Download Math 2201 Unit 3: Acute Triangle Trigonometry Read Learning

Math 2201
Unit 3: Acute Triangle Trigonometry
Read Learning Goals, p. 127 text.
Ch. 3 Notes
§3.1 Exploring Side-Angle Relationships in Acute Triangles (0.5 class)
Read Goal p. 130 text.
Outcomes:
1. Define an acute triangle. See notes
2. Make a conjecture about the relationship between the length of the sides of an acute triangle and
the sine of the opposite angles. p. 131
3. Derive the Law of Sines (Sine Law). p. 130
This chapter is about solving acute triangles. This means finding the lengths of the missing sides and/or the
measures of the missing angles. We will examine two ways to solve acute triangles – the Law of Sines
(Sine Law) and the Law of Cosines (Cosine Law).
In Math 1201, we labeled the sides of a right triangle with respect to the angles in the triangle. With respect
to acute angle A, these sides were called the adjacent side and the opposite side. The side opposite the right
angle was called the hypotenuse.
hypotenuse
A
opposite
adjacent
Recall from Math 1201 that you were introduced to three special ratios that were obtained from the lengths
of the three sides of a right triangle and that we gave them special names.
Ratio
opposite
hypotenuse
adjacent
hypotenuse
opposite
adjacent
Special Name
sine ratio (sin)
cosine ratio (cos)
tangent ratio (tan)
1
In the right triangle on page 1,
remember the name of the Indian below.
. You can remember these ratios if you
Def n Solving a triangle means finding the length(s) of the missing side(s) and/or the measure(s) of the
missing angle(s).
Up to now, we have used the Pythagorean Theorem and the primary trigonometric ratios to solve a right
triangle.
B
E.g.: Solve the right triangle below.
104cm
40cm
x
A
C
Since we know the lengths of two sides of the triangle and since it is a right triangle, we can use the
Pythagorean Theorem to find the length of the third side.
Now we must use the trig ratios to find one of the missing angles. We will use the lengths of the two given
sides just in case we made a mistake finding the length of the missing side. With respect to angle A, we are
given the lengths of the opposite side and the hypotenuse so we will use the sine ratio. Make sure your
graphing calculator is set to degree mode and not radian mode.
2
40
104
 40 
A  sin 1 

 104 
A  22.62
sin A 
Since m A  22.62, then m B  180  90  22.62  67.38
Solving a triangle using the Pythagorean Theorem and primary trigonometric ratios only works with a
RIGHT triangle. How do we solve triangles that do NOT have a right angle? Triangles that do not have a
right angle are either acute triangles or obtuse triangles. In this chapter we will deal with acute triangles
only.
Def n : An acute triangle is a triangle in which all the angles have a measure less than 90 .
Labeling Any Triangle
B
c
A
a
C
b
We label each vertex with an UPPERCASE letter and the side opposite the vertex with the corresponding
lowercase letter.
The Law of Sines (Sine Law) can be used to solve SOME acute triangles.
A Conjecture about the Relationship between the Length of the Sides of an Acute Triangle and the
Sine of the Opposite Angles.
Measure each side to the nearest tenth of a centimeter and each angle to the nearest tenth of a degree and
complete the table.
3
Measure   
(1 decimal place)
A
B
C
Side a
45.0
Side b
60.0
Side c
75.0
Length (cm)
Calculate
(1 decimal place) (4 decimal
places)
sin A
4.2
a
sin B
5.1
b
sin C
5.7
c
0.168358757
0.169---0.169 ---
Make a conjecture about the relationship between the length of the sides of an acute triangle and the sine of
the opposite angles.
Conjecture: In an acute triangle _______________________________________________________.
Derivation of the Law of Sines (Sine Law)
Let’s draw any a ute triangle with an altitude (h) drawn from a vertex.
A
c
h
B
D
a
Using the sine ratio we can write sin B 
sin C 
b
C
h
or h  c sin B . Similarly, we can write
c
h
or h  b sin C .
b
Since we have two expressions equal to h, we can substitute to get c sin B  b sin C .
Since b  c  0 , we can divide both sides by bc to get
c sin B b sin C

bc
bc
sin B sin C

b
c
4
Now let’s use the sa e triangle above but draw a different altitude.
A
c
h
B
D
a
Using the sine ratio we can write sin B 
sin A 
b
C
h
or h  a sin B . Similarly, we can write
a
h
or h  b sin A .
b
Since we have two expressions equal to h, we can substitute to get a sin B  b sin A .
Since a  b  0 , we can divide both sides by ab to get
a sin B b sin A

