Module - Solution of Right Triangles with Practical...

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LAST REVISED December, 2008 Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved. Trigonometry Module T09 Solutions of Right Triangles with Practical Applications

Transcript of Module - Solution of Right Triangles with Practical...

LAST REVISED December, 2008

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

Trigonometry Module T09

Solutions of Right Triangles with Practical Applications

Module T09 − Solutions of Right Triangles

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Solution of Right Triangles with Practical Applications Statement of Prerequisite Skills Complete all previous TLM modules before completing this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material? Right triangles are often used to measure distances that cannot be measured directly. Applications in surveying, projectiles, navigation, building construction, and many other technologies exist. The skills the student learns in this module will be applied both in the technology of their choice and in the trigonometry modules to follow.

Learning Outcome When you complete this module you will be able to… Solve right triangles.

Learning Objectives 1. Solve a right triangle given an acute angle and a side which is not the hypotenuse. 2. Solve a right triangle given an acute angle and the hypotenuse. 3. Solve a right triangle given the hypotenuse and one other side. 4. Solve a right triangle given two sides neither of which is the hypotenuse. 5. Solve applied right triangle problems. 6. Solve right triangle problems dealing with vectors and navigation.

Module T09 − Solutions of Right Triangles

Connection Activity Can you think of any examples of right triangles occurring in the world around you? The angle a building makes with the ground, the corner of a room, and the intersection of two streets are some examples. There are many instances that right triangles can be imposed on a situation to provide an easy way to measure distances. Consider the diagram below. How could this triangle be used to determine the width of the creek from B to C? A

B C

Creek

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Module T09 − Solutions of Right Triangles

OBJECTIVE ONE When you complete this objective you will be able to… Solve a right triangle given an acute angle and a side which is not the hypotenuse.

Exploration Activity SOLVE A RIGHT TRIANGLE In general, when you are asked to solve a triangle, this means to find all unknown sides and angles. The following review will be of assistance: 1. The two acute angles in a right triangle are complements of each other. That is, the

sum of the two acute angles is 90º. 2. The hypotenuse is greater than either of the other two sides. 3. The greater angle is opposite the greater side. 4. A side is opposite an angle if that side is not an arm of the angle.

EXAMPLE 1 Given ΔACB, side b = 30.0, ∠A = 25.0º and a right angle at C. Solve ΔACB. So we are to find all remaining sides and angles, which are: a, c, ∠B

A C

B

a c

b

Solve for ∠B since:

∠A + ∠B = 90.0º and ∠A = 25º therefore:

25.0º + ∠B = 90.0º solving:

∠B = 65.0º

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Module T09 − Solutions of Right Triangles

Solve for side a Here we must find a trig function that involves side "a" and a known side and a known angle.

sin A = ac

doesn't work since we don't know sides a and c.

cos A = bc

doesn't have an “a” in it.

tan A = ab

should work

tan 25.0º = 30.0

a

solving:

a = (30.0) (tan 25.0º) a = 13.9892

Note: Where possible, always use given values to find missing values.

That is, do not use tan B = ab because ∠B was a calculated angle.

Solve for side c By examining the functions, we decide to use the cosine,

cos A = bc

cos 25.0º = 30.0c

solving:

c = 30.0cos25.0°

c = 33.1013

Always check your solutions. 1. Use the Pythagorean Theorem to check the sides

(13.9892)2 + (30.0)2 = (33.1013)2 2. The sum of the three angles of a triangle is 180º

25º + 65º + 90º = 180º

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Module T09 − Solutions of Right Triangles

EXAMPLE 2 Solve the given triangle ΔPRQ. Given ∠Q = 48.1º and q = 1290 We have to find ∠P, r, and p.

P R

Q

r

q = 1290

p48.1º

Solve for ∠P

∠P + ∠Q = 90º ∠P = 90.0º – ∠Q ∠P = 90.0º – 48.1º ∠P = 41.9º

Solve for side r

sin Q = qr

sin 48.1º = 1290r

r = 1290sin 48.1°

r = 1733.1452 r = 1733

Solve for side p

tan Q = qp

tan 48.1º = 1290p

p = 1290tan 48.1°

Note: we could have used cosine of Q to solve for p which is cos Q = r

p.

