**Unformatted text preview: **12/3/2017 Week 9 Test MATH111 A002 Fall 17 Assessments 5.8 Applica ons of Trigonometric Func ons (Homework) Results Week 9 Test - Grade Report
Score: 94% (94 of 100 pts) Submitted: Dec 3 at 4:00am Question: 1 Grade: 1.0 / 4.0 Identify the quadrant in which the angle lies.
−1,209° lies in in Quadrant III (100%) Solution For negative angles, add 360° to the measure of the angle until the sum is between 0° and 360°.
−1,209° + 360° = −849°
and
−849° + 360° = −489°
and
−489° + 360° = −129°
and
−129° + 360° = 231°
Since an angle measuring 231° lies in Quadrant III, an angle measuring −1,209° also lies in Quadrant III. 1/22 12/3/2017
Question: 2 Week 9 Test
Grade: 1.0 / 4.0 Identify the graph that represents an angle in standard position with measure −300°. (100%) Solution Place the initial side of the angle on the positive x -axis. Rotate the terminal side 300° clockwise. (Since the angle is negative). 2/22 12/3/2017 Week 9 Test Question: 3 Grade: 1.0 / 4.0 Find the values of sine, cosine, tangent, cosecant, secant, and cotangent for the angle θ in standard position on the coordinate plane with the point
(4,5) on its terminal side. sin θ =
cos θ = −
−
5√41
41
−
−
4√41 (17%)
(17%) 41 tan θ = 5 (17%)
4 csc θ =
sec θ =
cot θ = −
−
√41
5
−
−
√41
4
4 (17%)
(17%) (17%) 5 Solution Draw the angle in standard position with the terminal side at (4,5). Then draw a line perpendicular to the x -axis through the point to form a right
triangle where r is the hypotenuse. Use r
r = −
−−
−
−
−
−
= to find r . √x2 + y 2 −
−−
−
−
−
−
−
−−−−−−−−
−
−
−
−−−−−
√x2 + y 2 = √(4)2 + (5)2 = √ 16 + 25 = √ 41 Evaluate the trigonmetric functions using x sin θ = cos θ = tan θ = csc θ = sec θ =
cot θ = y
r
x
r
y
x r
y
r
x
x
y = = = = =
= 5
−
− =
√ 41
4
−
− =
√ 41
5
4 5
−
−
√ 41
4
5 41
−
−
4√ 41
41 .
. . −
−
√ 41 4 −
−
5√ 41 −
−
= 4, y = 5, r = √ 41 . .
. . 3/22 12/3/2017 Week 9 Test Question: 4 Grade: 1.0 / 4.0 Suppose that θ is an acute angle and that sin θ = cos θ =
tan θ =
csc θ =
sec θ =
cot θ = –
4√6
11
–
5√6 5 (20%) (20%) –
11√6
24
–
4√6
5 . Use a right triangle to find all the other trigonometric functions of θ. (20%) 24
11 5
11 (20%)
(20%) Solution Begin by sketching a right triangle having θ for an angle.
Label the side opposite θ and the hypotenuse using the fact that
sin θ = opposite/hypotenuse= 5/11. −
− Then use the Pythagorean Theorem to find the unknown side of the triangle. In this case, the missing leg has length √96 .
Knowing all three sides of the right triangle, plug in the lengths of the sides into the trigonometric functions to find their values.
By definition, cosθ = By definition, tanθ = adjacent
hypotenuse By definition, cscθ = By definition, secθ = By definition, cotθ = opposite , so tan θ = adjacent hypotenuse
opposite
hypotenuse
adjacent
adjacent
opposite −
−
√ 96 , so cos θ = 11 . 5
−
−
√ 96 , so csc θ = 11
5 . 11
−
−
√ 96
−
−
√ 96 , so sec θ = , so cot θ = 5 . . . 4/22 12/3/2017
Question: 5 Week 9 Test
Grade: 1.0 / 4.0 If a windshield wiper covers an area of approximately 190 square inches when it rotates at an angle of 102°, find the length of the wiper to the
nearest tenth of an inch.
The length of the wiper is approximately 14.6 (100%) inches.
Solution Convert the angle to radians. 102° ⋅ π = 17π 180° 30 Substitute the radius and the angle measure into the formula for the area of a sector of a circle.
1 2
r θ A = 190 = 1 = 2
r = √ 190(60) 2 2 17 π
2
r
(
)
30 17π r −
−−−−−−
190(60)
17π Use a calculator to find the approximate value. r ≈ 14.6 inches. 5/22 12/3/2017 Week 9 Test Question: 6 Grade: 1.0 / 4.0 Consider the triangle ABC shown below. If θ = π
6 and c = 12 , what is the value of b? Round your answer to the nearest hundredth.
b ≈ 20.78 (100%) Solution The tan function relates the side adjacent an angle to the opposite side.
