Week 9 Test.pdf - Week 9 Test MATH111 A002 Fall 17...

This preview shows page 1 out of 22 pages.

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

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture