Download Module - Solution of Right Triangles with Practical

Trigonometry
Module T09
Solutions of Right
Triangles with
Practical
Applications
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED December, 2008
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.
2.
3.
4.
5.
6.
Solve a right triangle given an acute angle and a side which is not the hypotenuse.
Solve a right triangle given an acute angle and the hypotenuse.
Solve a right triangle given the hypotenuse and one other side.
Solve a right triangle given two sides neither of which is the hypotenuse.
Solve applied right triangle problems.
Solve right triangle problems dealing with vectors and navigation.
1
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
Creek
C
2
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
B
Given ΔACB, side b = 30.0, ∠A = 25.0º and a
right angle at C.
c
a
Solve ΔACB.
So we are to find all remaining sides and
angles, which are: a, c, ∠B
A
b
C
Solve for ∠B
since:
∠A + ∠B = 90.0º and ∠A = 25º
therefore:
25.0º + ∠B = 90.0º
solving:
∠B = 65.0º
3
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.
a
sin A =
doesn't work since we don't know sides a and c.
c
cos A =
b
doesn't have an “a” in it.
c
tan A =
a
should work
b
tan 25.0º =
a
30.0
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 =
b
because ∠B was a calculated angle.
a
Solve for side c
By examining the functions, we decide to use the cosine,
b
cos A =
c
30.0
cos 25.0º =
c
solving:
c=
30.0
cos 25.0°
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
4
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.
q = 1290
R
p
48.1º
P
r
Q
Solve for ∠P
∠P + ∠Q = 90º
∠P = 90.0º – ∠Q
∠P = 90.0º – 48.1º
∠P = 41.9º
Solve for side r
sin Q =
q
r
sin 48.1º =
r=
1290
r
1290
sin 48.1°
r = 1733.1452
r = 1733
Solve for side p
tan Q =
q
p
tan 48.1º =
p=
Note: we could have used cosine of Q to solve for p which is cos Q =
1290
p
p
r
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.
.
1290
tan 48.1°
p = 1157.4508
p = 1157
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Module T09 − Solutions of Right Triangles
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
6
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.
∠B = 27.1º
1. a = 11.5
2. a = 15.0
∠B = 74.9º
∠A = 74.9º
Show Me.
3. b = 424
4. b = 8.4
∠B = 77.6º
∠A = 82.9º
5. a = 11.2
A
c
B
b
a
C
Experiential Activity One Answers
1.
2.
3.
4.
5.
∠A = 62.9º
∠A = 15.1º
∠B = 15.1º
∠A = 12.4º
∠B = 7.1º
b = 5.88
b = 55.6
a = 1570
a = 1.85
b = 1.40
c = 12.9
c = 57.6
c = 1630
c = 8.60
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
B
In ΔABC, c = 45.3, and ∠A = 20.3º.
c = 45.3
Find b, a, and ∠B.
a
C
Solve for ∠B
Solve for side b
∠B = 90.0º – A
∠B = 90.0º – 20.3º
∠B = 69.7º
b
c
b = c ⋅ cos A
b = 45.3 ⋅ cos 20.3°
b = 42.49
cos A =
Check:
Sides
Angles
(15.72)2 + (42.49)2 = (45.3)2
A + B + C = 180º
2053 = 2053
20.3º + 69.7º + 90º = 180º
180º = 180º
8
Module T09 − Solutions of Right Triangles
20.3º
b
A
Solve for side a
a
c
a = c ⋅ sin A
a = 45.3 ⋅ sin 20.3°
a = 15.72
sin A =
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.
A
Round all sides to 3 significant digits.
Round all angles to one decimal place.
1.
2.
3.
4.
5.
c = 14.4
c = 14.6
c = 3.30
c = 7.30
c = 22.2
∠A = 30.3º
∠B = 36.1º
∠B = 38.7º
∠A = 13.8º
∠B = 58.1º
Show Me.
c
B
b
a
C
Experiential Activity Two Answers
1.
2.
3.
4.
5.
∠B = 59.7º
∠A = 53.9º
∠A = 51.3º
∠B = 76.2º
∠A = 31.9º
a = 7.27
a = 11.8
a = 2.58
a = 1.74
b = 18.8
b = 12.4
b = 8.60
b = 2.06
b = 7.09
a = 11.7
9
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
A
Given ΔABC where c = 38.3 and b = 23.1.
Find a, ∠A, and ∠B.
c
B
b
C
a
Solve for side a
Solve for ∠A
Solve for ∠B.
a2 + b2 = c2
a2 = c2 – b2
cos A =
b
c
23.1
cos A =
38.3
⎛ 23.1 ⎞
A = cos −1 ⎜
⎟
⎝ 38.3 ⎠
A = 52.91º
b
c
23.1
sin B =
38.3
⎛ 23.1 ⎞
B = sin −1 ⎜
⎟
⎝ 38.3 ⎠
B = 37.09º
a=
c2 − b2
a = (38.3) 2 − (23.1) 2
a = 30.55
sin B =
Check:
Angle
Sides
Since we now have checked our angles and know they
B = 90.0º – A
are correct, we will use them to solve for “a” again.
B = 90.0º –52.91º
a
cos B =
B = 37.09º
38.30
a = 38.3 cos B
Since we solved for B another way,
a = 38.3 cos 37.09º
and produced the same result, it should a = 30.55
be correct.
10
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.
2.
3.
4.
5.
c = 12.8
c = 11.5
c = 13.9
c = 14.2
c = 10.2
b = 7.80
a = 6.50
a = 8.90
a = 9.20
b = 5.20
A
c
Show Me
B
b
a
C
Experiential Activity Three Answers
1.
2.
3.
4.
5.
a = 10.1
b = 9.49
b = 10.7
b = 10.8
a = 8.77
∠B = 37.5º
∠A = 34.4º
∠B = 50.2º
∠A = 40.4º
∠B = 30.7º
∠A = 52.5º
∠B = 55.6º
∠A = 39.8º
∠B = 49.6º
∠A = 59.3º
11
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
B
Given ΔABC where a = 37.4 and b = 76.0.
c
a
Find c, ∠A, and ∠B
A
Solve for side c
Solve for ∠A
a2 + b2 = c2
a
b
37.4
tan A =
76.0
⎛ 37.4 ⎞
A = tan−1 ⎜
⎟
⎝ 76.0 ⎠
A = 26.2º
c=
a 2 + b2
c = (37.4) 2 + (76.0) 2
c = 84.70
tan A =
Check:
Angles
B = 63.80º
B = 90.0º – A
B = 90.0º – 26.20º
B = 63.80º
Sides
b
c
b = c sin B
b
c=
sin B
76.0
c=
sin 63.8°
c = 84.70
sin B =
12
Module T09 − Solutions of Right Triangles
b
Solve for ∠B
b
a
⎛ 76.0 ⎞
tan B = ⎜
⎟
⎝ 37.4 ⎠
⎛ 76.0 ⎞
B = tan−1 ⎜
⎟
⎝ 37.4 ⎠
B = 63.8º
tan B =
C
Experiential Activity Four
1. Solve all for all unknowns in right triangle
ABC where angle C is 90º.
A
Round all sides to 3 significant digits.
Round all angles to one decimal place.
a)
b)
c)
d)
e)
a = 27.8
a = 15.2
b = 35.1
b = 25.7
b = 7.7
c
b = 52.8
b = 31.4
a = 14.0
a = 8.7
a = 35.9
B
b
C
a
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
a)
Side b
3.327
Angle A
3.578
c)
7.833
d)
7.981
37.61º
8.042
e)
6.731
f)
4.647
g)
53.93º
7.351
4.830
8.958
h)
8.852
i)
1.653
Angle B
69.08º
b)
j)
Side c
79.75º
1.984
12.83º
9.272
65.98º
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Module T09 − Solutions of Right Triangles
3. Solve the following triangles.
a)
Round all sides to 3 significant digits. Round all angles to one decimal place.
