Download 4-1 Right Triangle Trigonometry

4-1 Right Triangle Trigonometry
Find the exact values of the six trigonometric functions of θ.
1. SOLUTION: The length of the side opposite θ is 8
is 18.
, the length of the side adjacent to θ is 14, and the length of the hypotenuse
2. SOLUTION: The length of the side opposite θ is 2
hypotenuse is 15.
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3. SOLUTION: , the length of the side adjacent to θ is 13, and the length of the
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4-1 Right Triangle Trigonometry
3. SOLUTION: The length of the side opposite θ is 9, the length of the side adjacent to θ is 4, and the length of the hypotenuse is
.
4. SOLUTION: The length of the side opposite θ is 12, the length of the side adjacent to θ is 35, and the length of the hypotenuse is
37.
5. SOLUTION: eSolutions
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The
length
of the side
opposite
hypotenuse is 29.
θ is
, the length of the side adjacent to θ is 26, and the length of the
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4-1 Right Triangle Trigonometry
5. SOLUTION: The length of the side opposite θ is
hypotenuse is 29.
, the length of the side adjacent to θ is 26, and the length of the
6. SOLUTION: The length of the side opposite θ is 30, the length of the side adjacent to θ is 5
hypotenuse is 35.
eSolutions
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SOLUTION: , and the length of the
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4-1 Right Triangle Trigonometry
7. SOLUTION: The length of the side opposite θ is 6 and the length of the hypotenuse is 10. By the Pythagorean Theorem, the
length of the side adjacent to θ is
= or 8.
8. SOLUTION: The length of the side opposite θ is 8 and the length of the side adjacent to θ is 32. By the Pythagorean Theorem,
the length of the hypotenuse is
= or 8
.
Use the given trigonometric function value of the acute angle θ to find the exact values of the five
remaining trigonometric function values of θ.
eSolutions
9. sinManual
θ = - Powered by Cognero
SOLUTION: Page 4
4-1 Right Triangle Trigonometry
Use the given trigonometric function value of the acute angle θ to find the exact values of the five
remaining trigonometric function values of θ.
9. sin θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because sin θ = = , label the opposite side 4 and the
hypotenuse 5.
By the Pythagorean Theorem, the length of the side adjacent to θ is
or 3.
10. cos θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because cos θ = = , label the adjacent side 6 and the
hypotenuse 7.
By the Pythagorean Theorem, the length of the side opposite θ is
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or .
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4-1 Right Triangle Trigonometry
10. cos θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because cos θ = = , label the adjacent side 6 and the
hypotenuse 7.
By the Pythagorean Theorem, the length of the side opposite θ is
or .
11. tan θ = 3
SOLUTION: Draw a right triangle and label one acute angle θ. Because tan θ = , label the side opposite θ 3 and
= 3 or
the adjacent side 1.
By the Pythagorean Theorem, the length of the hypotenuse is
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or .
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4-1 Right Triangle Trigonometry
11. tan θ = 3
SOLUTION: Draw a right triangle and label one acute angle θ. Because tan θ = , label the side opposite θ 3 and
= 3 or
the adjacent side 1.
By the Pythagorean Theorem, the length of the hypotenuse is
or .
12. sec θ = 8
SOLUTION: Draw a right triangle and label one acute angle θ. Because sec θ = = 8 or , label the hypotenuse 8 and
the adjacent side 1.
By the Pythagorean Theorem, the length of the side adjacent to θ is
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= or 3
.
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4-1 Right Triangle Trigonometry
12. sec θ = 8
SOLUTION: Draw a right triangle and label one acute angle θ. Because sec θ = = 8 or , label the hypotenuse 8 and
the adjacent side 1.
By the Pythagorean Theorem, the length of the side adjacent to θ is
= or 3
.
13. cos θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because cos θ = = , label the adjacent side 5 and the
hypotenuse 9.
By the Pythagorean Theorem, the length of the side opposite θ is
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or 2
.
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4-1 Right Triangle Trigonometry
13. cos θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because cos θ = = , label the adjacent side 5 and the
hypotenuse 9.
By the Pythagorean Theorem, the length of the side opposite θ is
or 2
.
14. tan θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because tan θ = = , label the side opposite θ 5 and
adjacent side 4.
By the Pythagorean Theorem, the length of the hypotenuse is
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or .
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4-1 Right Triangle Trigonometry
14. tan θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because tan θ = = , label the side opposite θ 5 and
adjacent side 4.
By the Pythagorean Theorem, the length of the hypotenuse is
or .
15. cot θ = 5
SOLUTION: Draw a right triangle and label one acute angle θ. Because cot θ = , label the side adjacent to θ as
= 5 or
5 and the opposite side 1.
By the Pythagorean Theorem, the length of the hypotenuse is
or .
16. csc θ = 6
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SOLUTION: Draw a right triangle and label one acute angle θ. Because csc θ = Page 10
= 6 or
, label the hypotenuse 6 and the
4-1 Right Triangle Trigonometry
16. csc θ = 6
SOLUTION: Draw a right triangle and label one acute angle θ. Because csc θ = = 6 or
, label the hypotenuse 6 and the
side opposite θ as 1.
By the Pythagorean Theorem, the length of the adjacent side is
or .
17. sec θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because sec θ = = , label the hypotenuse 9 and the side
opposite θ as 2.
By the Pythagorean Theorem, the length of the opposite side is
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or .
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4-1 Right Triangle Trigonometry
17. sec θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because sec θ = = , label the hypotenuse 9 and the side
opposite θ as 2.
By the Pythagorean Theorem, the length of the opposite side is
or .
18. sin θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because sin θ = = , label the side opposite θ as 8 and
the hypotenuse 13.
By the Pythagorean Theorem, the length of the adjacent side is
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or .
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4-1 Right Triangle Trigonometry
18. sin θ = SOLUTION: Draw a right triangle and label one acute angle θ. Because sin θ = = , label the side opposite θ as 8 and
the hypotenuse 13.
By the Pythagorean Theorem, the length of the adjacent side is
or .
Find the value of x. Round to the nearest tenth.
19. SOLUTION: An acute angle measure and the length of the hypotenuse are given, so the sine function can be used to find the
length of the side opposite θ.
