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