Use a calculator to evaluate the tan function, making sure that it is set in radian mode.
tan C = tan (
b = π
6 opposite
adjacent ) = = c
b 12
b 12
tan( π
6 ) b ≈ 20.78 6/22 12/3/2017
Question: 7 Week 9 Test
Grade: 1.0 / 4.0 Find the smallest positive angle that is coterminal with −805°.
275 (50%) °
Find the negative angle closest to 0° that is coterminal with −805°.
-85 (50%) ° Solution Add 360° to −805°.
−805° + 360° = −445°
Since −445° is less than −360°, add 360° again.
−445° + 360° = −85°
Since −85° is between 0° and −360°, −85° is the negative angle closest to 0° that is coterminal with −805°. θ =-805°:
Negative coterminal angle closest to 0° = −85° Add 360° to −85°.
−85° + 360° = 275°
Since 275° is between 0° and 360°, 275° is the smallest positive angle that is coterminal with −805°. θ =-805°:
Smallest positive coterminal angle = 275° 7/22 12/3/2017 Week 9 Test Question: 8 Grade: 1.0 / 4.0 Identify the point (x, y) on the unit circle that corresponds to t ( = − 5π
2 . 0 (50%), -1 (50%)) Solution t = − 5 π is not given on the unit circle.
2 However, since − 5 π +4π = From the unit circle, t = 2 Question: 9 3π
2 3π
2 , the point corresponding to t = 3π also corresponds to t 2 = −5π .
2 corresponds to ( 0 , −1 ). Grade: 1.0 / 4.0 Evaluate without using a calculator.
sin ( 7π
6 ) = -1/2 (100%) Solution Draw the angle and determine the reference angle. Given angle = 7π
6 Reference angle = π
6 The measure of the acute angle from the terminal side to the x -axis is Use the triangle to find the sin
sin θ = opposite
hypotenuse π
6 , so sin ( π
6 π
6 . Use a 30-60-90 triangle. .
) = 1
2 . The sine function is negative in quadrant III. Therefore, sin ( 7π
6 ) = − 1
2 . 8/22 12/3/2017 Week 9 Test Question: 10 Grade: 1.0 / 4.0 Find the values of sine, cosine, tangent, cosecant, secant, and cotangent for the angle θ in standard position on the coordinate plane with the point
(7, −4) on its terminal side. sin θ = − cos θ = 65
−
−
7√65 csc θ = − 4 (17%) 7 −
−
√65 4
−
−
√65 7
4 (17%) (17%) 7 cot θ =− (17%) (17%) 65 tan θ =− sec θ = −
−
4√65 (17%) Solution Draw the angle in standard position with the terminal side at (7, −4) . Then draw a line perpendicular to the x -axis to form a right triangle where
r is the hypotenuse. Use r
r −
−−
−
−
−
−
= √x2 + y 2 to find r . −
−
−−
−−−
−−
−−
−
−
−
−−−−−
2
2
= √(7)
+ (−4)
= √ 49 + 16 = √ 65 −
−
4√ 65
4
−
− = −
65
√ 65
−
−
7√ 65
x
7
cosθ =
=
−
− =
r
65
√ 65
sinθ = tanθ = cscθ = secθ =
cotθ = y
r y
x r
y
r
x
x
y = − = − = − = 4
7 −
−
√ 65 4
−
−
√ 65 = − 7
7
4 9/22 12/3/2017
Question: 11 Week 9 Test
Grade: 1.0 / 4.0 What is the measure of the supplementary angle for an angle of 41°?
Do not include the ° symbol in your answer. If the given angle has no complement, enter "none".
139 (100%)°
Solution The angle has a supplement if and only if the angle measure is positive and less than 180°.
Two angles are supplementary if the sum of their measures is 180°.
Solve the equation 41° + x = 180° to find the measure of the angle's supplement, x.
Therefore, the supplementary angle for an angle measuring 41° is 139°. 10/22 12/3/2017 Week 9 Test Question: 12 Grade: 1.0 / 4.0 Use a special right triangle to find the sine, cosine, tangent, cosecant, secant, and cotangent of sin π
4 cos π tan π csc π sec π cot 4
4
4
4
π
4 =
= –
√2
2 –
√2
2 π
4 . (17%)
(17%) = 1 (17%)
= –
√2 (17%) = –
√2 (17%) = 1 (17%) Solution An angle with measure π
4 is the one of the non-right angles in a 45-45-90 special right triangle. Use this triangle to find the sine, cosine, tangent, cosecant, secant, and cotangent of π .