b)
B
d
F
E
c
a = 7.515
e = 2.103
f = 3.194
28.28º
b
C
A
D
d)
c)
h = 6.936
G
I
j
L
K
g = 2.489
i
l = 2.881
k
70.85º
H
J
e)
f)
o = 6.996
M
R
N
q = 9.854
p
n = 2.883
m
O
P
g)
Q
r = 8.697
h)
U
X
50.64º
v = 3.567
t = 6.258
s
48.82º
T
u
w
W
S
i)
x
V
j)
AA
z
BB
Y
26.1º
cc = 2.317
y = 1.411
dd
aa
Z
DD
54.94º
bb
14
Module T09 − Solutions of Right Triangles
CC
Experiential Activity Four Answers
1. a)
b)
c)
d)
e)
∠B = 62.2º
∠B = 64.2
∠B = 68.3º
∠A = 18.7º
∠B = 12.1º
∠A = 27.8º
∠A = 25.8º
∠A = 21.7º
∠B = 71.3º
∠A = 77.9º
c = 59.7
c = 34.9
c = 37.8
c = 27.1
c = 36.7
2. a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
b = 1.27
a = 7.13
a = 6.03
a = 4.73
c = 9.97
b = 1.32
a = 1.62
c = 9.07
b = 7.26
a = 3.77
c = 3.56
∠A = 63.4º
c = 9.89
b = 6.50
∠A = 42.5º
∠A = 74.2º
c = 9.10
∠A = 77.4º
c = 7.44
b = 8.47
∠B = 20.9º
∠B = 26.6º
∠B = 52.4º
∠A = 36.1º
∠B = 47.5º
∠B = 15.8º
∠A = 10.3º
∠B = 12.6º
∠B = 77.2º
∠A = 24.0º
3. a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
b = 14.0
d = 2.40
i = 7.37,
k = 8.78
m = 7.57
p = 4.63
s = 4.71
x = 2.76
a = 3.21
bb = 1.33
c = 15.9
∠E = 41.2º
∠G = 19.7º
j = 8.30
∠N = 22.4º
∠P = 28.0º
u = 4.12
w = 2.26
z = 2.88
dd = 1.90
∠B = 61.7º
∠D = 48.8º
∠H = 70.3º
∠L = 19.2º
∠O = 67.6º
∠R = 62.0º
∠U = 41.2º
∠W = 39.4º
∠Z = 63.9º
∠BB = 35.1º
15
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.
Observation
point
Angle of elevation
2. Angle of depression: The angle between the horizontal and the line of sight looking
downward at an object
Observation
point
θ
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.
16
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 line
2.65 m
θ
36.15 m
To find θ
Use the tangent function
tan θ =
opposite
adjacent
tan θ =
2.65
36.15
Therefore:
⎛ 2.65 ⎞
θ = tan−1 ⎜
⎟
⎝ 36.15 ⎠
θ = 4.2º
17
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?
22.0 m
Solution:
opposite
adjacent
22.0
tan θ =
15.6
⎛ 22.0 ⎞
θ = tan−1 ⎜
⎟
⎝ 15.6 ⎠
θ = 54.7º
tan θ =
15.6 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.
Observation
Point
Bridge
15.5º
30.5 m
Building
(10 m)
Distance
Solution:
opposite
adjacent
x
tan 74.5º =
20.5
x = 20.5 tan 74.5º
x = 73.9 m
tan θ =
18
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
Creek
B
C
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.
19
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.
L1
L2
120.0
º
11 000 cm
15.5º
20
Module T09 − Solutions of Right Triangles
Experiential Activity Five Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
127 m
55.7º
54.1 m
11.9 m
a) 69.3º
b) 15.9 m
130 m
79.5 m
7.66º
9.10 km
19.3 m
2060 m
3020 m
1320 m
7.07 m
5200 mm
23.1 mm
L1 = 28 900 cm
L2 = 29 400 cm
21
Module T09 − Solutions of Right Triangles
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.
N
N
B
N
N
B
A
A
Bearing:
A
A
S
S
N 40º E
S 40º E
B
B
S
S 40º W
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.
22
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.
N
N
B
N
N
B
A
A
Bearing:
A
A
S
S
40º
140º
B
B
S
S
220º
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:
N
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.
B
C
d = rt
d = (20 km/hr)(4 hr)
d = 80 km
40º
A
AB = 80 km.
Solve for how far East (BC):
Using sin A we have:
BC
sin A =
AB
BC
sin 40º =
80
BC = 80 sin 40º
BC = 51.4 km.
E
Solve for how far North (AC):
Using cos A we have
AC
cos A =
AB
AC
cos 40º =
80
AC = 80 cos 40º
AC = 61.3 km.
The boat travels 61.3 km north and
51.4 km east.
23
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.
B
400
km/hr
D
θ
A
2750 km/hr
C
Solution:
In the right ΔADC:
Solve for the resultant velocity of the bullet:
AD2 = AC2 + CD2
AD = 4002 + 27502
AD = 2779 km/h
Solve for the direction of the bullet:
400
tan θ =
2750
⎛ 400 ⎞
θ = tan−1 ⎜
⎟
⎝ 2750 ⎠
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
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
4. x = –0.683, y = 0.427
7. 347 km/h, 1970 km/h
2. x = 5.04, y = 17.6
5. 79.4 km, S 70.9º E
8. 90.3 km E, 120 km S
3. x = −655, y = 459
6. 8.94, 26.6º
Practical Application Activity
Complete the Right Triangles assignment in TLM.
Summary
This module presented the student with some applications of solving right triangles.
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Module T09 − Solutions of Right Triangles