20. SOLUTION: An acute angle measure and the length of the side opposite θ are given, so the tangent function can be used to find
theManual
length- Powered
of the side
adjacent to θ.
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4-1 Right Triangle Trigonometry
20. SOLUTION: An acute angle measure and the length of the side opposite θ are given, so the tangent function can be used to find
the length of the side adjacent to θ.
21. SOLUTION: An acute angle measure and the length of the hypotenuse are given, so the cosine function can be used to find the
length of the side adjacent to θ.
22. SOLUTION: An acute angle measure and the length of the side adjacent to θ are given, so the cosine function can be used to
find the length of the hypotenuse.
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eSolutions
SOLUTION: Page 14
4-1 Right Triangle Trigonometry
23. SOLUTION: An acute angle measure and the length of the side opposite θ are given, so the sine function can be used to find the
length of the hypotenuse.
24. SOLUTION: An acute angle measure and the length of the side adjacent to θ are given, so the tangent function can be used to
find the length of the side opposite θ .
25. SOLUTION: An acute angle measure and the length of the side opposite θ are given, so the tangent function can be used to find
the length of the side adjacent to θ .
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SOLUTION: Page 15
4-1 Right Triangle Trigonometry
26. SOLUTION: An acute angle measure and the length of the side opposite θ are given, so the sine function can be used to find the
length of the hypotenuse.
27. MOUNTAIN CLIMBING A team of climbers must determine the width of a ravine in order to set up equipment to cross it. If the climbers walk 25 feet along the ravine from their crossing point, and sight the crossing
point on the far side of the ravine to be at a 35º angle, how wide is the ravine?
SOLUTION: An acute angle measure and the adjacent side length are given, so the tangent function can be used to find the
length of the opposite side.
Therefore, the ravine is about 17.5 feet wide.
28. SNOWBOARDING Brad built a snowboarding ramp with a height of 3.5 feet and an 18º incline.
a. Draw a diagram to represent the situation.
b. Determine the length of the ramp.
SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 18º and the opposite side 3.5 feet.
b. Because an acute angle measure and opposite side length are given, the sine function can be used to find the
Page 16
length of the hypotenuse.
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4-1 Right Triangle Trigonometry
Therefore, the ravine is about 17.5 feet wide.
28. SNOWBOARDING Brad built a snowboarding ramp with a height of 3.5 feet and an 18º incline.
a. Draw a diagram to represent the situation.
b. Determine the length of the ramp.
SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 18º and the opposite side 3.5 feet.
b. Because an acute angle measure and opposite side length are given, the sine function can be used to find the
length of the hypotenuse.
Therefore, the length of the ramp is about 11.3 feet.
29. DETOUR Traffic is detoured from Elwood Ave., left 0.8 mile on Maple St., and then right on Oak St., which intersects Elwood Ave. at a 32° angle.
a. Draw a diagram to represent the situation.
b. Determine the length of Elwood Ave. that is detoured.
SOLUTION: a. Draw a right triangle with acute angle 32º and opposite side length 0.8 mile. Label the side opposite the 32º angle
Maple St., the hypotenuse Oak St., and the adjacent side Elwood Ave.
b. Because an acute angle and the length of opposite side are given, the tangent function can be used to find the
adjacent side length.
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Therefore,
the length
of Elwood
Ave. that is detoured is about 1.3 miles.
Page 17
30. PARACHUTING A paratrooper encounters stronger winds than anticipated while parachuting from 1350 feet, 4-1 Right Triangle Trigonometry
Therefore, the length of the ramp is about 11.3 feet.
29. DETOUR Traffic is detoured from Elwood Ave., left 0.8 mile on Maple St., and then right on Oak St., which intersects Elwood Ave. at a 32° angle.
a. Draw a diagram to represent the situation.
b. Determine the length of Elwood Ave. that is detoured.
SOLUTION: a. Draw a right triangle with acute angle 32º and opposite side length 0.8 mile. Label the side opposite the 32º angle
Maple St., the hypotenuse Oak St., and the adjacent side Elwood Ave.
b. Because an acute angle and the length of opposite side are given, the tangent function can be used to find the
adjacent side length.
Therefore, the length of Elwood Ave. that is detoured is about 1.3 miles.
30. PARACHUTING A paratrooper encounters stronger winds than anticipated while parachuting from 1350 feet, causing him to drift at an 8º angle. How far from the drop zone will the paratrooper land?
SOLUTION: Because an acute angle and the length of the side that is adjacent to the angle are given, the tangent function can
be used to find the length of the opposite side.
So,Manual
the paratrooper
land
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Cognero
about 190 feet away from the drop zone.
Find the measure of angle θ. Round to the nearest degree, if necessary.
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4-1 Right Triangle Trigonometry
Therefore, the length of Elwood Ave. that is detoured is about 1.3 miles.
30. PARACHUTING A paratrooper encounters stronger winds than anticipated while parachuting from 1350 feet, causing him to drift at an 8º angle. How far from the drop zone will the paratrooper land?
SOLUTION: Because an acute angle and the length of the side that is adjacent to the angle are given, the tangent function can
be used to find the length of the opposite side.
So, the paratrooper will land about 190 feet away from the drop zone.
Find the measure of angle θ. Round to the nearest degree, if necessary.
31. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
32. SOLUTION: Because the length of the hypotenuse and side adjacent to θ are given, the cosine function can be used to find θ .
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Page 19
4-1 Right Triangle Trigonometry
32. SOLUTION: Because the length of the hypotenuse and side adjacent to θ are given, the cosine function can be used to find θ .
33. SOLUTION: Because the length of the hypotenuse and side opposite θ are given, the sine function can be used to find θ .
34. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
35. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
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4-1 Right Triangle Trigonometry
35. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
36. SOLUTION: Because the length of the hypotenuse and side adjacent to θ are given, the cosine function can be used to find θ .
37. SOLUTION: Because the length of the hypotenuse and side opposite θ are given, the sine function can be used to find θ .
38. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
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4-1 Right Triangle Trigonometry
38. SOLUTION: Because the lengths of the sides opposite and adjacent to θ are given, the tangent function can be used to find θ .