However, r is a calculated value and we should always use given values where possible. By using the tangent function, we are using only given values.

p = 1157.4508 p = 1157

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Module T09 − Solutions of Right Triangles 6

Check: Sides Since the answers calculated for sides r and p were rounded to 4 significant digits, the check of the solution for sides will reflect this approximation.

r2 = p2 + q2

(1733)2 = (1157)2 + (1290)2

3003289 = 1338649 + 1664100

3003289 = 3002749 (approximate ) This approximation can be improved by storing the original calculator values when calculating r and p and applying then Pythagorean theorem. Angles

P + Q + R = 180º

41.9º + 48.1º + 90º = 180º

180º = 180º

Module T09 − Solutions of Right Triangles

Experiential Activity One Solve the following right triangles for the unknown elements. Check each solution. Each is labeled as an ABC triangle with the right angle at ∠C. Round all sides to 3 significant digits. Round all angles to one decimal place. 1. a = 11.5 ∠B = 27.1º 2. a = 15.0 ∠B = 74.9º 3. b = 424 ∠A = 74.9º Show Me. 4. b = 8.4 ∠B = 77.6º 5. a = 11.2 ∠A = 82.9º

A

B C

c

a

b

Experiential Activity One Answers 1. ∠A = 62.9º b = 5.88 c = 12.9 2. ∠A = 15.1º b = 55.6 c = 57.6 3. ∠B = 15.1º a = 1570 c = 1630 4. ∠A = 12.4º a = 1.85 c = 8.60 5. ∠B = 7.1º b = 1.40 c = 11.3

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Module T09 − Solutions of Right Triangles

OBJECTIVE TWO When you complete this objective you will be able to… Solve a right triangle given an acute angle and the hypotenuse.

Exploration Activity

EXAMPLE 1 In ΔABC, c = 45.3, and ∠A = 20.3º. Find b, a, and ∠B.

Solve for ∠B ∠B = 90.0º – A ∠B = 90.0º – 20.3º ∠B = 69.7º

Solve for side b

cos A = bc

b = cosc A⋅ b = 45.3 cos20.3⋅ ° b = 42.49

Solve for side a

sin A = ac

a = sinc A⋅a = 45.3 sin 20.3⋅ °a = 15.72

Check: Sides (15.72)2 + (42.49)2 = (45.3)2 2053 = 2053

Angles A + B + C = 180º 20.3º + 69.7º + 90º = 180º 180º = 180º

C A

B

ac = 45.3

b 20.3º

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Module T09 − Solutions of Right Triangles

Experiential Activity Two Solve the following right angle triangles. Check your solutions. Solve all for all unknowns in right triangle ABC where angle C is 90º.

Round all sides to 3 significant digits. Round all angles to one decimal place.

1. c = 14.4 ∠A = 30.3º 2. c = 14.6 ∠B = 36.1º 3. c = 3.30 ∠B = 38.7º 4. c = 7.30 ∠A = 13.8º 5. c = 22.2 ∠B = 58.1º Show Me.

A

B C

c

a

b

Experiential Activity Two Answers 1. ∠B = 59.7º a = 7.27 b = 12.4 2. ∠A = 53.9º a = 11.8 b = 8.60 3. ∠A = 51.3º a = 2.58 b = 2.06 4. ∠B = 76.2º a = 1.74 b = 7.09 5. ∠A = 31.9º b = 18.8 a = 11.7

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Module T09 − Solutions of Right Triangles

OBJECTIVE THREE When you complete this objective you will be able to… Solve a right triangle given the hypotenuse and one other side.

Exploration Activity

EXAMPLE Given ΔABC where c = 38.3 and b = 23.1. Find a, ∠A, and ∠B.

