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Warm Up Chapter 4
1.
2.
3.
Find the inverse of the function
3
𝑓(𝑥) = 2 + 𝑥 − 5
Sketch the graph of g and g-1.Then find a rule for
g-1(x).
2
𝑔 𝑥 = 𝑥 + 2, 𝑥 ≥ 0
A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
a. Express the volume V of the box as a function of
the base width w.
b. What dimensions will produce a box with
maximum volume?
Warm Up Chapter 4
1.
3
Find the inverse of the function 𝑓(𝑥) = 2 + 𝑥 − 5,
𝑦 =2+
𝑥 =2+
𝑥−2=
𝑥−2
𝑥−2
3
3
3
3
3
𝑥−5
𝑦−5
𝑦−5
=𝑦−5
+5=𝑦
𝑓
−1
3
(𝑥) = 𝑥 − 2 + 5
Warm Up Chapter 4
2. Sketch the graph of g and g-1. 𝑔 𝑥 = 𝑥 2 + 2,
𝑥≥0
Then find a rule for g-1(x).
2
g
𝑥 =𝑦 +2
𝑥 − 2 = 𝑦2
g
𝑥−2=𝑦
g-1
𝑔
−1
𝑥 = 𝑥 − 2,
𝑥≥2
Warm Up Chapter 4
3. A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
a. Express the volume V of the box as a function of
the base width w.
V (w, h)  w h
2
𝑆𝐴: 𝑤 2 + 4𝑤ℎ = 400
4𝑤ℎ = 400 −𝑤 2
400 −𝑤 2
ℎ=
4𝑤
h
Write h in terms of w.
𝑉 𝑤 = 𝑤2
w w
400 −𝑤 2
4𝑤
400𝑤 −𝑤
𝑉 𝑤 =
4
3
Warm Up Chapter 4
3. A manufacturer wants to design an open top box
with a square base and a surface area of 400 in2.
b. What dimensions will produce a box with
maximum volume?
400𝑤 −𝑤
𝑉 𝑤 =
4
w = 11.55 𝑖𝑛
3
Calculate max value
on the calculator.
(11.55,769.80)
h
w
w
400 −𝑤 2 400 − 11.552
ℎ=
=
= 5.77 𝑖𝑛
4𝑤
4(11.55)
Section 9.3
Law of Sines
OBJECTIVE: TO USE THE LAW OF
SINES TO FIND UNKNOWN PARTS
OF A TRIANGLE
The Law of SINES
For any triangle (right, acute or obtuse), you
may use the following formula to solve for
missing sides or angles:
a
b
c


sin A sin B sin C
Use Law of SINES when you
are given...
 AAS - 2 angles and 1 adjacent side
 ASA - 2 angles and their included side
 SSA - 2 sides and 1 angle that is not
the included angle (ambiguous case)
Example 1
You are given a triangle, ABC,
with angle A = 70°, angle B = 80°
and side a = 12 cm. Find the
measures of angle C and sides b
and c.
Example 1
The angles in a ∆ total 180°,
so angle C = 30°.
B
80°
a = 12
c
A 70°
Set up the Law of Sines to
find side b:
b
C
12
b

sin 70 sin 80
12 sin 80  b sin 70
12 sin 80
b
 12.6cm
sin 70
Example 1
Set up the Law of
Sines to find side c:
B
80°
c
A 70°
a = 12
b = 12.6
30°
12
c

sin 70 sin 30
C
12 sin 30  c  sin 70
12 sin 30
c
 6.4cm
sin 70
Example 1
B
80°
A 70°
Angle C = 30°
a = 12
b = 12.6
30°
Side b = 12.6 cm
C
Side c = 6.4 cm
Example 2
You are given a triangle, ABC,
with angle C = 115°, angle B = 30°
and side a = 30 cm. Find the
measures of angle A and sides b
and c.
Example 2
To solve for the missing sides or
angles, we must have an angle and
opposite side to set up the first
equation.
B
30°
c
We MUST find angle A first because
the only side given is side a.
a = 30
C
115°
b
A
The angles in a ∆ total 180°, so angle
A = 35°.
Example 2
Set up the Law of
Sines to find side b:
B
30
b

sin 35 sin 30
30°
c
a = 30
115° 35°
C
b
30 sin 30  b sin 35
A
30 sin 30
b
 26.2cm
sin 35
Example 2
Set up the Law of
Sines to find side c:
B
30
c

sin 35 sin 115
30°
c
a = 30
30 sin 115  c  sin 35
C
115°
35°
b = 26.2
A
30 sin 115
c
 47.4cm
sin 35
Example 2
Angle A = 35°
B
Side b = 26.2 cm
30°
c = 47.4
Side c = 47.4 cm
a = 30
C
115°
35°
b = 26.2
A
Note: Use the Law of Sines
whenever you are given 2
angles and one side!
The Ambiguous Case (SSA)
When given SSA (two sides and
an angle that is NOT the included
angle), the situation is ambiguous.
The dimensions may not form a
triangle, or there may be 1 or 2
triangles with the given
dimensions.
The Ambiguous Case (SSA)
if the given angle is if opp < adj  no solution
if opp > adj  one solution
obtuse
if opp < h  no solution
adj
if the given angle is if h < opp < adj  2 solutions A
one angle is B acute, one angle B is
acute
obtuse
find the height,
h = adj ∙sinA
(supplementary)
if opp > adj  1 solution
If opp = h  1 solution
angle B is right
opp
h
𝐴 = given
angle
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
Given a triangle with angle A = 40°, side a = 12 cm
and side b = 15 cm, find the other dimensions.
Find the height:
C
15 = b
(adj)
A
40°
h
c
a = 12
(opp)
B
h  bsin A
h  15sin 40  9.6
Since h < opp (a) < adj (b),
there are 2 solutions and we
must find BOTH.
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
FIRST SOLUTION: Angle B is acute - this is the solution you get when
you use the Law of Sines!
C
a = 12
15 = b
A
h
40°
c
B
12
15

sin 40 sin B
15sin 40 
B  sin 1 
 53.5
 12

C  180  40  53.5  86.5
c
12

sin 86.5 sin 40
12sin 86.5
c
 18.6
sin 40
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
SECOND SOLUTION: Angle B is obtuse - use the first solution
to find this solution.
In the second set of possible
dimensions, angle B is obtuse,
C
because side ‘a’ is the same in both
1st ‘a’
solutions, the acute solution for
15 = b
a = 12
angle B & the obtuse solution for
40°
A
angle B are supplementary.
c B
1st ‘B’
Angle B = 180 - 53.5° = 126.5°
The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
SECOND SOLUTION: Angle B is obtuse
C
Angle C = 180°- 40°- 126.5° = 13.5°
15 = b
A
a = 12
40° 126.5°
c
Angle B = 126.5°
B
c
12

sin13.5 sin 40
12sin 13.5
c
 4.4
sin 40
The Ambiguous Case (SSA)
Situation II: Angle A is acute - Example
Angle B = 53.5°
Angle C = 86.5°
Side c = 18.6
Angle B = 126.5°
Angle C = 13.5°
Side c = 4.4
C
15 = b
A
40°
86.5°
13.5°
15 = b
a = 12
53.5°
B
c = 18.6
A
40° 126.5°
c = 4.4 B
C
a = 12
Homework
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