39. PARASAILING Kayla decided to try parasailing. She was strapped into a parachute towed by a boat. An 800foot line connected her parachute to the boat, which was at a 32º angle of depression below her. How high above
the water was Kayla?
SOLUTION: The angle of elevation from the boat to the parachute is equivalent to the angle of depression from the parachute to
the boat because the two angles are alternate interior angles, as shown below.
Because an acute angle and the hypotenuse are given, the sine function can be used to find x.
Therefore, Kayla was about 424 feet above the water.
40. OBSERVATION WHEEL The London Eye is a 135-meter-tall observation wheel. If a passenger at the top of
the wheel sights the London Aquarium at a 58º angle of depression, what is the distance between the aquarium and
the London Eye?
SOLUTION: The angle of depression from the top of the wheel to the aquarium is 58º, which means that the angle of elevation
from the aquarium to the top of the wheel is also 58º because they are alternate interior angles. Draw a diagram
of
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Page 22
a right triangle and label one acute angle 58º and the opposite side 135 m.
4-1 Right Triangle Trigonometry
Therefore, Kayla was about 424 feet above the water.
40. OBSERVATION WHEEL The London Eye is a 135-meter-tall observation wheel. If a passenger at the top of
the wheel sights the London Aquarium at a 58º angle of depression, what is the distance between the aquarium and
the London Eye?
SOLUTION: The angle of depression from the top of the wheel to the aquarium is 58º, which means that the angle of elevation
from the aquarium to the top of the wheel is also 58º because they are alternate interior angles. Draw a diagram of
a right triangle and label one acute angle 58º and the opposite side 135 m.
Because an acute angle and the side length opposite the angle are given, the tangent function can be used to find x.
Therefore, the distance between the aquarium and the London Eye is about 84 meters.
41. ROLLER COASTER On a roller coaster, 375 feet of track ascend at a 55º angle of elevation to the top before
the first and highest drop.
a. Draw a diagram to represent the situation.
b. Determine the height of the roller coaster.
SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 55º and the hypotenuse 375 feet.
b. Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find the
length of the opposite side.
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4-1 Right Triangle Trigonometry
Therefore, the distance between the aquarium and the London Eye is about 84 meters.
41. ROLLER COASTER On a roller coaster, 375 feet of track ascend at a 55º angle of elevation to the top before
the first and highest drop.
a. Draw a diagram to represent the situation.
b. Determine the height of the roller coaster.
SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 55º and the hypotenuse 375 feet.
b. Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find the
length of the opposite side.
Therefore, the height of the roller coaster is about 307 feet.
42. SKI LIFT A company is installing a new ski lift on a 225-meter-high mountain that will ascend at a 48º angle of
elevation.
a. Draw a diagram to represent the situation.
b. Determine the length of cable the lift requires to extend from the base to the peak of the mountain.
SOLUTION: a. Draw a diagram of a right triangle and label one acute angle 48º and the opposite side 225 meters.
b. Because an acute angle and the length of the side opposite the angle are given, you can use the sine function to
find the length of the hypotenuse.
So, the company will need about 303 meters of cable.
eSolutions
Manual - Powered Both Derek and Sam are 5 feet 10 inches tall. Derek looks at a 10-foot
by Cognero
43. BASKETBALL
basketball goal withPage
an 24
angle of elevation of 29°, and Sam looks at the goal with an angle of elevation of 43°. If Sam is directly in front of Derek, how far apart are the boys standing?
4-1 Right Triangle Trigonometry
So, the company will need about 303 meters of cable.
43. BASKETBALL Both Derek and Sam are 5 feet 10 inches tall. Derek looks at a 10-foot basketball goal with an
angle of elevation of 29°, and Sam looks at the goal with an angle of elevation of 43°. If Sam is directly in front of Derek, how far apart are the boys standing?
SOLUTION: Draw a diagram to model the situation. The vertical distance from the boys' heads to the rim is
10(12) – [5(12) + 10] or 50 inches. Label the horizontal distance between Sam and Derek as x and the horizontal
distance between Sam and the goal as y.
From the smaller right triangle, you can use the tangent function to find y.
From the larger right triangle, you can use the tangent function to find x.
Therefore, Derek and Sam are standing about 36.6 inches or 3.1 feet apart.
44. PARIS A tourist on the first observation level of the Eiffel Tower sights the Musée D’Orsay at a 1.4º angle of
depression. A tourist on the third observation level, located 219 meters directly above the first, sights the Musée D’Orsay at a 6.8º angle of depression.
a. Draw a diagram to represent the situation.
b. Determine the distance between the Eiffel Tower and the Musée D’Orsay.
SOLUTION: eSolutions Manual - Powered by Cognero
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4-1 Right Triangle Trigonometry
Therefore, Derek and Sam are standing about 36.6 inches or 3.1 feet apart.
44. PARIS A tourist on the first observation level of the Eiffel Tower sights the Musée D’Orsay at a 1.4º angle of
depression. A tourist on the third observation level, located 219 meters directly above the first, sights the Musée D’Orsay at a 6.8º angle of depression.
a. Draw a diagram to represent the situation.
b. Determine the distance between the Eiffel Tower and the Musée D’Orsay.
SOLUTION: a.
st
rd
b. Because the angles of depression from the 1 and 3 levels to the Musée D’Orsay are 1.4° and 6.8°,
st
rd
respectively, the angles of elevation from the Musée D’Orsay to the 1 and 3 levels are also 1.4° and 6.8°,
respectively.
Use the tangent function to write an equation for the smaller right triangle in terms of y.
Next, use the tangent function to write an equation for the larger right triangle in terms of y.
Set the equations that you found for each triangle equal to one another and solve for x.
Therefore, the distance between the Eiffel Tower and the Musée D’Orsay is about 2310 meters.
45. LIGHTHOUSE Two ships are spotted from the top of a 156-foot lighthouse. The first ship is at a 27º angle of
depression, and the second ship is directly behind the first at a 7º angle of depression.
a. Draw a diagram to represent the situation.
b. Determine the distance between the two ships.
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SOLUTION: a.
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4-1 Right Triangle Trigonometry
Therefore, the distance between the Eiffel Tower and the Musée D’Orsay is about 2310 meters.
