T46 – APPLICATIONS OF TRIGONOMETRY 1rashby-rem.weebly.com/uploads/1/9/2/6/19262051/fom...2: Choose the appropriate trigonometry rules/laws and solve the problem. II) SAMPLE PROBLEM

FOM 11 T46 – APPLICATIONS OF TRIGONOMETRY 1 1

© R. Ashby & R. Obayashi 2018. Duplication by permission only.

MATH SPEAK - TO BE UNDERSTOOD AND MEMORIZED 1) HORIZONTAL LINE OF SIGHT = an imaginary horizontal line form a person’s eyes looking straight ahead. 2) ANGLE OF ELEVATION = the angle formed between a person’s horizontal line of sight and the imaginary

line looking up at an object.

3) ANGLE OF DEPRESSION = the angle formed between a person’s horizontal line of sight and the imaginary line looking down at an object.

SOLVING PROBLEMS USING TRIGONOMETRY I) USE THESE STEPS TO SOLVE PROBLEMS USING TRIGONOMETRY

1: DRAW A DIAGRAM WITH A TRIANGLE!! The triangle does not have to be perfectly accurate. Its purpose is to organize the information to allow you to decide to use soh-cah-toa, the Pythagorean theorem, the cosine law or the sine law.

2: Choose the appropriate trigonometry rules/laws and solve the problem.

II) SAMPLE PROBLEM 1: Study these examples carefully. Be sure you understand and memorize the process used to complete them. 1. Each school day Helen walks a triangular route from home to school, from school to the corner store, then from

the corner store to home. She walks 350 m from school to the local corner store. While at the corner store she looks out the front door and estimates that the angle formed between the school and her home is 100º. She then walks 500 m home. a) What is the distance, to the nearest metre, she walked between her home and school in the morning? b) What is the total distance, to the nearest metre, she walked that day? 1: DRAW A DIAGRAM WITH A TRIANGLE!! The triangle does not have to be perfectly accurate. Its purpose is to

organize the information to allow you to decide to use soh-cah-toa, the Pythagorean theorem, the cosine law or the sine law.

C = Corner store H = Home S = School

∠C is the corner where the Corner store is. Helen is standing at the door of the Corner store looking between the School and her Home. This is the angle formed between the School and her Home thus, ∠C =100°

2: Choose the appropriate trigonometry rules/laws and solve the problem. Because we know SAS, use this cosine law pattern.

c2 = h2 + s2−2hscosC

c2 = 350( )2+ 500( )2

−2 350( ) 500( )cos 100!( )

c2 " 372500−−60776.86218

c2 " 372500 +60776.86218

c2 " 433276.8622

c2 " 433276.8622

c " 658.2376943 " 658 m

Continued on the next page.

C

S H

h = 350 m s = 500 m

c

100°

Horizontal

Object

Angle of Elevation

Horizontal

Object

Angle of Depression



a) What distance did she between her home and school did she walk in the morning?

Helen walked about 658 m between school and home.

b) What is the total distance she walked from home to school?

658+ 350+500 =1508

Helen walked a total of about 1508 m.

2. Mark and Rachel are 55 m apart. They are looking up at a UFO shaped cloud that is east of them. Rachel estimates that from her position her direct line of sight to the cloud forms an angle of elevation of 65º while Mark estimates that form his position the angle of elevation is 80º. What is the altitude of the cloud to the nearest tenth of a metre? 1: DRAW A DIAGRAM WITH A TRIANGLE!! The triangle does not have to be perfectly accurate. Its purpose is to


This question asks you to determine the altitude of a cloud in the sky. This means we draw a right triangle because the altitude is the vertical distance between the object and the ground directly beneath. Thus, the side of the triangle that represents the altitude of the UFO forms a 90º angle with the ground.

There are two people observing the cloud in the sky. They are both looking in the same direction (east). Thus, there are two direct lines of sight, one for each person. One person is closer to the cloud than the other. The closer we are to an object that we are looking up at, the greater the angle of elevation. The farther we are from an object that we are looking up at, the smaller the angle of elevation. Since the angle of elevation at Rachel’s position (R) is less than that of Mark (M), ∠R = 65° and ∠M = 80° , Rachel is farther from the cloud than is Mark.