Solve for side a a2 + b2 = c2 a2 = c2 – b2 a = 2 2c b−

a = 2 2(38.3) (23.1)− a = 30.55

Solve for ∠A

cos A = bc

cos A = 23.138.3

A = 1 23.1cos38.3

− ⎛ ⎞⎜ ⎟⎝ ⎠

A = 52.91º

Solve for ∠B.

sin B = bc

sin B = 23.138.3

B = 1 23.1sin38.3

− ⎛ ⎞⎜ ⎟⎝ ⎠

B = 37.09º

Check: Angle B = 90.0º – A B = 90.0º –52.91º B = 37.09º Since we solved for B another way, and produced the same result, it should be correct.

Sides Since we now have checked our angles and know they are correct, we will use them to solve for “a” again.

cos B = 38.30

a

a = 38.3 cos B a = 38.3 cos 37.09º a = 30.55

A

B C

c

a

b

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Module T09 − Solutions of Right Triangles

Experiential Activity Three Solve all for all unknowns in right triangle ABC where angle ∠C is 90º. Round all sides to 3 significant digits. Round all angles to one decimal place. 1. c = 12.8 b = 7.80 2. c = 11.5 a = 6.50 3. c = 13.9 a = 8.90 4. c = 14.2 a = 9.20 5. c = 10.2 b = 5.20 Show Me

A

B C

c

a

b

Experiential Activity Three Answers 1. a = 10.1 ∠B = 37.5º ∠A = 52.5º 2. b = 9.49 ∠A = 34.4º ∠B = 55.6º 3. b = 10.7 ∠B = 50.2º ∠A = 39.8º 4. b = 10.8 ∠A = 40.4º ∠B = 49.6º 5. a = 8.77 ∠B = 30.7º ∠A = 59.3º

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Module T09 − Solutions of Right Triangles

OBJECTIVE FOUR When you complete this objective you will be able to… Solve a right triangle given two sides neither of which is the hypotenuse.

Exploration Activity

EXAMPLE Given ΔABC where a = 37.4 and b = 76.0. Find c, ∠A, and ∠B

Solve for side c a2 + b2 = c2

c = 2 2a b+

c = 2 2(37.4) (76.0)+ c = 84.70

Solve for ∠A

tan A = ab

tan A = 37.476.0

A = tan−1 37.476.0

⎛ ⎞⎜ ⎟⎝ ⎠

A = 26.2º

Solve for ∠B

tan B = ba

tan B = 76.037.4

⎛ ⎞⎜ ⎟⎝ ⎠

B = tan−1 76.037.4

⎛ ⎞⎜ ⎟⎝ ⎠

B = 63.8º

Check: Angles B = 63.80º B = 90.0º – A B = 90.0º – 26.20º B = 63.80º

Sides

sin B = bc

b = c sin B

c = sin

bB

c = 76.0sin 63.8°

c = 84.70

B

C b

c a

A

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Module T09 − Solutions of Right Triangles

Experiential Activity Four 1. Solve all for all unknowns in right triangle

ABC where angle C is 90º.

Round all sides to 3 significant digits. Round all angles to one decimal place.

a) a = 27.8 b = 52.8 b) a = 15.2 b = 31.4 c) b = 35.1 a = 14.0 d) b = 25.7 a = 8.7 e) b = 7.7 a = 35.9

A

B C

c

a

b

2. Solve the following triangles, assuming angle C is 90º.

Round all sides to 4 significant digits. Round all angles to two decimal places.

Side a Side b Side c Angle A Angle B

a) 3.327 69.08º

b) 3.578 7.981

c) 7.833 37.61º

d) 8.042 53.93º

e) 6.731 7.351

f) 4.647 4.830

g) 8.958 79.75º

h) 8.852 1.984

i) 1.653 12.83º

j) 9.272 65.98º

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Module T09 − Solutions of Right Triangles

3. Solve the following triangles.

Round all sides to 3 significant digits. Round all angles to one decimal place. a)

C

B

c

b

a = 7.515

28.28º A

b)

D

F E

f = 3.194 e = 2.103

d

c)

I

H

G h = 6.936

g = 2.489i

d)

L K

J

k

j

l = 2.881

70.85º

e)

NM

O

n = 2.883

o = 6.996

m

f) R

Q

q = 9.854

r = 8.697 P

p

g)