45. LIGHTHOUSE Two ships are spotted from the top of a 156-foot lighthouse. The first ship is at a 27º angle of
depression, and the second ship is directly behind the first at a 7º angle of depression.
a. Draw a diagram to represent the situation.
b. Determine the distance between the two ships.
SOLUTION: a.
b. From the smaller right triangle, you can use the tangent function to find y.
From the larger right triangle, you can use the tangent fuction to find x.
Therefore, the distance between the two ships is about 964 feet.
46. MOUNT RUSHMORE The faces of the presidents at Mount Rushmore are 60 feet tall. A visitor sees the top of George Washington’s head at a 48º angle of elevation and his chin at a 44.76º angle of elevation. Find the height
of Mount Rushmore.
SOLUTION: From the smaller triangle, you can use the tangent function to find y.
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4-1 Right Triangle Trigonometry
Therefore, the distance between the two ships is about 964 feet.
46. MOUNT RUSHMORE The faces of the presidents at Mount Rushmore are 60 feet tall. A visitor sees the top of George Washington’s head at a 48º angle of elevation and his chin at a 44.76º angle of elevation. Find the height
of Mount Rushmore.
SOLUTION: From the smaller triangle, you can use the tangent function to find y.
From the larger triangle, you can use the tangent function to find y, too.
Next, set the two equations equal to one another to solve for x.
Therefore, Mount Rushmore is about 500 feet tall.
Solve each triangle. Round side measures to the nearest tenth and angle measures to the nearest
degree.
47. SOLUTION: Use trigonometric functions to find b and c.
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4-1 Right Triangle Trigonometry
Therefore, Mount Rushmore is about 500 feet tall.
Solve each triangle. Round side measures to the nearest tenth and angle measures to the nearest
degree.
47. SOLUTION: Use trigonometric functions to find b and c.
Because the measures of two angles are given, B can be found by subtracting A from
Therefore,
b
16.5, c
17.5.
48. SOLUTION: Use trigonometric functions to find y and z.
Because the measures of two angles are given, X can be found by subtracting Z from
Therefore,
y
37.1, z
32.5.
49. SOLUTION: Use the Pythagorean Theorem to find r.
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Page 29
Because the measures of two angles are given, X can be found by subtracting Z from
4-1 Right
Triangle Trigonometry
Therefore,
y
37.1, z
32.5.
49. SOLUTION: Use the Pythagorean Theorem to find r.
Use the tangent function to find P.
Because the measures of two angles are now known, you can find Q by subtracting P from
Therefore, P ≈ 43°, Q
47°, and r
34.0.
50. SOLUTION: Use the Pythagorean Theorem to find d.
Use the cosine function to find D.
Because the measures of two angles are now known, you can find E by subtracting D from
Therefore, D ≈ 77°, E
13°, and d
29.2.
51. eSolutions Manual - Powered by Cognero
SOLUTION: Use trigonometric functions to find j and k.
Page 30
Because the measures of two angles are now known, you can find E by subtracting D from
4-1 Right
Triangle Trigonometry
Therefore, D ≈ 77°, E
13°, and d
29.2.
51. SOLUTION: Use trigonometric functions to find j and k.
Because the measures of two angles are given, K can be found by subtracting J from
Therefore,
j
18.0, k
6.2.
52. SOLUTION: Use the Pythagorean Theorem to find y.
Use the sine function to find W.
Because the measures of two angles are now known, you can find Y by subtracting W from
Therefore,
and y
3.9.
53. SOLUTION: Use trigonometric functions to find f and h.
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Because the measures of two angles are now known, you can find Y by subtracting W from
4-1 Right
Triangle Trigonometry
Therefore,
and y
3.9.
53. SOLUTION: Use trigonometric functions to find f and h.
Because the measures of two angles are given, H can be found by subtracting F from
Therefore,
f
19.6, h
17.1.
54. SOLUTION: Use the Pythagorean Theorem to find t.
Use the tangent function to find R.
Because the measures of two angles are now known, you can find S by subtracting R from
Therefore,
and t
8.1.
55. BASEBALL Michael’s seat at a game is 65 feet behind home plate. His line of vision is 10 feet above the field.
a. Draw a diagram to represent the situation.
b. What is the angle of depression to home plate?
SOLUTION: a.
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b. Michael’s angle of depression to home plate is equivalent to the angle of elevation from home plate to where he
is sitting.
Because the measures of two angles are now known, you can find S by subtracting R from
4-1 Right
Triangle Trigonometry
Therefore,
and t
8.1.
55. BASEBALL Michael’s seat at a game is 65 feet behind home plate. His line of vision is 10 feet above the field.
a. Draw a diagram to represent the situation.
b. What is the angle of depression to home plate?
SOLUTION: a.
b. Michael’s angle of depression to home plate is equivalent to the angle of elevation from home plate to where he
is sitting.
Use the tangent function to find θ.
Therefore, the angle of depression to home plate is about
56. HIKING Jessica is standing 2 miles from the center of the base of Pikes Peak, and looking at the summit of the
mountain, which is 1.4 miles from the base.
a. Draw a diagram to represent the situation.
b. With what angle of elevation is Jessica looking at the summit of the mountain?
SOLUTION: a.
b. Use the tangent function to find the angle of elevation.
Therefore, the angle of elevation is about
Find the exact value of each expression without using a calculator.
57. sin 60°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
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Page 33
4-1 Right Triangle Trigonometry
Therefore, the angle of elevation is about
Find the exact value of each expression without using a calculator.
57. sin 60°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
The length of the side opposite the 60° angle is
x and the length of the hypotenuse is 2x.
58. cot 30°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
The length of the side adjacent to the 30º angle is
x and the length of the opposite side is x.
59. sec 30°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
The length of the hypotenuse is 2x and the length of the side adjacent to the 30º angle is
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x.
Page 34
4-1 Right Triangle Trigonometry
59. sec 30°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
The length of the hypotenuse is 2x and the length of the side adjacent to the 30º angle is
x.
60. cos 45°
SOLUTION: Draw a diagram of a 45º-45º-90º triangle.
The side length is x and the hypotenuse is
x.
61. tan 60°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
eSolutions
Manual
- Powered
by Cognero
The
length
of the side
opposite
the 60º is
x and the length of the adjacent side is x.