R = Rachel’s position M = Mark’s position U = UFO G = Ground position beneath UFO d = is the distance between Rachel and Mark x = direct line of sight between Rachel and the UFO y = direct line of sight between Mark and the UFO z = distance between Mark and the G (point beneath

the UFO

U

G M

a = altitude

x

d = 55m 65° R

80° z

y



2: Choose the appropriate trigonometry rules/laws and solve the problem. Calculate the measures of all unknown angles as this will help you determine which trigonometry rules/laws to use. The calculated angles are highlighted in bold.

In order to determine the altitude (a) of the UFO, we must determine the length of side x or side y (They are part of ∆RMU). As a result we will draw this triangle to isolate it and then determine which trigonometry rules/laws to use.

We know ASA, so we can use the sine law. There are two equally valid procedures that can be used to solve this problem. As a result, both will be shown side-by-side.

Solving for side x OR Solving for side y

xsin M = d

sin∠RUM

xsin 100!( ) = 55

sin 15!( )

sin 100 !( ) xsin 100!( )( ) = 55

sin 15!( )( )sin 100!( )

x = 55 • sin 100!( )sin 15!( )

x = 54.16442642sin 15!( )

x " 209.2752734 m

ysin R = d

sin∠RUM

ysin 65!( ) = 55

sin 15!( )

sin 65 !( ) ysin 65!( )( ) = 55

sin 15!( )( )sin 65 !( )

y = 55 • sin 65!( )sin 15!( )

y = 49.84692829sin 15!( )

y " 192.5937416 m

DO NOT ROUND THE SOLUTIONS UNTIL YOU ARE READY TO RECORD THE ANSWER TO THE PROBLEM!!!


U

G M

a = altitude x

d = 55m 65°

R 80°

z 100°

15° 10°

y

U

M R

y x

d = 55 m 65° 100°

15°



Now we know enough information in the two right triangles ∆RUG and of ∆MUG to be able to use soh-cah-toa to find the altitude (a).

Solving for the altitude (a) using side x OR Solving for the altitude (a) using side y

sin R = ax

sin65!= a209.2752734

209.2752734 sin65!( ) = a209.2752734( ) 209.2752734

a = 209.2752734 sin65!( )a " 189.6678099

sin M = ay

sin80!= a192.5937416

192.5937416 sin80!( ) = a192.5937416( )192.5937416

a = 192.5937416 sin80!( )a " 189.6678099

NOTICE that using either side x or side y the altitude (a) is the same distance of approximately 189.7 m

The altitude of the UFO shaped cloud is approximately 189.7 m.

3. An ocean going research ship is directly above and parallel to the wreck of a World War II aircraft carrier. The sonar operator measures aircraft carrier to be about 300 m long. The sonar operator measures the angle of depression to the front end of the aircraft carrier as 85º and the angle of depression to the rear end of the aircraft carrier as 78º. At what depth is the aircraft carrier to the nearest tenth of a metre? 1: DRAW A DIAGRAM WITH A TRIANGLE!! The triangle does not have to be perfectly accurate. Its purpose is to


S = research Ship’s position F = Front of aircraft carrier R = Rear of aircraft carrier l = length of aircraft carrier d = depth of the aircraft carrier A = point on aircraft carrier r = distance from research ship to front of the aircraft carrier f = distance from research ship to rear of the aircraft carrier

2: Choose the appropriate trigonometry rules/laws and solve the problem. The depth (d) is a leg in the two right triangles ∆FAS and ∆RAS. This means you must first determine the hypotenuse (r or f) of one of the right triangles in order to be able to use soh-cah-toa.

Solving for distance (r) using ∆FRS OR Solving for distance (f) using ∆FRS. The surface of the water is parallel to the aircraft carrier. This means that ∠F = 85° and ∠R = 78° due to alternate interior angles.

Continued on the next page

U

G

a = altitude

x ≐ 209.2752734 m

65° R

U

G

a = altitude y ≐ 192.5937416 m

80° M

r f

S

R F l = 300 m

78° 85° d

A

This line represents the surface of the water.

