Module 7 triangle trigonometry super final

GRADE 9 MATHEMATICS
QUARTER 4
Module 7
May 2014

Triangle Trigonometry
MODULE 7
May 20, 2014

In a right triangle, one of the
angles measures 90 𝑜
, and the
remaining two angles are acute and
complementary. The longest side of a
right triangle is known as the
hypotenuse and is opposite the right
angle. The other two sides are called
legs. The leg that is a side of an acute
angle is called the side adjacent to the
angle. The other leg is the side
opposite the angle.
Lesson 1 The Six Trigonometric Ratios: Sine,
Cosine, Tangent, Cosecant, Secant, and Cotangent
C
B
A
a
c
b
Hypotenuse
Side Opposite
Side Adjacent
3

If two angles of a triangle are congruent to two angles of another
triangle, the triangles are similar. If an acute angle of one right triangle is
congruent to an acute angle of another right triangle, the triangles are
similar, and the ratios of the corresponding sides are equal. Therefore, any
two congruent angles of different right triangles will have equal ratios
associated with them.
The ratios of the sides of the right triangles can be used to define the
trigonometric ratios. The ratio of the side opposite 𝜃 and the hypotenuse is
known as the sine. The ratio of the side adjacent 𝜃 and the hypotenuse is
known as the cosine. The ratio of the side opposite 𝜃 and the side adjacent 𝜃
is known as the tangent.
4

SOH-CAH-TOA is a mnemonic device commonly used for remembering
these ratios.
sin 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
tan 𝜃 =
5
Words Symbol Definition
Trigonometric
Ratios
sine 𝜃 sin 𝜃
sin 𝜃 =
𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
cosine 𝜃 cos 𝜃
cos 𝜃 =
𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
tangent 𝜃 tan 𝜃
tan 𝜃 =
C
B
A
a
c
b
Hypotenuse
Side Opposite
Side Adjacent

6
Words Symbol Definition
Reciprocal
Trigonometric
Ratios
cosecant 𝜃 csc 𝜃
𝑐𝑠𝑐 𝜃 =
1
sin 𝜃
𝑜𝑟
secant 𝜃 sec 𝜃
sec 𝜃 =
1
cos 𝜃
𝑜𝑟
cotangent 𝜃 cot 𝜃
cot 𝜃 =
1
tan 𝜃
𝑜𝑟
C
B
A
a
c
b
Hypotenuse
Side Opposite
Side Adjacent
These definitions are called the reciprocal identities.
CHO-SHA-CAO is a mnemonic device commonly used for remembering these ratios.
csc 𝜃 =
sec 𝜃 =
cot 𝜃 =
In addition to the trigonometric ratios sine, cosine, and tangent, there are
three other trigonometric ratios called cosecant, secant, and cotangent.
These ratios are the reciprocals of sine, cosine, and tangent, respectively.

7
Lesson 2 Trigonometric Ratios of Special Angles
Consider the special relationships among the sides
of 30 𝑜
- 60 𝑜
- 90 𝑜
and 45 𝑜
- 45 𝑜
- 90 𝑜
triangles.
x
x 3
2x
60 𝑜
30 𝑜
y
y
y 2
45 𝑜
45 𝑜
These special relationships can be used to
determine the trigonometric ratios for 30 𝑜
, 45 𝑜
, and 60 𝑜
.

𝜽 sin 𝜽 cos 𝜽 tan 𝜽 csc 𝜽 sec 𝜽 cot 𝜽
30 𝑜
45 𝑜
60 𝑜
Activity
Complete the table below that summarizes the values of the trigonometric
ratios of the angles 30 𝑜, 45 𝑜, and 60 𝑜.
8
x
x 3
2x
60 𝑜
30 𝑜
y
y
y 2
45 𝑜
45 𝑜

9
Answers:
30 𝑜 1
2
3
2
3
3
2 2 3
3
3
45 𝑜
2
2
2
2
1 2 2 1
60 𝑜
3
2
1
2
3 2 3
3
2 3
3

10
30 𝑜 1
2
3
2
3
3
2 2 3
3
3
45 𝑜
2
2
2
2
1 2 2 1
60 𝑜
3
2
1
2
3 2 3
3
2 3
3
Notice that sin 30 𝑜
= cos 60 𝑜
and cos 30 𝑜
= sin 60 𝑜
. This is an example
showing that the sine and cosine are cofunctions. That is, if 𝜃 is an
acute angle, sin 𝜃= cos (90 𝑜
−𝜃). Similar relationships hold true for
the other trigonometric ratios.

