Transcript Holt Geometry

Law of Sines 10.4
Holt Geometry
Calculator Review
1. What is the third angle measure in a triangle with
angles measuring 65° and 43°?
72°
Find each value. Round trigonometric
ratios to the nearest hundredth and angle
measures to the nearest degree.
2. sin 73° 0.96
3. cos 18° 0.95 4. tan 82° 7.12
5. sin-1 (0.34)
6. cos-1 (0.63)
20°
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51°
7. tan-1 (2.75)
70°
Objectives
Use the Law of Sines and the Law of
Cosines to solve triangles that are not
right triangles.
Solve a triangle by finding the
measures of all sides and angles.
Find the area of oblique (no right angles)
triangles.
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Finding Trigonometric Ratios
for Obtuse Angles
Use your calculator to find each trigonometric
ratio. Round to the nearest hundredth.
A. tan 103°
B. cos 165°
C. sin 93°
D. tan 175°
E. cos 92°
F. sin 160°
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You can use the Law of Sines to solve a triangle if you
are given
• two angle measures and any side length
(ASA or AAS) or
• two side lengths and a non-included angle measure
(SSA); called the ambiguous case and requires extra
care.
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Using the Law of Sines AAS
Find the length of FG. Round lengths to
the nearest tenth and angles to the
nearest degree.
AAS, not the ambiguous case,
no extra care required.
Law of Sines
Substitution.
FG sin 39° = 40 sin 32°
Cross Multiply.
Divide both sides by sin 39.
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Using the Law of Sines AAS
Find the length of NP. Round lengths
to the nearest tenth and angles to the
nearest degree.
AAS, not the ambiguous case,
no extra care required.
Law of Sines
Substitution.
NP sin 39° = 22 sin 88° Cross Multiply.
Divide both sides by sin 39°.
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Using the Law of Sines ASA
Find the length of AC. Round lengths to
the nearest tenth and angles to
69
the nearest degree.
ASA, not the ambiguous case,
no extra care required.
mA + 67° + 44° = 180°
mA = 69°
Prop of ∆.
Subtraction.
Law of Sines
Substitution.
AC sin 69° = 18 sin 67°
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Cross Multiply
Divide both sides
by sin 69°.
Assignment
Geometry:
10-4 Law of Sines
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Using the Law of Sines SSA
Find mL.
K = 51°, k = 10, and l = 6.
SSA, the ambiguous case, extra care required.
Since the angle is obtuse and the opposite side
is greater than
then the adjacent side, there is one
triangle.
Law of Sines
Substitute the given values.
10 sin L = 6 sin 125°
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Cross Products Property
Use the inverse sine
function to find mL.
Using the Law of Sines SSA
Find mB with
A = 51°, a = 3.5, and b = 5.
SSA, the ambiguous case, extra care required.
Law of Sines
sin B sin 51

5
3.5
3.5 sin B = 5 sin 51°
 5 sin 51 
mB  sin 

 3.5 
1
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Substitute the given values.
Cross Products Property
Using the inverse function
on the calculator indicates
no such triangle exists.
EXAMPLE 3 Using the Law of Sines SSA
Find mB with
A = 40°, a = 13,
and b = 16.
SSA, the ambiguous case, extra care required.
Law of Sines
sin B sin 40

16
13
Substitute the given values.
Cross Products Property
Use the inverse function on
16
sin
40


1 
mB  sin 
  52.3 the calculator, then check to
 13 
see if another triangle exists.
180 - 52.3 = 127.7
See if 127.7+ 40<180
If so, then 2 triangles
exist, as in this case.
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13 sin B = 16 sin 40°
Using the Law of Sines SSA
Find mQ with P = 51°,
p = 8, and q = 6.
SSA, the ambiguous case,
extra care required.
Law of Sines
Substitute the given
values.
Multiply both sides by 6.
180 - 36 = 144
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Use the inverse function on
the calculator, then check to
see if another triangle exists.
See if sin 144 + 51 < 180.
If not, then only one triangle
exists, as in this case.
Using the Law of Sines Mixed Review
Find mB
a. A = 105°, b = 13, a = 6
Since the angle is obtuse and the
Sketch and classify
C
6
13
c
SSA obtuse
B
b. A = 110°, b = 100, a = 125
Sketch and classify
C
c
SSA obtuse
B
B
c
Since the angle is obtuse and the
opposite side is greater than the
adjacent side, one triangle can be
formed.
C
b = 100
110
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A
125
100
A
a=6
105
b = 13
105
A
opposite side is less than the
adjacent side, a triangle cannot be
formed.
C
a = 125
110
A
c
B
sin B sin 110 mB  48.7

100
125
Using the Law of Sines Mixed Review
Find mB
sin B sin 70
c. A = 70°, a = 3.2, b = 3
mB  61.8

Sketch and classify
3
3.2
C
180 - 61.8 = 118.2
C
3.2
3
A
70
c
SSA acute
B
B
d. A = 76, a = 18, b = 20
Sketch and classify
C
A
76
c
SSA acute
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sin B sin 76

20
18
No such
triangle exists.
18
20
Since
118.2 + 70 > 180
a = 3.2
only one triangle
exists.
B
b=3
70
c
A
C
b = 20
a = 18
76
B
A
Using the Law of Sines Mixed Review
Find mB
sin B sin 58

12.8
11.4
e. A = 58, a = 11.4, b = 12.8
Sketch and classify
C
12.8
A
180 - 72.2 = 107.8
Since
107.8 + 58 < 180
two triangles exist.
11.4
58
c
SSA acute
B
C
C
b = 12.8
a = 11.4
58
B
c
b = 12.8
a = 11.4
58
A
mB  72.2
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mB  72.2
B
c
mB  107.8
A
Using the Law of Sines Mixed Review
Find the length of AB
f. B = 60°, C = 10, a = 4.5 ASA just use the Law of Sines
Sketch and classify
The missing angle is 110
C
C
sin 110 sin 10
10
4.5
10
60
A
B
4.5
AB  .8

AB
a = 4.5
b
ASA acute
A
g. A = 65, B = 80, a = 17
Sketch and classify
C
17
A
65
80
c
AAS acute
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60
B
AAS just use the Law of Sines
The missing angle is 35
sin 65 sin 35

17
AB
AB  10.8
c
B
Assignment
Geometry:
Ambiguous Case 1
Law of Sines
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Area of an Oblique Triangle
1 bc
1 ab
1 bc
1Area
1abac
sin
sin
CB
Area
 11 bc
sin
A
 1
Area
Area
 1 bc
sinsin
AA
 1 ab
sin
Area
sin
CC

ac
ac
sinsin
A
BB
1 1bcab
sinAC
sin
22
22 2
2
22
22
22
C
Example:
Find the area of the triangle.
A = 74, b = 103 inches, c = 58 inches
Area = 1 bc sin A
2
= 1 (103)(58) sin 74
2
 2871 square inches
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103 in
a
b
A
74
c
58 in
B
Assignment
Geometry:
10-4 Area
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