S

R F l = 300 m

78° 85°

85° 78° 85° 78°

S

R F l = 300 m

78° 85° r

f



Draw the only triangle ∆FRS. Then determine the measure of the ∠S.

Use the sine law to determine side length (r). OR Use the sine law to determine side length (f).

rsin R = l

sinS

rsin 78!( ) = 300

sin 17!( )

sin 78 !( ) rsin 78!( )( ) = 300

sin 17!( )( )sin 78!( )

r = 300 • sin 78!( )sin 17!( )

r = 293.4442802sin 17!( )

r " 1003.668534 m

fsin F = l

sinS

fsin 85!( ) = 300

sin 17!( )

sin 85 !( ) fsin 85!( )( ) = 300

sin 17!( )( )sin 85!( )

f = 300 • sin 85!( )sin 17!( )

f = 298.8584094sin 17!( )

f " 1022.1865 m

Draw the only obtuse triangle ∆FRS. Then determine the measure of the ∠S.

Side length (r) is a leg in right triangle ∆FAS OR Side length (f) is a leg in right triangle ∆RAS.

Depth (d) is a leg in the two right triangles ∆FAS and ∆RAS. This means you can use soh-cah-toa.

sinF = df

sin85!= d1003.668534

1003.668534 sin85!( ) = d1003.668534( )1003.668534

d = 1003.668534 sin85!( )d " 999.8492722

sin R = dr

sin78!= d1022.1865

1022.1865 sin78!( ) = d1022.1865( )1022.18653

d = 1022.1865 sin78!( )d " 999.8492725

NOTICE that using either side r or side y the depth (d) is the same distance of approximately 999.8 m

The depth of the aircraft carrier is approximately 999.8 m.

85° 78° 85° 78° r f

S

R F l = 300 m

17°

S

R F l = 300 m

17° r ≐

1003

.6685

34 m

S

R F l = 300 m

17°

85° 78°

f ≐ 1022.1865 m

S

R F l = 300 m

17°

78° 85°

S

A

d r ≐

1003.668534 m

85° F

S

R

d

f ≐ 1022.1865 m 78° A



4. Read the problem and answer the question (red sentence to the nearest degree) found in LEARN ABOUT the Math on page 140 of your text. 1: DRAW A DIAGRAM WITH A TRIANGLE!! The triangle does not have to be perfectly accurate. Its purpose is to

organize the information to allow you to decide to use soh-cah-toa, the Pythagorean theorem, the cosine law or the sine law. The problem describes a room in a museum so we begin by drawing a rectangle to represent the room. The walls of the room are 12 m apart which means the top and bottom sides of the rectangle are 12 m long.

There are two cameras directly across from each other located 6 m above the floor. Assuming the cameras are at the two top corners of the room (rectangle), the sides of the rectangle are 6 m long. Also, the cameras are pointing to a 1.5 m tall display case. The top-center of the display case is 4.8 m away from the left-hand camera. The display case is closer to the left-hand camera because 4.8 m is less than half of the 12 m width of the room.

L = Left hand camera

R = Right hand camera

D = top-centre of Display case

Draw a line from the Right hand camera (R) to the top of the Display case (D).

L = Left hand camera

R = Right hand camera


The question asks us to determine the angle of depression for each camera. These angles are

∝ = Left camera angle of depression

ß = Right camera angle of depression



12 m

12 m

6 m 6 m

L R

4.8 m

1.5 m D

12 m

6 m 6 m

L R

4.8 m 1.5 m D

12 m

6 m 6 m

L R

4.8 m

1.5 m D

∝ ß



2: Choose the appropriate trigonometry rules/laws and solve the problem. Draw ∆LDR to determine if the problem can be solved using soh-cah-toa, the Pythagorean theorem, the must be solved using the cosine law or the sine law.

