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4.3 Right Triangle Trigonometry Day 2 Objective: Use right triangle trigonometry to solve real life problems.

# 4.3 Right Triangle Trigonometry Day 2 Objective: Use right triangle trigonometry to solve real life problems

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4.3 Right Triangle Trigonometry Day

2

Objective: Use right triangle trigonometry

to solve real life problems.

Warm-up

15Find the exact value of sec

4

Quiz Wed. on sections 4.1-4.4

There is another way to state the size of an angle, one that subdivides a degree into smaller pieces.

In a full circle there are 360 degrees. Each degree can be divided into 60 parts, each part being 1/60 of a degree. These parts are called minutes.

Each minute can divided into 60 parts, each part being 1/60 of a minute. These parts are called seconds.

Degrees, Minutes, and Seconds

Degrees, Minutes, and Seconds Conversions

•To convert decimal degrees into DMS, multiply decimal degrees by 60

•To convert from DMS to decimal degrees, divide minutes by 60, seconds by 3600

•OR use the Angle feature of your calculator

•Examples

Express 52 28 22 in decimal degrees

Express 152.65 in degrees, minutes, and seconds

152 39

52.47277

Vocabulary

• Angle of Depression and Angle of Elevation are Equal Angle of Depression

Angle of Elevation

Horizontal Line

Horizontal Line

Example: An airplane pilot sights a life raft at a 26o angle of depression. The airplane’s altitude is 3 km. What is the airplane’s horizontal distance d from the raft?

26

26

3 km

d

3Tan 26

d

3

Tan 26d

km 2.6d

Example: A six-foot person walks from the base of a streetlight directly toward the tip of the shadow cast by the streetlight. When the person is 20 feet from the streetlight and 8 feet from the tip of the streetlight’s shadow, the person’s shadow starts to appear beyond the streetlight’s shadow.

What is the height of the streetlight?

rights reserved.

12

21 .ft

Example: A helicopter is hovering 560 feet above a straight and level road that runs east and west. On the road east of the helicopter are a car and a truck. From the helicopter the angle of depression to the car is 48º and the angle of depression to the truck is 38º. Find the distance between the car and the truck.

Not drawn to scale.Question developed by V. Borlaug WSCC, 2008

560

H

C T48 38X

YTan 48 = 560/x

X = 504.2

Tan 38 = 560/y Y = 716.8

Distance = 716.8 – 504.2 = 212.6

Example: A helicopter takes off from ground level and goes 853 feet with an angle of elevation of 23º. The helicopter then changes course and goes 719 feet with and angle of elevation of 49º and then it hovers in this position. Find the height of the hovering helicopter.

23°

49°853

719

X

Y

Sin 23 = X/853 X = 333.3

Sin 49 = Y/719 Y = 542.6

333.3 + 542.6 = 875.9

Example: A short building is 200 feet away from a taller building. Jessie is on the roof of the short building. To see the top of the taller building requires Jessie to look up with a 38 angle of elevation. To see the bottom of the taller building requires Jessie to look down with 12 angle of depression. Find the height of the taller building.(You may assume that Jessie’s height is negligible.)

(not dawn to scale)

38°12°

200X

Y

Tan 38 = X/200

X = 156.3

Tan 12 = Y/200 Y = 42.5

156.3 + 42.5 = 198.8

Example: In traveling across flat land you notice a mountain directly in front of you. Its angle of elevation to the peak is 3.5 degrees. After you drive 13 miles closer to the mountain, the angle of elevation is 9 degrees.Approximate the height of the mountain.

Homework

• Right Triangle Applications Worksheet

Example 4.) Derek is scuba diving in still water. Starting from the surface he dives in a straight line with a 65º angle of depression. Derek is traveling at a constant rate of 25 feet per minute along this straight line path. a.) Find the distance Derek has traveled along this straight line path three minutes into the dive.b.) Find Derek’s vertical depth three minutes into the dive.c.) Find the rate at which Derek’s vertical depth is changing as he descends. Question developed by V. Borlaug

WSCC, 2008

65°

75X

4) a) 3 x 25 = 75

b) Sin 65 = XX = 68.0

c) 68.0 / 3 = 22.7 ft per min

75

Example 5.) A river runs between a tree and a lamp post. The tree is directly north of the lamp post. The surveyor is 324 feet east of the lamp post. He measures an angle of 11.3º from the tree to the lamp post. Find the distance from the surveyor to the tree.

Not drawn to scale.Question developed by V. Borlaug

WSCC, 2008

S L

TX

11.3°

324

5) Cos 11.3 = 324 X X = 330.4

Example 3.) A crane has a 100 foot arm with a hook at the end of the arm. This crane is designed so that the angle of elevation of the arm can be changed. When the crane’s hook is attached to an object on the ground, the arm’s angle of elevation is 5º. The crane’s arm then rotates upward and raises the object to a position that gives the arm a 25º angle of elevation. Find the height the crane has lifted the object.HINT: Find the tip of the arm’s initial distance from horizontal. Then find the distance from horizontal after the object has been lifted. Use these to find the height the object has been lifted.

Not drawn to scale.

Question developed by V. Borlaug WSCC, 2008

5°100 X

3) Sin 5 ° = X/100 X = 8.7

Sin 25 = Y/100

Y = 42.3

Height = 42.3 – 8.7

Height = 33.6 Feet

25°

100 Y

Example 6.) Jessica is standing 75 feet from the base of a vertical tree. She is 5 ½ feet tall and her eyes are 4 inches from the top of her head. It takes a 38º angle of elevation for Jessica to look at the top of the tree. (Use four decimal place accuracy.)

a.) Find the height of the tree in feet.b.) Find the height of the tree in meters.

Not drawn to scale.Question developed by V. Borlaug

WSCC, 2008

NOTE: One meter is the length equal to 1,650,763.73 wavelengths in a vacuum of the orange-red radiation of krypton 86. One meter also equals 39.37 inches. Ref: The American Heritage Dictionary of the English Language, American Publishing Co., Inc., 1969

38°X

75 5 ft 2 inches

6) Cos 38 = 75/x

X = 95.1764

5’ 2” = 5.1667ft

Height =95.1764 + 5.1667 =100.3431 feet or100.3431 / 39.37 =2.5488 meters

Example 9.) A straight 67 inch ramp is being designed for a skate board course.a.) Find the vertical height required to give the skate board ramp a 75º angle of depression.b.) Find the vertical height required to give the skate board ramp a 65º angle of depression.c.) Find the vertical height required to give the skate board ramp a 55º angle of depression.

Question developed by V. Borlaug WSCC, 2008

75°

67 X

9)

Sin 65 = X/67

X = 60.7

Sin 55 = X/67

X = 54.9

Sin 75 = X 67 X = 64.7

Example 12: Solar panels are being installed on a roof. The roof has an angle of elevation of 9.2º. Each solar panel is 7.20 feet long and 4.80 feet wide. The longer edge of each panel will be horizontally installed on the roof. The shorter edge of the panel will be installed with an angle of elevation 29.6º. a.) Find the vertical height of each solar panel relative to horizontal.b.) Find the vertical height of each solar panel relative to the roof directly the solar panel’s highest edge.

Hint for part “b”: Using the two angles of elevation there here are two right triangles. First solve the larger right triangle completely.

9.2°29.6°

4.80 X

YZ

12) Sin 29.6 = X 4.8 X = 2.4

cos 29.6 = Y/4.8

Y = 4.17

tan 9.2 = Z/4.17

Y = 0.7

2.4 – 0.7 = 1.7

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