Page 35
4-1 Right Triangle Trigonometry
61. tan 60°
SOLUTION: Draw a diagram of a 30º-60º-90º triangle.
The length of the side opposite the 60º is
x and the length of the adjacent side is x.
62. csc 45°
SOLUTION: Draw a diagram of a 45º-45º-90º triangle.
The length of the hypotenuse is
x and the side length is x.
Without using a calculator, find the measure of the acute angle θ that satisfies the given equation.
63. tan θ = 1
SOLUTION: Because tan θ = 1 and tan θ = , it follows that
= 1. In the opposite an acute angle is 1 and the adjacent side length is also 1. So,
eSolutions
Manual - Powered
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Therefore,
θ = 45°.
triangle below, the side length
= 1.
Page 36
4-1 Right Triangle Trigonometry
Without using a calculator, find the measure of the acute angle θ that satisfies the given equation.
63. tan θ = 1
SOLUTION: Because tan θ = 1 and tan θ = , it follows that
= 1. In the triangle below, the side length
= 1.
opposite an acute angle is 1 and the adjacent side length is also 1. So,
Therefore, θ = 45°.
64. cos θ = SOLUTION: Because cos θ = and cos θ = length adjacent to the 30º angle is
, it follows that
= and the hypotenuse is 2. So,
. In the
= triangle below, the side
.
Therefore, θ = 30°.
65. cot θ =
SOLUTION: Because cot θ =
and cot θ = , it follows that
= . In the
length that is adjacent to the 60º angle is 1 and the length of the opposite side is
eSolutions
Manual - Powered
by Cognero
Therefore,
θ = 60°.
triangle below, the side
. So,
= or .
Page 37
4-1 Right Triangle Trigonometry
Therefore, θ = 30°.
65. cot θ =
SOLUTION: Because cot θ =
and cot θ = , it follows that
= . In the
triangle below, the side
length that is adjacent to the 60º angle is 1 and the length of the opposite side is
= . So,
or .
Therefore, θ = 60°.
66. sin θ = SOLUTION: Because sin θ = and sin θ = , it follows that
length opposite an acute angle is 1 and the hypotenuse is
= . In the
. So, = triangle below, the side
or .
Therefore, θ = 45°.
67. csc θ = 2 SOLUTION: Because csc θ = 2 and csc θ = , it follows that
= 2 or . In the
hypotenuse is 2 and the side length that is opposite the 30º angle is 1. So,
triangle below, the
= or 2.
Therefore, θ = 30°.
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68. sec θ = 2
SOLUTION: Page 38
4-1 Right Triangle Trigonometry
Therefore, θ = 45°.
67. csc θ = 2 SOLUTION: Because csc θ = 2 and csc θ = , it follows that
= 2 or . In the
hypotenuse is 2 and the side length that is opposite the 30º angle is 1. So,
triangle below, the
= or 2.
Therefore, θ = 30°.
68. sec θ = 2
SOLUTION: Because sec θ = 2 and sec θ = , it follows that
= 2 or . In the
hypotenuse is 2 and the side length that is adjacent to the 60º angle is 1. So,
triangle below, the
= or 2.
Therefore, θ = 60°.
Without using a calculator, determine the value of x.
69. SOLUTION: Because an acute angle, the hypotenuse, and the length of the side adjacent to the angle are given, the cosine
function can be used to find x. Using the properties of
triangles, you can find that cos 30º =
.
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Page 39
4-1 Right Triangle Trigonometry
Therefore, θ = 60°.
Without using a calculator, determine the value of x.
69. SOLUTION: Because an acute angle, the hypotenuse, and the length of the side adjacent to the angle are given, the cosine
function can be used to find x. Using the properties of
triangles, you can find that cos 30º =
.
70. SOLUTION: Because the triangle is a
triangle, the legs are the same length.
71. SCUBA DIVING A scuba diver located 20 feet below the surface of the water spots a shipwreck at a 70º angle
of depression. After descending to a point 45 feet above the ocean floor, the diver sees the shipwreck at a 57º
angle of depression. Draw a diagram to represent the situation, and determine the depth of the shipwreck.
SOLUTION: First, draw a diagram to represent the situation.
eSolutions Manual - Powered by Cognero
Page 40
4-1 Right Triangle Trigonometry
71. SCUBA DIVING A scuba diver located 20 feet below the surface of the water spots a shipwreck at a 70º angle
of depression. After descending to a point 45 feet above the ocean floor, the diver sees the shipwreck at a 57º
angle of depression. Draw a diagram to represent the situation, and determine the depth of the shipwreck.
SOLUTION: First, draw a diagram to represent the situation.
Label the horizontal distance from the shipwreck to the point on the ocean floor below the diver as x. Label the
vertical distance from the point 20 feet below sea level to the point 45 feet below sea level as y. Find the
complementary angle for each angle of depression.
Use the tangent function to write an equation for the smaller right triangle in terms of x.
Use the tangent function to write an equation for the larger right triangle in terms of x.
Set the two equations equal to one another to solve for y.
Therefore, the depth of the shipwreck is 20 + 35 + 45 or about 100 feet.
Find the value of cos θ if θ is the measure of the smallest angle in each type of right triangle.
72. 3-4-5
SOLUTION: Draw a diagram of a 3-4-5 triangle. The smallest angle will be the angle opposite the side with a length of 3.
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Page 41
4-1 Right Triangle Trigonometry
Therefore, the depth of the shipwreck is 20 + 35 + 45 or about 100 feet.
Find the value of cos θ if θ is the measure of the smallest angle in each type of right triangle.
72. 3-4-5
SOLUTION: Draw a diagram of a 3-4-5 triangle. The smallest angle will be the angle opposite the side with a length of 3.
73. 5-12-13
SOLUTION: Draw a diagram of a 5-12-13 triangle. The smallest angle will be the angle opposite the side with a length of 5.
74. SOLAR POWER Find the total area of the panel shown below.
SOLUTION: The area of the panel is given by
, where the width is 10 feet and the length is unknown.
Notice that the length of the panel is also the hypotenuse of a right triangle with an acute angle of 55º and an
opposite side length of 3.5 feet. The sine function can be used to find the hypotenuse of the triangle.