NOTICE we are not told that ∆LDR is a right triangle so we cannot use soh-cah-toa or Pythagorean theorem. It is an obtuse triangle so we will try to use the cosine law or the sine law. We don’t have enough information to use cosine law or the sine law. This means we must make at least one right triangle in order to be able to use soh-cah-toa or Pythagorean theorem in order to determine one of the side lengths or angles of ∆LDR. To create the right triangle, we draw a straight line up from D, the top-centre of the display case to the top of the rectangle (T) creating a 90º angle at the point labeled T. This line (DT) is 4.5 m long (wall height – Display case height = 6 m – 1.5 m = 4.5 m)

NOTICE that we now have two right triangles: ∆LDT and ∆RDT. Draw ∆LDT so we can determine which of soh-cah-toa to use.

In ∆LDT we want to determine the measure of ∝ and we know the length of the hypotenuse (RD) and the side-length opposite ∝ (TD). This means we can use sine (soh).

sinα = TDRD

sinα = 4.54.8

sinα = 0.9375

α = sin−1 0.9375( )

α ! 69.63586519"

Thus, ∝ ≐ 70º. The angle of depression of the left hand camera is about 70º. Continued on the next page.

12 m

4.8 m

L R

D

∝

ß

4.5 m 4.8 m

R

D

∝

T

12 m

4.8 m

L R

D

∝

ß

T

4.5 m



Now we know that ∝ = 69.63586519º. We can draw ∆LDR including all known side lengths and angle measures to determine what calculations must be done in order to determine the measure of ß.

We know SAS so we can use the cosine law to determine the side length DR (l).

l2 = 4.8( )2+ 12( )2

−2 4.8( ) 12( )cos 69.63586519!( )

l2 " 167.04−40.08790342

l2 " 126.95209066

l2 " 126.95209066

l " 11.2673021 m

Now we can use the sine law to determine ß.

sinβ4.8 !

sin 69.63586519"( )11.2673021

4.8 sinβ4.8( ) = sin 69.63586519"( )

11.2673021( )4.8

sinβ = 4.8 • sin 69.63586519"( )11.2673021 ↗

sinβ = 4.511.2673021

sinβ ! 0.399385758

β ! sin−1 0.399385758( )

β ! 23.5397849"

Thus, ß ≐ 24º. The angle of depression of the right hand camera is about 24º.

ANS: The angle of depression for the left hand camera is about 70º and that of the right hand camera is about 24º.

III) REQUIRED PRACTICE 1: Page 125: Questions 4 & 5. Page 138: Questions 8, 9 & 12. Page 147 - 149: Questions 4, 5, 6, 7, 8, 9 & 11(a). {Page 125 answers are on page 540; Page 138 and Page 147 - 149 answers are on page 541 of the text.}

12 m

4.8 m

L R

D

69.63586519º

ß

l

12 m

4.8 m

L R

D

69.63586519º

ß

l ≐ 11.2673021 m



ASSIGNMENT: PRINT THIS INFORMATION ON YOUR OWN GRID PAPER

LAST then FIRST Name T46 – Applications of Trigonometry 1 Block:

Show the process required to complete each problem to avoid receiving a zero grade. Neatness Counts!!! (Marks indicated in italicized brackets.)

REMEMBER TO USE GRID PAPER FOR ALL ASSIGNMENTS!!!

∞ For each of these questions be sure to draw a diagram and record answers to nearest degree, metre or kilometre.

1) Max and Jenny are 600 m apart. They see a UFO is hovering above and between them. Max estimates the angle of elevation of the UFO to be 75°, while Jenny estimates its of elevation to be 70°. Determine the altitude of the UFO. (10)

2) An Architect is looking at a hawk perched at the top of a building she designed. She measured the angle of elevation to the hawk to be 53°. The then walked 75 m directly away from the building and measured the angle of elevation to be 48°. What is the height of the building? (11)

3) Alex is standing on his deck. His horizontal line of sight is 10 m above the ground. The angle of elevation to the top of a tree is 11° and the angle of depression to the base of the tree is 23°. Determine the height of the tree to the nearest tenth of a metre. (8)

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T46 – APPLICATIONS OF TRIGONOMETRY 1rashby-rem.weebly.com/uploads/1/9/2/6/19262051/fom...2: Choose the appropriate trigonometry rules/laws and solve the problem. II) SAMPLE PROBLEM