11
30 𝑜 1
2
3
2
3
3
2 2 3
3
3
45 𝑜
2
2
2
2
1 2 2 1
60 𝑜
3
2
1
2
3 2 3
3
2 3
3
sin 𝜃= cos (90 𝑜−𝜃) cos 𝜃= s𝐢𝐧 (90 𝑜−𝜃)
Cofunctions tan 𝜃= cot (90 𝑜−𝜃) cot 𝜃= tan (90 𝑜−𝜃)
sec 𝜃= csc (90 𝑜
−𝜃) csc 𝜃= sec (90 𝑜
−𝜃)

There are many applications that
require trigonometric solutions.
For example, surveyors use
special instruments to find the
measures of angles of elevation
and angles of depression.
12
Lesson 3 Angles of Elevation and Angles of Depression

13
An angle of elevation is the
angle between a horizontal
line and the line of sight from
an observer to an object at a
higher level.

14
An angle of depression is the
angle between a horizontal
line and the line of sight from
the observer to an object at a
lower level.

15
The angle of elevation and the
angle of depression are equal
in measure because they are
alternate interior angles.

16
Lesson 4 Word Problems Involving Right Triangles
Trigonometric functions can be used
to solve word problems involving right
triangles. The most common functions
used are the sine, cosine, and
tangent.
Moreover, you can use trigonometric
functions and inverse relations to
solve right triangles. To solve a
triangle means to find all the
measures of its sides and angles.
Usually, two measures are given. Then
you can find the remaining measures.

17
Example 1. A ladder is 12 feet long.
a) If the ladder is placed against a wall so that its
base is 2 feet from the wall, find, to the nearest
degree, the acute angle the ladder makes with
the ground.
b) Suppose the base of the ladder is 𝑥 feet from the
wall. Find an expression for 𝜃 , the angle the
ladder makes with the ground.

18
a) If the ladder is placed against a wall so that its base
is 2 feet from the wall, find, to the nearest degree,
the acute angle the ladder makes with the ground.
Solution.
cos 𝜃 =
2
12
𝜃 = 𝑐𝑜𝑠−1 1
6
𝜃 ≈ 80 𝑜

19
b) Suppose the base of the ladder is 𝑥 feet from the
wall. Find an expression for 𝜃, the angle the ladder
makes with the ground.
Solution.
cos 𝜃 =
𝑥
12
𝜃 = 𝑐𝑜𝑠−1 𝑥
12

20
Example 2. Latashi and Markashi are flying kites on
a windy day. Latashi has released 250
feet of string, and Markashi has
released 225 feet of string. The angle
that Latashi’s kite string makes with
the horizontal is 35 𝑜
. The angle that
Markashi’s kite string makes with the
horizontal is 42 𝑜
. Which kite is higher
and by how much?

21
Solution.
For Latashi’s kite:
sin 35 𝑜
=
ℎ
250 𝑓𝑡
ℎ = 250 𝑓𝑡 sin 35 𝑜
ℎ = 143.3941091 𝑓𝑡
Latashi’s kite has a height about 143.39 ft.
250 ft
35 𝑜
Height = ?

22
Solution.
For Markashi’s kite:
sin 42 𝑜
=
ℎ
225 𝑓𝑡
ℎ = 225 𝑓𝑡 sin 42 𝑜
ℎ = 150.5543864 𝑓𝑡
Markashi’s kite has a height about 150.55 ft.
225 ft
42 𝑜
Height = ?