ab
ab
sin B sin A

b
a
Since
sin B sin C
sin B sin A


and
then
b
c
b
a
*************
sin A sin B sin C


*****************
a
b
c
This is the Sine Law.
Sine Law
sine angle 1
sine angle 2
sine angle 3


length of side opposite angle 1 length of side opposite angle 2 length of side opposite angle 3
5
§3.2 Applying the Sine Law (2 classes)
Read Goal p. 132 text.
Outcomes:
1. Use the Sine Law to find the length of a missing side in an acute triangle. p. 134
2. Use the Sine Law to find the measure of a missing angle in an acute triangle. p. 136
3. Solve problems using the Sine Law. pp. 135-137
E.g.: Solve the triangle below.
C
52
b
39
11.2cm
A
c
B
Substituting into
sin A sin C

gives,
a
c
sin 39 sin 52

11.2
c
Cross multiplying and solving gives
c sin 39  11.2sin 52
c sin 39 11.2sin 52

sin 39
sin 39
c  14.0cm
Since m A  39 and m C  52 , then m B  180  39  52  89
Substituting into
sin A sin B

gives,
a
b
sin 39 sin 89

11.2
b
Cross multiplying and solving gives
b sin 39  11.2sin 89
b sin 39 11.2sin 89

sin 39
sin 39
b  17.8cm
6
Generally, there are two cases when you can use the Sine Law.
E.g.: Find the value of x in the diagram to the right.
Substituting into
sin A sin B

gives
a
b
sin 63 sin 49

10
x
x sin 63  10sin 49
x sin 63 10sin 49

sin 63
sin 63
x  8.5cm
Do #’s 2 a, 3 a, b, c, 8, pp. 138-140 text in your homework booklet.
E.g.: Find the value of x in the diagram to the right.
Substituting into
sin A sin B

gives
a
b
7
sin 50 sin x

11
7
7 sin 50  11sin x
7 sin 50 11sin x

11
11
sin x  0.4874828274.....
x  29.2
Do #’s 2 b, 3 d, e, f, 7, 11, pp. 138-140 text in your homework booklet.
Problem Solving Using the Sine Law.
We need to find the values of a and b.
Substituting into
sin B sin C

gives
b
c
sin 47 sin 68

a
46
46sin 47  a sin 68
46sin 47 a sin 68

sin 68
sin 68
a  36.3m
Substituting into
sin A sin C

gives
a
c
8
sin 65 sin 68

b
46
46sin 65  b sin 68
46sin 65 b sin 68

sin 68
sin 68
b  45.0m
So 46  36.3  45.0  127.3m of chain-link fence is needed to enclose the entire park.
Substituting into
sin Q sin R
gives

q
r
m P  180  60  35.3  84.7
sin 60 sin R

184.5
123
123sin 60  184.5sin R
h
184.5
h  184.5sin 84.7  183.7ft
sin 84.7 
123sin 60 184.5sin R

184.5
184.5
sin R  0.5773502692.....
R  35.3
9
Find the value of x.
First let’s look at the triangle to the right.
10
sin Z sin C

Substituting into
gives
z
c
z
sin 30 sin10

z
50
50sin 30  z sin10
50sin 30 z sin10

sin10
sin10
z  144.0ft
30
50 feet
Using right triangle trigonometry, we can write
x
144
x  144 cos 40  110.3ft
cos 40 
Do #’s 4, 10, 13-15, pp. 139-141 text in your homework booklet.
Do #’s 4-7, 9 a, p. 143 text in your homework booklet.
10
§3.3 Deriving and Applying the Cosine Law (2 classes)
Read Goal p. 144 text.
Outcomes:
1. Illustrate and explain situations where the Sine Law cannot be used to find a missing side or a
missing angle in a triangle. p. 144
2. Derive the Law of Cosines (Cosine Law). p. 144
The Law of Sines CAN be used given two sides and the non-included angle.
The Law of Sines CANNOT be used in the following situations:
1. Given two sides and the included angle.
C
sin 52 sin A

c
19.50
19.50cm
52
11.2cm
c
A
This is impossible to solve since there are two unknowns in the equation.
C
2. Given three sides.
19.50
B
15.39
sin B sin A