4 By definition, sinθ = By definition, cosθ = By definition, tanθ = By definition, cscθ = By definition, secθ = By definition, cotθ = opposite
hypotenuse
adjacent
hypotenuse
opposite 4 opposite
hypotenuse
adjacent π , so sec π π
4 = 4 , so csc , so cot = π 4 hypotenuse opposite , so cos π , so tan π = adjacent adjacent , so sin 4 4 = 1
1 =
=
1
1 1
–
√2 = 1
–
√2
= 1 = . 2
–
√2 . 2 . –
√2 –
= √2 1
–
√2 = 1
= –
√2 1 –
√2 .
. . 11/22 12/3/2017 Week 9 Test Question: 13 Grade: 0.0 / 4.0 A building 1,800 yards from an observer subtends an angle of 2°. Use the arc length formula to estimate the height of building.
Round the answer to the nearest yard, if necessary.
The height of the building is approximately (0%) yards. Solution Think of the height of the building as an arc of a circle of radius 1,800 yards.
First convert 2° to radians
2° = 2°⋅ ( π
180° ) radians Now use the arc length formula and round to the nearest yard
s = rθ = (1,800 yards) ( 2
180 π) ≈ 63 yards The building is approximately 63 yards tall. 12/22 12/3/2017 Week 9 Test Question: 14 Grade: 1.0 / 4.0 A holiday tree is formed by extending single strands of lights from the top of a pole to an anchor point on the ground. Each strand's anchor point is
on the circumference of a circle with radius 54 ft, where the bottom of the pole is the center of the circle. If the angle formed between the ground
and a strand of lights is 5π
18 , find the length of a single strand of lights. Round to the nearest hundredth of a foot, if necessary.
The length of the light strand is 84.01 (100%) feet Solution Since the adjacent side to the angle θ is given and the hypotenuse is needed (the length of the light strand), the cosine function is used. Let the
length of the hypotenuse be h . ⅆ ⅇ
ⅇ a ȷac nt cos θ = ⅇ hypot nus
cos 5π
18 54 = θ = h 5 π
18 Solve for h . h ⋅ cos
h ⋅ cos
h = 5 π
18
5π =
= 18 54
h h 54 54
cos 5π
18 h ≈ 84.01 ft 13/22 12/3/2017 Week 9 Test Question: 15 Grade: 1.0 / 4.0 The height of a tower is 70 ft and the height of a building is 35 ft.
If the angle of depression from the top of the tower to the top of the building is 37.4°, what is the distance between the two, to the nearest foot?
46 (100%) feet
Solution A right triangle can be drawn between the structures, as shown in the picture, where θ is the angle of depression, x is the distance between the
structures, and d is the difference in the structure's heights. Subtract the height of the building from the height of the tower to find d .
d = 70 ft − 35 ft = 35 ft. is the side opposite of θ and x is the side adjacent to θ .
Set up the trigonometric ratio using tangent since tanθ = opposite/adjacent.
d tan 37.4° x tan 37.4°
x 35 = x
= 35 35 = tan37.4°
x ≈ 45.78
x ≈ 46 ft 14/22 12/3/2017 Week 9 Test Question: 16 Suppose that cosθ = Grade: 1.0 / 4.0 −
−
√ 48
8 . What is the value of θ? Give your answer in radians and degrees.
Assume that θ is an acute angle.
π
6
30 (50%) radians
(50%) ° Solution cosθ = adjacent
hypotenuse = −
−
√ 48
8 = –
4√ 3
8 = –
√3
2
– Set the adjacent side of the triangle equal to √3 and the hypotenuse equal to 2.
This should stand out as the long leg and the hypotenuse of a 30-60-90 triangle.
Since the long leg is adjacent to θ, θ must be the smaller angle. Since the smaller angle is 30°, θ = 30°. Equivalently, θ = π
6 . 15/22 12/3/2017 Week 9 Test Question: 17 Grade: 0.5 / 4.0 Find csc θ and sec θ, given that sin θ = −7/25 and cos θ > 0.
csc θ = sec θ = 25
24 25
7 (0%) (50%) Solution θ lies in quadrant IV because sin θ < 0 and cos θ > 0. Since
x y −7
=
y = −7
r = 25.
