /3 =
• Tan /3 =
• Cos 7 /4 =
• Tan 7/4 =
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Solution
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• Verify that (1/3, - 2 sqrt(2) / 3) is a point on the unit circle.
• Find the value of all the trig functions associated with
this point
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Solution
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80
• A camera crew rids a cart on a circular arc. The radius of the
arc is 75 ft and can sweep an angle of 172.5º in 20 sec.
– Find the length of the track in feet
– Find the angular velocity of the cart
– Find the linear velocity of the cart.
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Solution
• A camera crew rids a cart on a circular arc. The radius of the
arc is 75 ft and can sweep an angle of 172.5º in 20 sec.
– Find the length of the track in feet
• S=r = 75 ft (172.5)(/180) = 225.8 ft
– Find the angular velocity of the cart
• ω = /time = 172.5(/180) /20 = 0.15 rad/sec
– Find the linear velocity of the cart
• V = r ω = (0.15 rad/sec) (75ft) = 11.29 ft/sec
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• Find t, between 0 and 2 if
Sin t = -1/2 in Q3
Sec t = 2 sqrt(3)/3 in Q4
Tan t = -1 in Q2
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Solution
• Find t, between 0 and 2 if
Sin t = -1/2 in Q3: t= 7 /6
Sec t = 2 sqrt(3)/3 in Q4: t = 11 /6
Tan t = -1 in Q2: t = 3 /4
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Example
• If (20/29, 21/29) is a point on the central unit circle, use
symmetry to find 3 other points on the circle.
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Solution
• If (20/29, 21/29) is a point on the central unit circle, use
symmetry to find 3 other points on the circle.
(-20/29, 21/29)
(-20/29, -21/29)
(20/29, -21/29)
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• Convert to radians:
300
-7239’
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• Convert to radians:
300: 5 /3
-7239’= -1.26; First change the minutes to seconds
= 0.0548
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• Convert to degrees:
9.29
-3 /2
45
88
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Solution
• Convert to degrees:
9.29 = 532.3
-3 /2 = -270
45 = 2578.3
Remember, to convert to degrees, multiply by 180 /
89
90
• Assume Memphis is in directly north of New Orleans, at
90 º W longitude. Find the distance between cities, in km, if
the radius of the earth is 6000 km, Memphis is at 35º north,
and New Orleans is at 29.6º.
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Solution
• Assume Memphis is in directly north of New Orleans, at
90 º W longitude. Find the distance between cities, in km, if
the radius of the earth is 6000 km, Memphis is at 35º north,
and New Orleans is at 29.6º.
Angle is 35º - 29.6 º = 5.4º
5.4º (/180) = 0.094 rad
Radius is 6000 km,
so arclength is 6000 ( 0.094) = 565 km
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Chapter 3 Summary Test
With additional Chapter 3 problems
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1. What angles in [0, 2) make this true?
Sin t = - 3 / 2
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Solution
1. What angles in [0, 2) make this true?
Sin t = - 3 / 2
t=7/6, 11/6
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2. Given (3/4, - sqrt(7)/4) is a point on the unit circle, find all 6 trig
functions.
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Solution
2. Given (3/4, - sqrt(7)/4) is a point on the unit circle, find all 6 trig
functions.
x =3/4, y = -sqrt(7)/4, r = 1
cos = 3/4 sec = 4/3
sin = -sqrt(7)/4 csc = -4/sqrt(7)
tan = -sqrt(7)/3 cot = -3/sqrt(7)
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3. Find the complement of 67 22’ 39”
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Solution
3. Find the complement of 67 22’ 39”;
Need 90 – the angle.
Also, 67 = 66 + 59’ + 60” – need this to subtract
22 37’ 21”
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4. Given cot (/8) = sqrt(2) + 1, find tan2 (3 /8)
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Solution
4. Given cot (/8) = sqrt(2) + 1, find tan2 (3 /8)
Note that /8 + 3 /8 = /2. These are cofunctions!