S T

U

t = 6.258 s

u 48.82º

h)

W V

X

50.64º v = 3.567 w

i)

AA Y

Z

aa

z

y = 1.411

26.1º

j)

DD CC

BB

54.94º

cc = 2.317 dd

bb

x

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Module T09 − Solutions of Right Triangles

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Experiential Activity Four Answers 1. a) ∠B = 62.2º ∠A = 27.8º c = 59.7 b) ∠B = 64.2 ∠A = 25.8º c = 34.9 c) ∠B = 68.3º ∠A = 21.7º c = 37.8 d) ∠A = 18.7º ∠B = 71.3º c = 27.1 e) ∠B = 12.1º ∠A = 77.9º c = 36.7

2. a) b = 1.27 c = 3.56 ∠B = 20.9º b) a = 7.13 ∠A = 63.4º ∠B = 26.6º c) a = 6.03 c = 9.89 ∠B = 52.4º d) a = 4.73 b = 6.50 ∠A = 36.1º e) c = 9.97 ∠A = 42.5º ∠B = 47.5º f) b = 1.32 ∠A = 74.2º ∠B = 15.8º g) a = 1.62 c = 9.10 ∠A = 10.3º h) c = 9.07 ∠A = 77.4º ∠B = 12.6º i) b = 7.26 c = 7.44 ∠B = 77.2º j) a = 3.77 b = 8.47 ∠A = 24.0º

3. a) b = 14.0 c = 15.9 ∠B = 61.7º b) d = 2.40 ∠E = 41.2º ∠D = 48.8º c) i = 7.37, ∠G = 19.7º ∠H = 70.3º d) k = 8.78 j = 8.30 ∠L = 19.2º e) m = 7.57 ∠N = 22.4º ∠O = 67.6º f) p = 4.63 ∠P = 28.0º ∠R = 62.0º g) s = 4.71 u = 4.12 ∠U = 41.2º h) x = 2.76 w = 2.26 ∠W = 39.4º i) a = 3.21 z = 2.88 ∠Z = 63.9º j) bb = 1.33 dd = 1.90 ∠BB = 35.1º

Module T09 − Solutions of Right Triangles

OBJECTIVE FIVE When you complete this objective you will be able to… Solve applied right triangle problems.

Exploration Activity PROBLEM SOLVING The following definitions will be useful in this objective: 1. Angle of elevation: The angle between the horizontal and the line of sight looking

upward at an object.

Angle of elevation

Observation point

2. Angle of depression: The angle between the horizontal and the line of sight looking

downward at an object

θ

Observation point

Angle of depression

Line of sight

Rules for solving applied right triangle problems: 1. Construct a diagram and label the three sides and angles given in the problem. 2. Select the most convenient trigonometric function(s) to solve for the unknown sides

and/or angles. Where possible, choose the function that will derive the unknown parts directly, without having to find other parts first.

3. Round off side lengths and angle measurements after the final calculation as

indicated.

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Module T09 − Solutions of Right Triangles

Example 1: A transmission line rises 2.65 m in a run of 36.15 m. What is the angle the line makes with the horizontal? Solution:

Transmission line2.65 m

36.15 m θ

To find θ Use the tangent function

tan θ = adjacentopposite

tan θ = 15.3665.2

Therefore:

θ = tan−1 ⎟⎠⎞

⎜⎝⎛

15.3665.2

θ = 4.2º

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Module T09 − Solutions of Right Triangles

Example 2 A 22.0 m high tree casts a shadow 15.6 m long. What is the angle of elevation of the sun? Solution:

tan θ = adjacentopposite

tan θ = 6.150.22

θ = tan−1 22.015.6

⎛ ⎞⎜ ⎟⎝ ⎠

θ = 54.7º

15.6 m

22.0 m

Example 3: From the top of a bridge 30.5 m high, the angle of depression of an adjacent building is 15.5º. Determine the horizontal distance to the building from a point directly beneath this observation point.