23
Solution.
Let’s subtract the height of Markashi’s kite and the
height of Latashi’s kite.
150.5543864 𝑓𝑡 − 143.3941091 𝑓𝑡 = 7.160277343 𝑓𝑡
Markashi’s kite is higher than Latashi’s kite by about
7.16 ft.

24
Lesson 5 Oblique Triangles
Trigonometry enables sides and angle
measures to be found in triangles
other than right triangles.
An oblique triangle is one that does
not contain a right angle.
Oblique triangles may be classified
into two---acute and obtuse.
An acute triangle is one that has
three acute angles.
An obtuse triangle is one that has
one obtuse angle.

25
Activity
Identify the acute and obtuse triangles.

26
Activity
Identify the acute and obtuse triangles.

27
Lesson 5.1 The Law of Sines and Its Applications
Law of Sines
Let ∆𝐴𝐵𝐶 be any triangle with 𝑎, 𝑏, and
𝑐 representing the measures of the sides
opposite the angles with measures 𝐴, 𝐵,
and 𝐶, respectively.
Then, the following are true.
sin 𝐴
𝑎
=
sin 𝐵
𝑏
=
sin 𝐶
𝑐
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶

28
From geometry, you know that a unique triangle can be formed
if you know
a) the measures of two angles and the included side (ASA) or
b) the measures of two angles and the non-included side (AAS).
Therefore, there is one unique solution when you use the Law
of Sines to solve a triangle given the measures two angles and
one side.

29
From geometry, you know that
c) the measures of two sides and a non-included angle (SSA)
do not necessarily define a unique triangle. However, one
of the following will be true.
1. No triangle exists.
2. Exactly one triangle exists.
3. Two triangles exist.
In other words, there may be no solution, one solution, or two
solutions. A situation with two solutions is called the ambiguous case.

30
Suppose you know the
measures 𝑎 , 𝑏 , and 𝐴 .
Consider the following cases.

31
ASA Case
Example 1. Cartography To draw a map,
a cartographer needed to find
the distances between point 𝑍
across the lake and each of
point 𝑋 and 𝑌 on another side.
The cartographer found
𝑋𝑌 ≈ 0.3 miles, 𝑚∠𝑋 ≈ 50 𝑜
,
and 𝑚∠𝑌 ≈ 100 𝑜
. Find the
distances from 𝑋 to 𝑍 and from
𝑌 to 𝑍.
𝑋
𝑍
𝑌

32
ASA Example 1
Solution.
𝑚∠𝑍 ≈ 180 𝑜
− 100 𝑜
− 50 𝑜
≈ 30 𝑜
sin 100 𝑜
𝑋𝑍
=
sin 30 𝑜
0.3
𝑋𝑍 ∙ sin 30 𝑜 = 0.3 ∙ sin 100 𝑜
𝑋𝑍 =
0.3 sin 100 𝑜
sin 30 𝑜
𝑋𝑍 ≈ 0.59
The distance from 𝑋 to 𝑍 is about
0.59 miles.
𝑋
𝑍
𝑌
0.3 mi
50 𝑜 100 𝑜

33
ASA Example 1 (Continuation)
Solution.
𝑚∠𝑍 ≈ 180 𝑜 − 100 𝑜 − 50 𝑜 ≈ 30 𝑜
sin 50 𝑜
𝑌𝑍
=
sin 30 𝑜
0.3
Y𝑍 ∙ sin 30 𝑜 = 0.3 ∙ sin 50 𝑜
Y𝑍 =
0.3 sin 50 𝑜
sin 30 𝑜
Y𝑍 ≈ 0.46
The distance from 𝑌 to 𝑍 is about
0.46 miles.
𝑋
𝑍
𝑌
0.3 mi
50 𝑜 100 𝑜

34
AAS Case
Example 2. A hill slopes upward at an angle of 5 𝑜 with the
horizontal. A tree grows vertically on the hill.
When the angle of elevation of the sun is 30 𝑜
,
the tree casts a shadow downhill that is 32
meters long. If the shadow is entirely on the
hill, how tall is the tree?