So, the length of the panel is 4.27 feet. Find the area.
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Page 42
4-1 Right Triangle Trigonometry
74. SOLAR POWER Find the total area of the panel shown below.
SOLUTION: The area of the panel is given by
, where the width is 10 feet and the length is unknown.
Notice that the length of the panel is also the hypotenuse of a right triangle with an acute angle of 55º and an
opposite side length of 3.5 feet. The sine function can be used to find the hypotenuse of the triangle.
So, the length of the panel is 4.27 feet. Find the area.
Therefore, the area of the panel is 42.7 square feet.
Without using a calculator, insert the appropriate symbol >, <, or = to complete each equation.
75. sin 45° cot 60°
SOLUTION: Use a graphing calculator to find sin 45º and cot 60º. To find cot 60º, find
.
sin 45º ≈ 0.707
cot 60º ≈ 0.577
Therefore, sin 45º > cot 60º.
76. tan 60° cot 30°
SOLUTION: Use a graphing calculator to find tan 60º and cot 30º. To find cot 30º, find
.
tan 60º ≈ 1.732
cot 30º ≈ 1.732
Therefore, tan 60° = cot 30°.
77. cos 30° csc 45°
SOLUTION: Use a graphing calculator to find cos 30º and csc 45º. To find csc 45º, find
cos 30º ≈ 0.866
csc 45º ≈ 1.414
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Therefore, cos 30° < csc 45°.
78. cos 30° sin 60°
.
Page 43
Use a graphing calculator to find tan 60º and cot 30º. To find cot 30º, find
4-1
.
tan 60º ≈ 1.732
cot 30º Triangle
≈ 1.732
Right
Trigonometry
Therefore, tan 60° = cot 30°.
77. cos 30° csc 45°
SOLUTION: Use a graphing calculator to find cos 30º and csc 45º. To find csc 45º, find
.
cos 30º ≈ 0.866
csc 45º ≈ 1.414
Therefore, cos 30° < csc 45°.
78. cos 30° sin 60°
SOLUTION: Use a graphing calculator to find cos 30º and sin 60º.
cos 30º ≈ 0.866
sin 60º ≈ 0.866
Therefore, cos 30° = sin 60°.
79. sec 45° csc 60°
SOLUTION: sec 45° ○ csc 60°
Use a graphing calculator to find sec 45º and csc 60º. To find sec 45º find
, and to find csc 60º find
.
sec 45º ≈ 1.414
csc 60º ≈ 1.155
Therefore, sec 45º > csc 60°.
80. tan 45° sec 30°
SOLUTION: tan 45° ○ sec 30°
Use a graphing calculator to find tan 45º and sec 30º. To find sec 30º, find
.
tan 45º = 1
sec 30º ≈ 1.155
Therefore, tan 45° < sec 30°.
81. ENGINEERING Determine the depth of the shaft at the large end d of the air duct shown below if the taper of
the duct is 3.5º.
SOLUTION: Draw a diagram of the side-view of the air duct. Label the lengths of the top and bottom of the duct 48 in. because
4 feet = 4 × 12 or 48 inches.
eSolutions Manual - Powered by Cognero
Page 44
Use a graphing calculator to find tan 45º and sec 30º. To find sec 30º, find
4-1
.
tan 45º = 1
sec 30ºTriangle
≈ 1.155
Right
Trigonometry
Therefore, tan 45° < sec 30°.
81. ENGINEERING Determine the depth of the shaft at the large end d of the air duct shown below if the taper of
the duct is 3.5º.
SOLUTION: Draw a diagram of the side-view of the air duct. Label the lengths of the top and bottom of the duct 48 in. because
4 feet = 4 × 12 or 48 inches.
Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find x.
Find the length of the large end of the duct.
Therefore, the length of the large end of the duct is about 17.9 inches.
82. MULTIPLE REPRESENTATIONS In this problem, you will investigate trigonometric functions of acute
angles and their relationship to points on the coordinate plane.
a. GRAPHICAL Let P(x, y) be a point in the first quadrant. Graph the line through point P and the origin. Form a
right triangle by connecting the points P, (x, 0), and the origin. Label the lengths of the legs of the triangle in terms
of x or y. Label the length of the hypotenuse as r and the angle the line makes with the x-axis θ.
b. ANALYTICAL Express the value of r in terms of x and y.
c. ANALYTICAL Express sin θ, cos θ, and tan θ in terms of x, y, and/or r.
d. VERBAL Under what condition can the coordinates of point P be expressed as (cos θ, sin θ)?
e. ANALYTICAL Which trigonometric ratio involving θ corresponds to the slope of the line?
f. ANALYTICAL Find an expression for the slope of the line perpendicular to the line in part a in terms of θ.
SOLUTION: a.
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Page 45
d. VERBAL Under what condition can the coordinates of point P be expressed as (cos θ, sin θ)?
e. ANALYTICAL Which trigonometric ratio involving θ corresponds to the slope of the line?
f. ANALYTICAL Find an expression for the slope of the line perpendicular to the line in part a in terms of θ.
4-1 Right
Triangle Trigonometry
SOLUTION: a.
2
2
2
b. By the Pythagorean Theorem, x + y = r .
So, r =
.
c. Use the definitions of the sine, cosine, and tangent functions and the diagram from part a to find sin θ, cos θ, and
tan θ.
d. Sample answer: Because cos θ =
and sin θ =
, when r = 1, cos θ = or x and sin θ = or y.
Therefore, a point P(x, y) can be written as (cos θ, sin θ) when r = 1.
e. Find the slope of the line using the points (0, 0) and (x, y).
From part c,
. Therefore, tan θ corresponds to the slope of the line.
f. The slope of the line perpendicular to a line with slope it follows that
is the negative reciprocal or
. Because
,
. Therefore, an expression for the slope of the line perpendicular to the line in part a is –cot
θ.
83. PROOF Prove that if θ is an acute angle of a right triangle, then tan θ = and cot θ =
.