15.5º Bridge

Building (10 m)

Observation Point

Distance

30.5 m

Solution:

tan θ = oppositeadjacent

tan 74.5º = 20.5

x

x = 20.5 tan 74.5º x = 73.9 m

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Module T09 − Solutions of Right Triangles

Experiential Activity Five Solve the following problems. Round all answers to 3 significant digits 1. A tower casts a shadow 143 m long, and at the same time the angle of elevation of

the sun is 41.7º. What is the height of the tower? 2. An antenna 120 m tall casts a shadow 82 m long. What is the angle of elevation of

the sun? 3. At a horizontal distance of 90 m from the foot of a water tower, the angle of elevation

of the top to the tower is found to be 31º. How high is the tower? 4. A light pole 14.0 m high is to be guy wired from its middle, and the guy wire is to

make an angle of 45º with the ground. Allowing 2 m extra for splicing, how long must the guy wire be?

5. An extension ladder 17 m long rests against a vertical wall with its base 6 m from the

wall. a) What angle does the ladder make with the ground? b) How far up the wall does the ladder reach?

6. From the top of a hill 60 m high, the angle of depression of a boat is 27.5º. How far

away is the boat from the viewing point at the top of the hill? 7. In order to find the width BC of a creek, a

distance AB was laid off along the bank, the point B being directly opposite a steel pole C on the opposite side, as shown in Figure 1. If the angle BAC was observed to be 59.4º and AB was measured to be 47 m, find the width of the creek.

A

B C

Creek

8. An aircraft flying at 50 m/s climbs a vertical distance of 400 m in one minute. At

what angle is the airplane climbing? 9. An airplane traveling horizontally at 250 km/h forms an angle of depression of 20º

with a ship sailing below. Six minutes later, the airplane is directly above the ship, which is stationary. Determine how high the airplane is flying. Show Me.

10. An electric light pole is braced with a wire 20 m long. One end of the wire is

fastened to the pole 1 m from the top and the other end to a stake in the ground some distance from the foot of the pole. If the wire makes an angle of 66 degrees with the ground, find the height of the pole.

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Module T09 − Solutions of Right Triangles

11. A man observes the angle of elevation of a hovering helicopter is 70.0º when he is

standing 750 m from its shadow while the sun is directly overhead. How high is the helicopter flying above the level ground?

12. A pilot observes an object on the ground to have an angle of depression of 35º and its

distance along the line of sight to be 4000 m. Two minutes later the same object has an angle of depression of 55º and is 450 m away. Assuming that the object was in front of the aircraft in both observations, what is the horizontal distance the aircraft traveled between the observation points? Show Me.

13. A helicopter rises vertically from level ground and is observed from a point

150 m from a point directly beneath the helicopter to have an angle of elevation of 35º. Two minutes later, the helicopter has an angle of elevation of 84º. How far did the helicopter travel upward between sightings?

14. A room is 5 m long, 4 m wide, and 3 m high. What is the diagonal distance across

the room from a lower corner to an upper corner?

15. A circular base conical storage tank has a slant height of 5350 mm and is 2540 mm in diameter. Determine the vertical height of the tank.

16. A hexagonal head bolt measures 20

mm between parallel faces. Find the distance “y” across the corners.

20 mm

y 17. Determine the distances L1, and L2 in

the metal frame shown below.

15.5º

120.0º

L2

11 000 cm

L1

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Module T09 − Solutions of Right Triangles

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Experiential Activity Five Answers 1. 127 m 2. 55.7º 3. 54.1 m 4. 11.9 m 5. a) 69.3º

b) 15.9 m 6. 130 m 7. 79.5 m 8. 7.66º 9. 9.10 km 10. 19.3 m 11. 2060 m 12. 3020 m 13. 1320 m 14. 7.07 m 15. 5200 mm 16. 23.1 mm 17. L1 = 28 900 cm

L2 = 29 400 cm

Module T09 − Solutions of Right Triangles

B

OBJECTIVE SIX When you complete this objective you will be able to… Solve right triangle problems dealing with vectors and navigation.