35
AAS Case (Illustration)
Example 2. A hill slopes upward at an angle of 5 𝑜
with the
horizontal. A tree grows vertically on the hill.
When the angle of elevation of the sun is 30 𝑜
,
the tree casts a shadow downhill that is 32
meters long. If the shadow is entirely on the
hill, how tall is the tree?

36
AAS Example 2
Solution.
a) Right Triangle Involving the Hill
Alone
𝑚∠3𝑟𝑑 = 180 𝑜
− 90 𝑜
− 5 𝑜
= 85 𝑜
b) Right Triangle Involving the Tree
𝑚∠3𝑟𝑑 = 180 𝑜 − 90 𝑜 − 30 𝑜 = 60 𝑜
c) Straight Angle Involving the Hill
and Tree
180 𝑜 − 85 𝑜 = 95 𝑜
d) Oblique Triangle Above the Hill
𝑚∠3𝑟𝑑 = 180 𝑜
− 95 𝑜
− 60 𝑜
= 25 𝑜

37
AAS Example 2 (Continuation)
Solution.
sin 25 𝑜
𝑡
=
sin 60 𝑜
32
t ∙ sin 60 𝑜
= 32 ∙ sin 25 𝑜
t =
32 sin 25 𝑜
sin 60 𝑜
t ≈ 15.6
The tree is about 15.6 meters tall.

38
SSA Case
Let’s have to cases.
a) Case 1: 𝐴 < 90 𝑜
b) Case 2: 𝐴 ≥ 90 𝑜
Example 3. Determine the number of possible solutions for
each triangle.
a) 𝐴 = 30 𝑜
, 𝑎 = 8, 𝑏 = 10
b) 𝑏 = 8, 𝑐 = 10, 𝐵 = 118 𝑜

39
SSA Case
Example 3. Determine the number
of possible solutions
for each triangle.
a) 𝐴 = 30 𝑜
, 𝑎 = 8, 𝑏 = 10
Since 30 𝑜
< 90 𝑜
, consider Case 1.
𝑏 𝑠𝑖𝑛 𝐴 = 10 sin 30 𝑜
𝑏 𝑠𝑖𝑛 𝐴 = 10 0.5
𝑏 𝑠𝑖𝑛 𝐴 = 5
Since 5 < 8 < 10 , there are two
solutions for the triangle.

40
SSA Case (Continuation)
Example 3. Determine the number of
possible solutions for
each triangle.
b) 𝑏 = 8, 𝑐 = 10, 𝐵 = 118 𝑜
Since 118 𝑜
≥ 90 𝑜
, consider Case 2.
In this triangle, 8 ≤ 10 , so there
are no solutions.

41
SSA Case
Once you have determined that there is/are one or two solution(s)
for a triangle given the measures of two sides and a non-included
angle, you can use the Law of Sines to solve the triangle.
Example 4. Find all solutions for the triangle. If no
solutions exist, write none.
𝐴 = 51 𝑜
, 𝑎 = 40, 𝑐 = 50
Since 51 𝑜
< 90 𝑜
, consider Case 1.
𝑐 𝑠𝑖𝑛 𝐴 = 50 sin 51 𝑜
𝑐 sin 𝐴 ≈ 38.85729807
Since 38.9 < 40 < 50, there are two solutions
for the triangle.

42
Solution.
Use the Law of Sines to find 𝐶.
𝑎
sin 𝐴
=
𝑐
sin 𝐶
40
sin 51 𝑜 =
50
sin 𝐶
40 ∙ sin 𝐶 = 50 ∙ sin 51 𝑜
sin 𝐶 =
50 sin 51 𝑜
40
𝐶 = sin−1 50 sin 51 𝑜
40
C ≈ 76.27180414 𝑜
SSA Case
Example 4 (Continuation)
Given:
𝐴 = 51 𝑜
, 𝑎 = 40, 𝑐 = 50

43
SSA Case
Example 4 (Continuation)
So, 𝐶 ≈ 76.3 𝑜
. Since we know there are two solutions, there
must be another possible measurement for 𝐶.
In the second case, 𝐶 must be less than 180 𝑜
and have the
same sine value.
Since we know that if 𝛼 < 90 𝑜, sin 𝛼 = sin 180 𝑜 − 𝛼 , 180 𝑜 −
76.3 𝑜
or 103.7 𝑜
is another possible measure for 𝐶.