All that remains is to square [sqrt(2)+1]
we get 2 sqrt(2)+3;
since (a+b)2 = a2 + b2 + 2ab
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5. Given A = 9x, B = (6x+4) , and C = 7x , find the measures
of the angles A, B, C in triangle ABC
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Solution
5. Given A = 9x, B = (6x+4) , and C = 7x , find the measures
of the angles A, B, C in triangle ABC
9x + 6x + 4 + 7x = 180; the sum of the angles is 180
22 x = 180 – 4
x = 176/22 = 8
A = 72 , B = 52 , c = 56 .
Check 99+52+56 = 180
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6. You stand 457 m from a tower. The angle of elevation to the
top of the tower is 30 deg. How tall is the tower?
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Solution
6. You stand 457 m from a tower. The angle of elevation to the
top of the tower is 30 deg. How tall is the tower?
tan 30 = H/457, where H is height.
tan 30 = 1/sqrt(3), so H = 457 sqrt(3)
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7. One angle in a right triangle is 3115’18”, find the other two in
radians
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Solution
7. One angle in a right triangle is 3115’18”, find the other two
One angle is a right angle, /2. The other two must sum to 90
Again, need to convert to decimal degrees, then calculate
(90 – angle) ( / 180) = 1.025
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8. Find the value of the trig functions if the point (-5, -8) is on the
terminal side of the angle
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Solution
8. Find the value of the trig functions if the point (-5, -8) is on the
terminal side of the angle.
Pythagorean thm: 25 + 64 = 89, hyp = sqrt (89)
We are in quadrant 3, so only tan and cot are positive
Cos = -5/sqrt(89)
Sin = -8/sqrt(89)
Tan = 8/5
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10. t = / 6; find the trig functions
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Solution
10. t = / 6; find the trig functions
/ 6 corresponds to a 30 deg angle.
Cos = sqrt(3)/2
Sin = 1/2
Tan = 1/(sqrt(3))
From these you can get the other three
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13. A conveyor belt moves on rollers 2 in radius, turning at
252 rpm. Find the angular velocity of the rollers.
How fast are your groceries moving on the belt?
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Solution
13. A conveyor belt moves on rollers 2 in radius, turning at
252 rpm. Find the angular velocity of the rollers.
How fast are your groceries moving on the belt?
Angular velocity = angle/time; 252 rpm is 252 (2)/min
linear velocity is just r (angular velocity) = 2(252)(2)in/min
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15. Verify (sin x cos x + cos x) / (sin x + sin2 x) = cot x
114
Solution
15. Verify (sin x cos x + cos x) / (sin x + sin2 x) = cot x
Take a cos x out of the left numerator, and sin x out of the
denominator:
cos x (sin x + 1)/ [sin x (sin x + 1)] = cot x
remove (sin x + 1) from top and bottom, leaving
cos x / sin x = cot x
115
16. If t = 5.37, in what quadrant does it terminate?
116
Solution
16. If t = 5.37, in what quadrant does it terminate?
2 is 6.28.. t is close to that, and certainly greater than 3 /2, so
is in the 4th quadrant
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18. Find the values of sin, cos, tan, if = 225
118
Solution
18. Find the values of sin, cos, tan, if = 225
Reference angle is 225 – 180 = 45 , Q 3.
Cos = -1/sqrt(2)
Sin = -1/sqrt(2)
Tan = 1
119
23. Given that (sin A)(tanA) >0 and (cos2A)(sin A) < 0, in which
quadrant is the terminal side of A?
120
Solution
23. Given that (sin A)(tanA) >0 and (cos2A)(sin A) < 0, in which
quadrant is the terminal side of A?
If ab > 0, the a and b are either both > 0 or both < 0. Therfore,
Need sin and tan both > or both < 0, but (cos2A)> 0, which
means that sin <0, so tan < 0.
Sin and Tan are < 0 in Q4.