Exploration Activity DEFINITIONS OF DIRECTION MEASUREMENT The following definitions will be useful in this objective:\ 1. The bearing of a point B from a point A is defined as the acute angle made by the

line drawn from A through B with the north-south line through A. The bearing is then read from the north or south line toward the east or west.

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Bearing:

N

A

S

N 40º E

N

B

A

S

S 40º E

N

B

A

S

S 40º W

N B

A

S

N 40º W

This is a surveyor's way of reading bearings, the first letter is always N or S, and the last letter is always E or W, with an acute angle between them.

Module T09 − Solutions of Right Triangles

2. The bearing of B from A in aeronautics is given as the angle made by the line AB

with the north line through A, measured clockwise from north.

Bearing:

B

N

A

S

40º

N

B

A

S

140º

N

B

A

S

220º

N B

A

S

320º

Example 1: A motorboat moves in the direction N40ºE from A for 4 hr at 20 km/h. How far north and how far east does it travel? Solution: Solve distance traveled (AB): Since the problem gave us the velocity of the boat and the time traveled, we need to determine the total distance traveled (AB) first. d = rt d = (20 km/hr)(4 hr) d = 80 km AB = 80 km. Solve for how far East (BC): Using sin A we have:

sin A = BCAB

sin 40º = 80BC

BC = 80 sin 40º BC = 51.4 km.

40º

N

C

A

B

E

Solve for how far North (AC): Using cos A we have

cos A = ACAB

cos 40º = 80AC

AC = 80 cos 40º AC = 61.3 km. The boat travels 61.3 km north and 51.4 km east.

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Module T09 − Solutions of Right Triangles

Example 2: An airplane is moving at 400 km/h when a bullet is shot with a speed of 2750 km/h at right angles to the path of the airplane. Find the resultant speed and direction of the bullet. In the diagram, vector AB represents the velocity of the airplane, vector AC represents the velocity of the bullet, and vector AD represents the resultant velocity of the bullet.

400 km/hr

2750 km/hr

D

C A

B

θ

Solution: In the right ΔADC: Solve for the resultant velocity of the bullet:

AD2 = AC2 + CD2

AD = 2 2400 2750+ AD = 2779 km/h

Solve for the direction of the bullet:

tan θ = 4002750

θ = tan−1 4002750

⎛ ⎞⎜ ⎟⎝ ⎠

θ = 8.3º The angle between the path of the bullet and the plane is ∠BAC. To get this angle we subtract θ from 90° to get 81.7°.

Thus, the bullet travels at 2779 km/h along a path making an angle of 81.7° with the path of the plane.

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Module T09 − Solutions of Right Triangles

25

Experiential Activity Six Solve the given vector problems. Round all angles to one decimal place. Round all other values to 3 significant digits. In questions 1 to 4, find the “x” and “y” components of the indicated vectors by the use of trigonometric ratios. 1. Vector A, magnitude 5.50, directed 28º above positive direction of “x” axis. 2. Vector B, magnitude 18.3, directed 16º to right of positive direction of “y” axis. 3. Vector C, magnitude 800, directed 35º above negative direction of “x” axis. 4. Vector D, magnitude 0.805, directed 32º above negative direction of “x” axis. 5. A motorboat travels 75 km due east from a port and then turns south for 26 km.

What is the displacement of the boat from the port? Show Me. 6. A ship which travels 8.0 km/h in still water heads directly across a stream that flows

at 4.0 km/h. What is the resultant velocity of the ship? 7. A rocket is fired at an angle of 80º with the horizontal, with a speed of 2000 km/h;

find the horizontal and vertical components of the velocity. 8. An airplane flies 150 km in the direction 143º. How far south and how far east of the

starting point is it?

Experiential Activity Six Answers 1. x = 4.86, y = 2.58 2. x = 5.04, y = 17.6 3. x = −655, y = 459 4. x = –0.683, y = 0.427 5. 79.4 km, S 70.9º E 6. 8.94, 26.6º 7. 347 km/h, 1970 km/h 8. 90.3 km E, 120 km S

Practical Application Activity Complete the Right Triangles assignment in TLM.

Summary This module presented the student with some applications of solving right triangles.