44
SSA Case Example 4 (Continuation)
Now solve the triangle for each possible measure of 𝐶.
Solution I.
𝐵 ≈ 180 𝑜 − 51 𝑜 + 76.3 𝑜
𝐵 ≈ 52.7 𝑜
𝑎
sin 𝐴
=
𝑏
sin 𝐵
40
sin 51 𝑜 =
𝑏
sin 52.7 𝑜
b sin 51 𝑜 ≈ 40 ∙ sin 52.7 𝑜
b ≈
40 sin 52.7 𝑜
sin 51 𝑜
𝑏 ≈ 40.94332444
One solution is 𝐵 ≈ 52.7 𝑜 , 𝐶 ≈ 76.3 𝑜 , and 𝑏 ≈ 40.9.
A
B
C
51 𝑜
76.3 𝑜
50
b
40

45
SSA Case Example 4 (Continuation)
Solution II.
𝐵 ≈ 180 𝑜
− 51 𝑜
+ 103.7 𝑜
𝐵 ≈ 25.3 𝑜
𝑎
sin 𝐴
=
𝑏
sin 𝐵
40
sin 51 𝑜 =
𝑏
sin 25.3 𝑜
b sin 51 𝑜
≈ 40 ∙ sin 25.3 𝑜
b ≈
40 sin 25.3 𝑜
sin 51 𝑜
𝑏 ≈ 21.99627275
Another solution is 𝐵 ≈ 25.3 𝑜 , 𝐶 ≈ 103.7 𝑜 , and 𝑏 ≈ 22.0.
A
B
C
51 𝑜
103.7 𝑜
50
b
40

46
B
C
A
a
h
b
c
The area of any triangle can be expressed in
terms of two sides of a triangle and the
measure of the included angle.
Suppose you know the measures of 𝐴𝐶 and 𝐴𝐵
and the measure of the included angle ∠𝐴 in
∆𝐴𝐵𝐶.
Let 𝐾 represent the measure of the area of
∆𝐴𝐵𝐶, and let ℎ represent the measure of the
altitude from 𝐵 . Then 𝐾 =
1
2
𝑏ℎ . But, sin 𝐴 =
ℎ
𝑐
or ℎ = 𝑐 sin 𝐴 . If you substitute 𝑐 sin 𝐴 for ℎ ,
the result is the following formula.
𝐾 =
1
2
𝑏𝑐 sin 𝐴

47
If you drew altitudes from A and C, you could also
develop two similar formulas.
Area of Triangles
Let ∆𝐴𝐵𝐶 be any triangle with 𝑎 , 𝑏 , and 𝑐
representing the measures of the sides opposite
the angles with measurements 𝐴 , 𝐵 , and 𝐶 ,
respectively. Then the area 𝐾 can be determined
using one of the following formulas.
𝐾 =
1
2
𝑏𝑐 sin 𝐴
𝐾 =
1
2
𝑎𝑐 sin 𝐵
𝐾 =
1
2
𝑎𝑏 sin 𝐶
B
C
A
a
h
b
c

48
Example 5. Find the area of ∆𝐴𝐵𝐶 if 𝑎 =
4.7 , 𝑐 = 12.4 , and 𝐵 =
47 𝑜
20′
.
𝐾 =
1
2
𝑎𝑐 sin 𝐵
𝐾 =
1
2
4.7 12.4 sin 47 𝑜
20′
𝐾 ≈ 21.42690449
The area of ∆𝐴𝐵𝐶 is about 21.4 square
units.
AB
C
4.7
12.4
47 𝑜20′

49
Deriving the Law of Cosines
• Write an equation
using Pythagorean
theorem for shaded
triangle.
b h a
k c - k
A B
C
c
Abk
Abh
cos
sin


   
 
Abccba
AbccAAba
AbAbccAba
AbcAba
cos2
cos2cossin
coscos2sin
cossin
222
22222
222222
222