SOLUTION: Sample answer: For an acute angle θ of a right triangle, sin θ =
. Using these definitions,
84. ERROR
eSolutions
Manual - ANALYSIS Powered by Cognero
Jason
= = , cos θ =
= tan θ. Similarly,
, tan θ =
=
= , and cot θ =
= cot θ.
and Nadina know the value of sin θ = a and are asked to find csc θ . Jason saysPage
that46
this is not possible, but Nadina disagrees. Is either of them correct? Explain your reasoning.
SOLUTION: . Using these definitions,
= = = tan θ. Similarly,
= =
= cot θ.
4-1 Right Triangle Trigonometry
84. ERROR ANALYSIS Jason and Nadina know the value of sin θ = a and are asked to find csc θ . Jason says that
this is not possible, but Nadina disagrees. Is either of them correct? Explain your reasoning.
SOLUTION: Nadina; the cosecant function is the reciprocal function of the sine function. Therefore, if sin θ = a, then csc θ =
0.
, where a
85. Writing in Math Explain why the six trigonometric functions are transcendental functions.
SOLUTION: The trigonometric functions are transcendental functions because they cannot be expressed in terms of algebraic
operations. For example, there is no way to find the value of θ in y = cos θ by adding, subtracting, multiplying, or
dividing a constant and θ or raising θ to a rational power.
86. CHALLENGE Write an expression in terms of θ for the area of the scalene triangle shown. Explain.
SOLUTION: Sample answer: If you draw the height of the triangle, it forms two right triangles, as shown.
Use the sine function to find h.
Substitute the expression for h into the formula for the area of a triangle, A =
Therefore, an expression in terms of θ for the area of the triangle is A =
bh.
.
87. PROOF Prove that if θ is an acute angle of a right triangle, then (sin θ)2 + (cos θ)2 = 1.
SOLUTION: eSolutions
Manual
- Poweredof
byaCognero
Draw
a diagram
right triangle
with an acute angle θ.
Page 47
4-1 Right
Triangle
Trigonometry
Therefore,
an expression
in terms of θ for the area of the triangle is A =
.
87. PROOF Prove that if θ is an acute angle of a right triangle, then (sin θ)2 + (cos θ)2 = 1.
SOLUTION: Draw a diagram of a right triangle with an acute angle θ.
From the definitions of the sine and cosine functions, sin θ =
2
and cos θ = . By the Pythagorean Theorem, c
2
2
=a +b .
= + = = = 1
2
2
Therefore, (sin θ) + (cos θ) = 1.
REASONING If A and B are the acute angles of a right triangle and m A < m
each statement is true or false . If false, give a counterexample.
88. sin A < sin B
B, determine whether
SOLUTION: Find sin A and sin B for various angle measures of a right triangle, where m A < m B.
sin A
sin B
m A
m B
0.0175
0.9998
0.2588
0.9659
0.6947
0.7193
Sample answer: From the table, it appears that when m A < m B, sin A < sin B. Therefore, the statement is
true.
89. cos A < cos B
SOLUTION: False; sample answer: In ABC, the m B < m
cos B > cos A, and thus the statement is false.
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90. tan A < tan B
A, cos B
0.7986, and cos A
0.6018. Therefore, Page 48
0.6947
4-1
0.7193
Sample answer: From the table, it appears that when m A < m
Right
true. Triangle Trigonometry
B, sin A < sin B. Therefore, the statement is
89. cos A < cos B
SOLUTION: False; sample answer: In ABC, the m B < m
cos B > cos A, and thus the statement is false.
A, cos B
0.7986, and cos A
0.6018. Therefore, 90. tan A < tan B
SOLUTION: Find tan A and tan B for various angle measures of a right triangle, where m A < m B.
tan A
tan B
m A
m B
0.0175
57.29
0.2679
3.732
0.9657
1.036
Sample answer: From the table, it appears that when m A < m B, tan A < tan B. Therefore, the statement is
true.
91. Writing in Math Notice on a graphing calculator that there is no key for finding the secant, cosecant, or
cotangent of an angle measure. Explain why you think this might be so.
SOLUTION: Sample answer: Since cosine is the inverse of secant, sine is the inverse of cosecant, and tangent is the inverse of
cotangent, you can find the value of the secant, cosecant, or cotangent by finding the value of its inverse and using
the reciprocal key on your calculator.
92. ECONOMICS The Consumer Price Index (CPI) measures inflation. It is based on the average prices of goods and services in the United States, with the annual average for the years 1982-1984 set at an index of 100. The
table shown gives some annual average CPI values from 1955 to 2005. Find an exponential model relating this data
(year, CPI) by linearizing the data. Let x = 0 represent 1955. Then use your model to predict the CPI for 2025.
SOLUTION: Linearize the data by finding (x, ln y).
0 by Cognero
10
20
eSolutionsxManual - Powered
ln y
3.29
3.45
3.99
30
4.68
40
5.03
50
5.27
Page 49
4-1 Right
Triangle Trigonometry
SOLUTION: Linearize the data by finding (x, ln y).
x
0
10
20
ln y
3.29
3.45
3.99
Calculate the linear regression.
30
4.68
40
5.03
50
5.27
Please note that the regression was calculated using the actual values of ln y as opposed to the rounded values
shown in the table above.
The rounded regression equation is
Graph the linearized data.
Replace with ln y and solve for y.
To find the CPI for 2025, find y when x = 2025 – 1955 or 70.
According to this model, the CPI in 2025 will be about 523.2.
Solve each equation. Round to the nearest hundredth.
93. e5x = 24
SOLUTION: x–7
94. 2e Manual– -6Powered
= 0 by Cognero
eSolutions
SOLUTION: x–7
Page 50
4-1 Right Triangle Trigonometry
94. 2ex – 7 – 6 = 0
SOLUTION: 2e
x–7
– 6=0
Sketch and analyze the graph of each function. Describe its domain, range, intercepts, asymptotes, end
behavior, and where the function is increasing or decreasing.
95. f (x) = −3x − 2
SOLUTION: Evaluate the function for several x-values in its domain.
x
f (x)
–2
–0.01
–1
–0.04
0
–0.11
1
–0.33
2
–1
3
–3
4
–9
Use a smooth curve to connect each of the ordered pairs.