15.39 19.50
11.20
A
This is impossible to solve since there are two unknowns in the equation.
C
a
52
3. Given three angles.
35
c
sin 93 sin 52

a
c
93
A
This is impossible to solve since there are two unknowns in the equation.
Therefore, we need another formula that will enable us to solve a triangle in two of those situations. This
law is the Law of Cosines.
11
Derivation of the Law of Cosines (Cosine Law)
B
a
c
h
A
P
C
b
Using right triangle APB and the Pythagorean Theorem, we can write
Similarly, using right triangle BPC and the Pythagorean Theorem, we can write
Since the expressions on the left hand side (LHS) of equation 1 and equation 2 equal
to each other. By substitution,
Solving for
gives
But m is not a side of the triangle.
Using right triangle APB again and the trig ratios, we can write
Therefore, substituting for m in
we get,
This is the Law of Cosines.
******* Note that the Law of Cosines can be written in three ways.
1.
2.
3.
12
, they must be equal
E.g.: Solve the following triangle.
B
This is not a right triangle and we cannot use the Law of Sines so we must use the Law of Cosines to solve
the triangle.
Now we have the choice of using the Law of Sines or the Law of Cosines to find the measure of angle B.
We will use the Law of Cosines again for practice.
The measure of angle A is
Like the Sine Law, we can use the Cosine Law to find the length of a missing side or the measure of a
missing angle.
13
E.g.: How long is the tunnel in the diagram to the right?
Substituting into a2  b2  c2  2bc cos A gives
BC 2  3002  2002  2  300  200  cos80
BC 2  90000  40000  120000 cos80
BC 2  130000  120000 cos80
BC 2  130000  20837.78132.....
BC 2  109162.2187...
BC   109162.2187...
BC  330.4
BC  330.4ft
So the length of the tunnel is about 330.4ft long.
Draw a triangle which illustrates the information in the workings to the
right.
Do #’s 2, 4 a, 6 a, p. 151 text in your homework booklet.
E.g.: Find the value of A .
Substituting into a2  b2  c2  2bc cos A gives
162  92  192  2  9 19  cos A
256  81  361  342 cos A
256  442  342 cos A
256  442  442  442  342 cos A
186  342 cos A
186 342 cos A

342
342
cos A  0.5438596491...
A  cos 1  0.5438596491...
A  57.1
Do #’s 3, 5 a, 6 c, 7 a, c, pp. 151-152 text in your homework booklet.
14
Problem Solving Using the Cosine Law.
E.g.: Aircraft 1 flies at 400km/h and aircraft 2 flies at 350km/h. If the angle between their paths is 49 ,
how far apart are the aircraft after 2h?
Substituting into a2  b2  c2  2bc cos A gives
d 2  8002  7002  2  800  700  cos 49
d 2  640000  490000  1120000 cos 49
d 2  1130000  1120000 cos 49
d 2  1130000  734786.1125.....
d 2  395213.8875...
d   395213.8875...
d  628.66
d  628.66km
The aircraft are about 628.66km apart.
E.g.: How far apart are the two trees in the diagram to the right.
Substituting into a2  b2  c2  2bc cos A gives
d 2  752  1002  2  75 100  cos 32
d 2  5625  10000  15000 cos 32
d 2  15625  15000 cos 32
d 2  15625  12720.72144.....
d 2  2904.278558...
d   2904.278558...
d  53.89
d  53.89m
The trees are about 53.89m apart.
15
E.g.: What is the angle between the 22ft and the 18ft sides?
Substituting into a2  b2  c2  2bc cos B gives
122  182  222  2 18  22  cos B
144  324  484  792 cos B
144  808  792 cos b
144  808  808  808  792 cos B
664  792 cos B
664 792 cos B

792
792
cos B  0.83

B  cos 1 0.83

B  33.0
The angle between the 22ft and 18ft sides is about 33.0
16
Errors Using the Cosine Law
E.g.: Find the error in the solution below and correct it.
Step 1: BC 2  3752  4002  2  375  400  cos 35
Step 2: BC 2  140625  160000  300000 cos 35
Step 3: BC 2  300625  300000 cos 35
Step 4: BC 2  625cos 35
Step 5: BC 2  511.9700277...
Step 6: BC   511.9700277...
Step 7: BC  22.63
Step 8: BC  22.63m
The error occurs in Step ____ where ________________________________________________________.
In this step, ____________________________________________________________________________.
Do #’s 9, 13, pp. 152-153 text in your homework booklet.
Do #’s 1, 5, 6, 7, 9, 13, pp. 161-163 text in your homework booklet.
Do #’s 1-7, 9, 10 p. 168 text in your homework booklet.
Pythagorean Theorem
Primary Trigonometric Ratios
Sine Law
Formulae
c 2  a 2  b2
o
a
o
sin   ; cos   ; tan  
h
h
a
sin A sin B sin C


a
b
c
2
2
2
a  b  c  2bc cos A
b 2  a 2  c 2  2ac cos B
Cosine Law
c 2  a 2  b 2  2ab cos C
17