Transcript Math 112 – Elementary Functions

Math 112
Elementary Functions
Chapter 7 – Applications of Trigonometry
Section 1
The Law of Sines
Solving Right Triangles
– Revisited!
Solving Triangles?

Using given information, determine the lengths of the
sides and measures of the angles.
What must be known to solve a right triangle?
Lengths of two sides.
 Length of a side and the measure of an acute angle.

How do you solve the triangle?

sin  = opp/hyp
cos  = adj/hyp
tan  = opp/adj
Solving Oblique Triangles
– Five Cases
Given 1 side and 2 angles

20
AAS and ASA
40°
60°
40°
60°
25
Solving Oblique Triangles
– Five Cases
Given 1 side and 2 angles

AAS and ASA
Given 2 sides and 1 angle

20
60°
25
SSA and SAS
20
40°
25
Solving Oblique Triangles
– Five Cases
Given 1 side and 2 angles

AAS and ASA
Given 2 sides and 1 angle

SSA and SAS
Given 3 sides

SSS
20
13
25
Solving Oblique Triangles
– Five Cases
Law of Sines (this section)

Used to solve AAS, ASA, and SSA triangles.
20
20
40°
40°
60°
60°
60°
25
25
Law of Cosines (next section)

Used to solve SAS and SSS triangles.
20
40°
25
20
13
25
The Law of Sines
– Acute Triangle
C
b
h
A
c
h
sin A 
b

h  b sin A
h
a

h  a sin B
a
sin B 
B
b sin A  a sin B

a
b

sin A sin B
The Law of Sines
– Obtuse Triangle
C
h
sin A 
b
b
a
A
c
h



h
sin 180  B 
a

B
b sin A  a sin B

a
b

sin A sin B
h  b sin A

h  a sin B
The Law of Sines
C
b
A
a
c
a
b
c


sin A sin B sin C
B
Solving Oblique Triangles
– AAS w/ the Law of Sines
1.
Find the third angle.
 = 180° - (60° + 40°) = 80°
2.
Use the law of sines to find a
second side.
x/sin 40° = 20/sin 60°
x  14.8
3.
Use the law of sines to find the
third side.
y/sin 80° = 20/sin 60°
y  22.7

20
x
40°
60°
y
NOTE:
Always use
EXACT values
if possible.
Solving Oblique Triangles
– ASA w/ the Law of Sines
1.
Find the third angle.
 = 180° - (60° + 40°) = 80°
2.
Use the law of sines to find a
second side.
x/sin 40° = 25/sin 80°
x  16.3
3.
Use the law of sines to find the
third side.
y/sin 60° = 25/sin 80°
y  22.0

y
x
40°
60°
25
NOTE:
Always use
EXACT values
if possible.
Solving Oblique Triangles
– SSA w/ the Law of Sines
With AAS and ASA, the given data will
determine a unique triangle.
With SSA, the given data could determine …
no triangle
 one triangle
 two triangles

?
60°
20
25
Solving Oblique Triangles
– SSA w/ the Law of Sines
Case 1: No Solution

The side opposite the given angle is not long enough to reach
the other side of the angle.
22
8

sin B sin 40
C
8
22
40°
A
B?

22 sin 40
sin B 
 1.77
8

B  sin 1 1.77

No Solution!
Solving Oblique Triangles
– SSA w/ the Law of Sines
Case 2a: One Solution

The side opposite the given angle is just barely long enough to
reach the other side of the angle.
C
22
11
22
11

sin B sin 30

22 sin 30
sin B 
 1
11

B  sin 1 1  90

Right Triangle!
30°
A
B?
NOTE: Angle C and side c still
need to be determined.
Solving Oblique Triangles
– SSA w/ the Law of Sines
Case 3: Two Solutions

The side opposite the given angle is more than long enough to
reach the other side of the angle but is shorter than the other
given side.
22
20

sin B sin 40
C
22
20
20

22 sin 40
sin B 
 0.707
20

B  sin 1 0.707 

B  45 or 135
40°
A
B?
B?
NOTE: Angle C and side c still need to
be determined for BOTH solutions.
Solving Oblique Triangles
– SSA w/ the Law of Sines
Case 2b: One Solution

The side opposite the given angle is more than long enough to
reach the other side of the angle but is longer than the other
given side.
C
22
22
25

sin B sin 40
25

22 sin 40
sin B 
 0.566
25

B  sin 1 0.566

B  34.4 or 145.6
40°
A
NOTE: Angle C and side c still
need to be determined.
B?
Since 145.6 + 40  180,
this solution is invalid.
The Area of a Triangle
area = ½bh
sin C = h / a  h = a sin C
a
h
Therefore, …
area = ½ ab sin C
b
C
In general, the area of a triangle is half the
product of two sides times the sine of the
included angle.