96. f (x) = 23x − 4 + 1
SOLUTION: Evaluate the function for several x-values in its domain.
eSolutions Manual
x - Poweredfby
(x)Cognero
–2
–1
1.00
1.01
Page 51
4-1 Right Triangle Trigonometry
96. f (x) = 23x − 4 + 1
SOLUTION: Evaluate the function for several x-values in its domain.
x
f (x)
1.00
–2
1.01
–1
0
1.06
1
1.5
2
5
3
33
4
257
Use a smooth curve to connect each of the ordered pairs.
97. f (x) = – 4–x + 6
SOLUTION: Evaluate the function for several x-values in its domain.
x
f (x)
4
–16
5
–4
6
–1
7
–0.25
8
–0.06
9
–0.02
10
–0.004
Use a smooth curve to connect each of the ordered pairs.
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Page 52
4-1 Right Triangle Trigonometry
97. f (x) = – 4–x + 6
SOLUTION: Evaluate the function for several x-values in its domain.
x
f (x)
4
–16
5
–4
6
–1
7
–0.25
8
–0.06
9
–0.02
10
–0.004
Use a smooth curve to connect each of the ordered pairs.
Solve each equation.
98. =
–
SOLUTION: The LCD is (x + 4)(2x – 1).
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99. Page 53
4-1 Right Triangle Trigonometry
Solve each equation.
98. =
–
SOLUTION: The LCD is (x + 4)(2x – 1).
99. + =
SOLUTION: The LCD is (x + 1)(x – 5).
Because the original equation is not defined when x = 5, you can eliminate this extraneous solution. Therefore, the
only solution is 4.
100. =
–
SOLUTION: =
–
The LCD is (3x + 2)(x + 1).
eSolutions Manual - Powered by Cognero
Page 54
BecauseTriangle
the originalTrigonometry
equation is not defined when x = 5, you can eliminate this extraneous solution. Therefore, the
4-1 Right
only solution is 4.
100. =
–
SOLUTION: =
–
The LCD is (3x + 2)(x + 1).
101. NEWSPAPERS The circulation in thousands of papers of a national newspaper is shown.
a. Let x equal the number of years after 2001. Create a scatter plot of the data.
b. Determine a power function to model the data.
c. Use the function to predict the circulation of the newspaper in 2015.
SOLUTION: a. Enter the data into a graphing calculator and create a scatter plot.
b. Use the power regression function on the graphing calculator to find values for a and b.
–0.149
Rounding to the nearest thousandth, a power function that can be used to model the data is f (x) = 904.254x
.
c. Graph the regression equation using a graphing calculator. To predict the circulation of the newspaper in 2015,
use the TRACE function on the graphing calculator. Find the value of y when x = 2015 – 2001 or 14.
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Page 55
4-1 Right Triangle Trigonometry
101. NEWSPAPERS The circulation in thousands of papers of a national newspaper is shown.
a. Let x equal the number of years after 2001. Create a scatter plot of the data.
b. Determine a power function to model the data.
c. Use the function to predict the circulation of the newspaper in 2015.
SOLUTION: a. Enter the data into a graphing calculator and create a scatter plot.
b. Use the power regression function on the graphing calculator to find values for a and b.
–0.149
Rounding to the nearest thousandth, a power function that can be used to model the data is f (x) = 904.254x
.
c. Graph the regression equation using a graphing calculator. To predict the circulation of the newspaper in 2015,
use the TRACE function on the graphing calculator. Find the value of y when x = 2015 – 2001 or 14.
So, the circulation in 2015 will be about 611,068.
102. SAT/ACT In the figure below, what is the value of z?
A 15
B 15
C 15
D Manual
30
eSolutions
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E 30
Page 56
4-1 Right Triangle Trigonometry
So, the circulation in 2015 will be about 611,068.
102. SAT/ACT In the figure below, what is the value of z?
A 15
B 15
C 15
D 30
E 30
SOLUTION: First, find x.
Notice that this is a 45º-45º-90º triangle. Therefore, the legs of the triangle are the same length. Solve for y.
So, the length of each side is 3(5) or 15. Use the Pythagorean Theorem to find z.
Therefore, the correct answer is B.
103. REVIEW Joseph uses a ladder to reach a window 10 feet above the ground. If the ladder is 3 feet away from the
wall, how long should the ladder be?
F 9.39 ft
G 10.44 ft
H 11.23 ft
J 12.05 ft
SOLUTION: Draw a diagram where x represents the length of the ladder.
Use the Pythagorean Theorem to find x.
Therefore, the correct answer is G.
104. A person holds one end of a rope that runs through a pulley and has a weight attached to the other end. Assume
eSolutions
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thatManual
the weight
is by
at Cognero
the same
weight?
height as the person’s hand. What is the distance from the person's hand to the Page 57
Use the Pythagorean Theorem to find x.
4-1 Right Triangle Trigonometry
Therefore, the correct answer is G.
104. A person holds one end of a rope that runs through a pulley and has a weight attached to the other end. Assume
that the weight is at the same height as the person’s hand. What is the distance from the person's hand to the
weight?
A 7.8 ft
B 10.5 ft
C 12.9 ft
D 14.3 ft
SOLUTION: Draw a diagram where x represents the distance from the person's hand to the weight.
Because an acute angle and the length of the side adjacent to the angle are given, you can use the tangent function
to find x.
The distance from the person's hand to the weight is about 7.8 feet. Therefore, the correct answer is A.
105. REVIEW A kite is being flown at a 45° angle. The string of the kite is 120 feet long. How high is the kite above
the point at which the string is held?
F 60 ft
G 60
ft
H 60
ft
J 120 ft
SOLUTION: Draw a diagram where x represents the distance from where the string is being held to the kite.
Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find x.
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Page 58
4-1 Right Triangle Trigonometry
The distance from the person's hand to the weight is about 7.8 feet. Therefore, the correct answer is A.
105. REVIEW A kite is being flown at a 45° angle. The string of the kite is 120 feet long. How high is the kite above
the point at which the string is held?
F 60 ft
G 60
ft
H 60
ft
J 120 ft
SOLUTION: Draw a diagram where x represents the distance from where the string is being held to the kite.
Because an acute angle and the length of the hypotenuse are given, you can use the sine function to find x.
The kite is 60
ft above the point where the string is being held. Therefore, the correct answer is G.
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Page 59