1
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.
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 Special Name
opposite
hypotenuse sine ratio (sin)
adjacent
hypotenuse cosine ratio (cos)
opposite
adjacent tangent ratio (tan)
A adjacent
hypotenuse opposite
2
In the right triangle on page 1,
. You can remember these ratios if you
remember the name of the Indian below.
nDef 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.
E.g.: Solve the right triangle below.
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.
104cm
40cm
A
B
C x
3
1
40sin
104
40sin
104
22.62
A
A
A
Since m 22.62 , then m 180 90 22.62 67.38A B
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.
nDef :An acute triangle is a triangle in which all the angles have a measure less than 90 .
Labeling Any Triangle
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.
c
b A
B
C
a
4
Measure
(1 decimal place)
Length (cm)
(1 decimal place)
Calculate
(4 decimal
places)
A 45.0
Side a 4.2
sin A
a
0.168358757
B 60.0
Side b 5.1
sin B
b
0.169----
C 75.0
Side c 5.7
sinC
c
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.
Using the sine ratio we can write sin or sinh
B h c Bc
. Similarly, we can write
sin or sinh
C h b Cb
.
Since we have two expressions equal to h, we can substitute to get sin sinc B b C .
Since 0b c , we can divide both sides by bc to get
c sin B
b c
b
sin C
b
sin sin
c
B C
b c
a
b c
D B C
A
h
5
Now let’s use the sa e triangle above but draw a different altitude.
Using the sine ratio we can write sin or sinh
B h a Ba
. Similarly, we can write
sin or sinh
A h b Ab
.
Since we have two expressions equal to h, we can substitute to get sin sina B b A .
Since 0a b , we can divide both sides by ab to get
a sin B
a
b
b
sin A
a b
sin sinB A
b a
Since sin sinB C
b c and
sin sinB A
b a then
*************sin sin sinA B C
a b c *****************
This is the Sine Law.
a
b c
D B C
A
h
6
§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.
Substituting into sin sinA C
a c gives,
sin39 sin52
11.2 c
Cross multiplying and solving gives
sin39 11.2sin52
sin39 11.2sin52
sin39 sin39
14.0cm
c
c
c
Since m 39A and m 52C , then m 180 39 52 89B
Substituting into sin sinA B
a b gives,
sin39 sin89
11.2 b
Cross multiplying and solving gives
sin39 11.2sin89
sin39 11.2sin89
sin39 sin39
17.8cm
b
b
b
b
39
52
11.2cm A
C
B
c
7
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 sinA B
a b gives
sin 63 sin 49
10
sin 63 10sin 49
sin 63 10sin 49
sin 63 sin 63
8.5cm
x
x
x
x
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 sinA B
a b gives
8
sin50 sin
11 7
7sin50 11sin
7sin50 11sin
11 11
sin 0.4874828274.....
29.2
x
x
x
x
x
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 sinB C
b c gives
sin 47 sin 68
46
46sin 47 sin 68
46sin 47 sin 68
sin 68 sin 68
36.3m
a
a
a
a
Substituting into sin sinA C
a c gives
9
sin 65 sin 68
46
46sin 65 sin 68
46sin 65 sin 68
sin 68 sin 68
45.0m
b
b
b
b
So 46 36.3 45.0 127.3m of chain-link fence is needed to enclose the entire park.
Substituting into sin sinQ R
q r gives
sin 60 sin
184.5 123
123sin 60 184.5sin
123sin 60 184.5sin
184.5 184.5
sin 0.5773502692.....
35.3
R
R
R
R
R
m P 180 60 35.3 84.7
sin84.7184.5
184.5sin84.7 183.7ft
h
h
10
Find the value of x.
First let’s look at the triangle to the right.
Substituting into sin sinZ C
z c gives
sin30 sin10
50
50sin30 sin10
50sin30 sin10
sin10 sin10
144.0ft
z
z
z
z
Using right triangle trigonometry, we can write
cos40144
144cos40 110.3ft
x
x
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.
30
50 feet
10
z
11
§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.
sin52 sin
19.50
A
c
This is impossible to solve since there are two unknowns in the equation.
2. Given three sides.
sin sin
15.39 19.50
B A
This is impossible to solve since there are two unknowns in the equation.
3. Given three angles.
sin93 sin52
a c
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.
19.50cm
52
11.2cm
C
A
c
a
35
52
C
A
c
93
11.20
B
19.50 C
15.39
A
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Derivation of the Law of Cosines (Cosine Law)
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 , they must be 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.
A
B
C
a
b
c
P
h
13
E.g.: Solve the following triangle.
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.
B
14
E.g.: How long is the tunnel in the diagram to the right?
Substituting into 2 2 2 2 cosa b c bc A gives
2 2 2
2
2
2
2
300 200 2 300 200 cos80
90000 40000 120000cos80
130000 120000cos80
130000 20837.78132.....
109162.2187...
109162.2187...
330.4
330.4ft
BC
BC
BC
BC
BC
BC
BC
BC
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 2 2 2 2 cosa b c bc A gives
2 2 2
1
16 9 19 2 9 19 cos
256 81 361 342cos
256 442 342cos
256 442 442 442 342cos
186 342cos
186 342cos
342 342
cos 0.5438596491...
cos 0.5438596491...
57.1
A
A
A
A
A
A
A
A
A
Do #’s 3, 5 a, 6 c, 7 a, c, pp. 151-152 text in your homework booklet.
15
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 2 2 2 2 cosa b c bc A gives
2 2 2
2
2
2
2
800 700 2 800 700 cos49
640000 490000 1120000cos49
1130000 1120000cos49
1130000 734786.1125.....
395213.8875...
395213.8875...
628.66
628.66km
d
d
d
d
d
d
d
d
The aircraft are about 628.66km apart.
E.g.: How far apart are the two trees in the diagram to the right.
Substituting into 2 2 2 2 cosa b c bc A gives
2 2 2
2
2
2
2
75 100 2 75 100 cos32
5625 10000 15000cos32
15625 15000cos32
15625 12720.72144.....
2904.278558...
2904.278558...
53.89
53.89m
d
d
d
d
d
d
d
d
The trees are about 53.89m apart.
16
E.g.: What is the angle between the 22ft and the 18ft sides?
Substituting into 2 2 2 2 cosa b c bc B gives
2 2 2
1
12 18 22 2 18 22 cos
144 324 484 792cos
144 808 792cos
144 808 808 808 792cos
664 792cos
664 792cos
792 792
cos 0.83
cos 0.83
33.0
B
B
b
B
B
B
B
B
B
The angle between the 22ft and 18ft sides is about 33.0
17
Errors Using the Cosine Law
E.g.: Find the error in the solution below and correct it.
2 2 2
2
2
2
2
Step 1: 375 400 2 375 400 cos35
Step 2: 140625 160000 300000cos35
Step 3: 300625 300000cos35
Step 4: 625cos35
Step 5: 511.9700277...
Step 6: 511.9700277...
Step 7: 22.63
Step 8
BC
BC
BC
BC
BC
BC
BC
: 22.63mBC
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.
Formulae
Pythagorean Theorem 2 2 2c a b
Primary Trigonometric Ratios sin ; cos ; tano a o
h h a
Sine Law sin sin sinA B C
a b c
Cosine Law
2 2 2
2 2 2
2 2 2
2 cos
2 cos
2 cos
a b c bc A
b a c ac B
c a b ab C
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