r
25
−
−−
−
−
−
−
−
−
−
−
−
−
−
−
−−−−−−−−−
−
−−−
−−−
2
r = √x2 + y 2 , x = ±√r2 − y 2 = ±√252 − (−7)
= ±√ 625 − 49 = Sincesinθ = , let and > 0 because θ lies in quadrant IV , so x By definition cscθ = r
y and secθ = r
x ±24 . = 24. , therefore cscθ = − 25
7 and sec θ = 25
24 . 16/22 12/3/2017 Week 9 Test Question: 18 Grade: 1.0 / 4.0 Evaluate the tangent and cotangent of t = − 7π
6 If the trigonometric function is undefined, enter "u". tan(− cot(− 7π
6
7π
6 ) = − –
√3 (50%) 3
–
) = −√3 (50%) Solution Since tangent is an odd function, tan(−t)
Therefore, tan(− 7π )
6 From the unit circle, t tan(− 7π ) = − tan = 7π 7π 6 7π = − tan 6 6 6 = − tan(t) corresponds to the point (− = − y
x −
= −
− 1 2
–
√3 6 = − cot 7π
6 = − =− –
√3 ,− 2 1
2 ) . So x = − –
√3
2 and y = − 1
2 . –
√3
3 2 Since cotangent is an odd function, cot(−t)
Therefore, cot(− 7π ) . . x
y = − cot(t) − . –
√3 = −
− 2
1 –
=−√ 3 2 17/22 12/3/2017 Week 9 Test Question: 19 Suppose that Grade: 1.0 / 4.0 secθ = 12
6 . What is the value of θ? Give your answer in radians and degrees.
Assume that θ is an acute angle in a right triangle.
π
3 (50%) radians 60 (50%) ° Solution hypotenuse secθ = adjacent = 12
6 = 2
1 Set the adjacent side of the triangle equal to 1 and the hypotenuse equal to 2.
This should stand out as the short leg of the hyptenuse of a 30-60-90 triangle.
Since the short leg is adjacent θ, θ must be the larger angle. Since the larger angle is 60°, θ = 60°. Equivalently, θ Question: 20 Convert
23π = π
3 . Grade: 1.0 / 4.0 to degree measure. 23π
5 radians = 828 (100%) ° 5 Solution Multiply the radian measure by
23π
5 ⋅ 180°
π = 4,140°π
5π 180°
π and simplify. = 828° 18/22 12/3/2017 Week 9 Test Question: 21 Grade: 1.0 / 4.0 Evaluate the sine, cosine, tangent, cosecant, secant, and cotangent of t = −3π .
2 If the trigonometric function is undefined, enter "u".
sin (− 3π
2 cos (− sec (− 1 (17%) ) = 0 (17%) 2
3π tan (− csc (− ) = 3π ) = 2
3π
2
3π (17%) ) = 1 (17%) ) = u (17%) ) = 0 (17%) 2
cot (− u 3π
2 Solution By the unit circle, t sin (− csc (− 3π
2
3π
2 ) = ) = y 1
y = = = − 3 π corresponds to the point ( 0 , 1 ) because − 3 π
2 1
1
1 2 cos (− = 1 sec (− 3π
2
3π
2 ) = x = ) = 1
x = +2 π = 0 π
2 . So x = tan (− 1
0 = undefined cot (− 0 and y 3π
2
3π
2 ) = ) = = y
x
x
y 1. = = 1
0
0
1 = undefined = 0 19/22 12/3/2017
Question: 22 Week 9 Test
Grade: 1.0 / 4.0 Identify the graph that represents an angle in standard position with measure −535°. (100%)
Solution Place the initial side of the angle on the positive x -axis. Rotate the terminal side 535° clockwise (since the angle is negative). 20/22 12/3/2017 Week 9 Test Question: 23 Grade: 1.0 / 4.0 Find the value of cot 30° without using a calculator.
If the answer is not defined enter "u".
– cot 30° = √3 (100%) Solution Since 30° is an important angle, use the chart for trigonometric functions for important angles. θ (deg) θ (rad) sin θ cos θ tan θ 0° 0 0 1 0 –
√3 –
√3
3 30° π
6 2 2 45° π –
√2 –
√2 4 2 2 π –
√3 1 3 2 2 π 1 0 60°
90° From the chart, tan cot 30 ° 30° 1 = = tan 30° –
–
3√ 3
√3
3
3
– =
– ⋅
– =
3
√3
√3
√3 1
–
√3 und . 3 1
–
√3 Rationalize the denominator. Question: 24 –
√3 = 2 1 3
–
√3 = 3 – = √3 Grade: 1.0 / 4.0 If a 5 foot blade of a propeller completes 22 revolutions per second, find the angular speed of the blade in radians per second.
Round your answer to the nearest hundredth.
Angular speed = 138.23 (100%) radians/second
Solution The blade completes 22 revolutions per second. Since each revolution is 2π radians, the blade turns through 44 π radians per second. Round your answer to the nearest hundredth.
angular speed = θ
t = 44 π radians
1 second = 44π radians/second = 138.23 radians/second 21/22 12/3/2017 Week 9 Test Question: 25 Grade: 1.0 / 4.0 Convert 660° to radians.
Enter the answer in terms of π, as needed. Do not enter a decimal approximation.
Instructions ( (1).html) for entering π
11π 660° = 3 (100%) radians Solution Multiply the degree measure by
660° ⋅ π
180° = 660°
180° π = π
180° 11
3 and simplify. π 22/22 ...

View
Full Document