50
Law of Cosines
• Similarly
• Note the pattern
Cabbac
Baccab
Abccba
cos2
cos2
cos2
222
222
222




51
Lesson 5.2 The Law of Cosines and Its Applications
Law of Cosines
Let ∆𝐴𝐵𝐶 be any triangle with 𝑎, 𝑏, and
𝑐 representing the measures of the sides
opposite the angles with measures 𝐴, 𝐵,
and 𝐶, respectively.
Then, the following are true.
𝑐2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏 cos 𝐶
𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵
𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 cos 𝐴
A
B C
b
a
c
D
a - x x
h

52
From geometry, you know that a unique triangle can be formed
if
a) the measures of two sides and an included angle are known
(SAS) or
b) the measures of three sides of a triangle are known and the
sum of any two measures is greater than the remaining
measure (SSS).

53
SAS Case
Example 1. Landscaping Suppose you want
to fence a triangular lot as shown
at the right. If two sides measure
84 feet and 78 feet and the angle
between the two sides is 102 𝑜,
what is the length of the fence to
the nearest foot?
84 ft 78 ft
102 𝑜

54
SAS Example 1
Solution.
𝑐2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏𝑐𝑜𝑠 𝐶
𝑐2
= 842
+ 782
− 2 84 78 𝑐𝑜𝑠102 𝑜
𝑐2 = 7056 + 6084 + 2724.47496
𝑐2 = 15864.4748
𝑐 ≈ 125.9542568 𝑓𝑡
Let’s add the three lengths.
84 𝑓𝑡 + 78 𝑓𝑡 + 125.95 𝑓𝑡 ≈ 288 𝑓𝑡
The length of the fence is about 288 ft.
84 ft 78 ft
102 𝑜

55
SSS Case
Example 2. The sides of a triangle are 18 inches, 21 inches,
and 14 inches. Find the measure of the angle
with the greatest measure.

56
18 in
21 in
14 in
SSS Case (Continuation)
Example 2. The sides of a triangle are 18 inches, 21 inches,
and 14 inches. Find the measure of the angle
with the greatest measure.

57
Formulas in Finding for the Angles
𝐶 = cos−1
𝑎2 + 𝑏2 − 𝑐2
2𝑎𝑏
𝐵 = cos−1
𝑎2 + 𝑐2 − 𝑏2
2𝑎𝑐
𝐴 = cos−1
𝑏2
+ 𝑐2
− 𝑎2
2𝑏𝑐

58
18 in
21 in
14 in
∠ with the greatest
measure
SSS Example 2 (Continuation)
Solution.
𝐶 = cos−1
𝑎2
+ 𝑏2
− 𝑐2
2𝑎𝑏
𝐶 = cos−1 142+182 −212
2 14 18
𝐶 = cos−1 196+324−441
504
𝐶 = cos−1 79
504
𝐶 ≈ 80.98192557 𝑜
The largest angle has a measure about
81 𝑜
.

59
References
Holliday, B. et al. (2004) Glencoe ADVANCED
Mathematical Concepts. The McGraw-Hill
Companies, Inc., United States of America
Senk, S. et al. (1998) UCSMP Functions, Statistics,
and Trigonometry. Addison Wesley Longman, Inc.,
United States of America
Website Link for Video
http://www.youtube.com/watch?v=geDSwx2TuiE

60
Website Links for Images
1. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc
h&sa=X&ei=h -lxU9voOYOllQXK 6IHYBQ&sqi=2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555
h&sa=X&ei=h-
lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=solving+right+triangles&t
bm=isch
h&sa=X&ei=h-
lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=oblique+triangles&tbm=i
sch
h&sa=X&ei=h-
lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=acute+triangles+in+the+r
eal+world&tbm=isch
h&sa=X&ei=h-
lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=obtuse+triangles+in+the+
real+world&tbm=isch
h&sa=X&ei=h-
lxU9voOYOllQXK6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=law+of+sines&tbm=isch

Module 7 triangle trigonometry super final

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