holt geometry law of sines 10.4. holt geometry calculator review 1. what is the third angle measure...
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Holt Geometry
Law of Sines 10.4
Holt Geometry
Calculator Review1. What is the third angle measure in a triangle with
angles measuring 65° and 43°?
Find each value. Round trigonometric
ratios to the nearest hundredth and angle
measures to the nearest degree.
2. sin 73° 3. cos 18° 4. tan 82°
5. sin-1 (0.34) 6. cos-1 (0.63) 7. tan-1 (2.75)
72°
0.96 0.95 7.12
20° 51° 70°
Holt Geometry
Use the Law of Sines and the Law of Cosines to solve triangles that are not right triangles.
Objectives
Solve a triangle by finding the measures of all sides and angles.
Find the area of oblique (no right angles) triangles.
Holt Geometry
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°
Holt Geometry
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.
Holt Geometry
Holt Geometry
Find the length of FG. Round lengths to the nearest tenth and angles to the nearest degree.
Law of Sines
Substitution.
Cross Multiply.
Divide both sides by sin 39.
FG sin 39° = 40 sin 32°
Using the Law of Sines AAS
AAS, not the ambiguous case, no extra care required.
Holt Geometry
Law of Sines
Substitution.
Cross Multiply.
Divide both sides by sin 39°.
NP sin 39° = 22 sin 88°
Using the Law of Sines AAS
AAS, not the ambiguous case, no extra care required.
Find the length of NP. Round lengths to the nearest tenth and angles to the nearest degree.
Holt Geometry
mA + 67° + 44° = 180° Prop of ∆.
Subtraction.
Using the Law of Sines ASA
ASA, not the ambiguous case, no extra care required.
Find the length of AC. Round lengths to the nearest tenth and angles to the nearest degree.
Law of Sines
Substitution.
Cross Multiply
Divide both sides by sin 69°.
AC sin 69° = 18 sin 67°
mA = 69°
69
Holt Geometry
AssignmentGeometry:
10-4 Law of Sines
Holt Geometry
Find mL. K = 51°, k = 10, and l = 6.
Law of Sines
Substitute the given values.
Cross Products Property
Use the inverse sine function to find mL.
10 sin L = 6 sin 125°
SSA, the ambiguous case, extra care required. Since the angle is obtuse and the opposite side is greater then the adjacent side, there is one triangle.
Using the Law of Sines SSA
SSA, the ambiguous case, extra care required. Since the angle is obtuse and the opposite side is greater than the adjacent side, there is one triangle.
Using the Law of Sines SSA
Holt Geometry
Law of Sines
Substitute the given values.
Cross Products Property
Using the inverse function on the calculator indicates no such triangle exists.
3.5 sin B = 5 sin 51°
SSA, the ambiguous case, extra care required.
Using the Law of Sines SSA
Find mB with A = 51°, a = 3.5, and b = 5.
5.3
51sin
5
Bsin
5.3
51sin5sinBm 1
Holt Geometry
EXAMPLE 3
Law of Sines
Substitute the given values.
Cross Products Property
Use the inverse function on the calculator, then check to see if another triangle exists.
13 sin B = 16 sin 40°
SSA, the ambiguous case, extra care required.
Using the Law of Sines SSA
Find mB with A = 40°, a = 13, and b = 16.
13
40sin
16
Bsin
3.5213
40sin16sinBm 1
180 - 52.3 = 127.7 See if 127.7+ 40<180 If so, then 2 triangles exist, as in this case.
Holt Geometry
Using the Law of Sines SSA
Law of Sines
Substitute the given values.
Multiply both sides by 6.
SSA, the ambiguous case, extra care required.
Find mQ with P = 51°, p = 8, and q = 6.
Use the inverse function on the calculator, then check to see if another triangle exists.
180 - 36 = 144 See if sin 144 + 51 < 180. If not, then only one triangle exists, as in this case.
Holt Geometry
a. A = 105°, b = 13, a = 6
Using the Law of Sines Mixed Review Find mB
b. A = 110°, b = 100, a = 125
Sketch and classify C
BA
613
c105
SSA obtuse
C
BA
b = 13
c
a = 6105
Since the angle is obtuse and the opposite side is less than the adjacent side, a triangle cannot be formed.
C
BA
125100
c110
Sketch and classify
SSA obtuse
C
BA
b = 100
c
a = 125110
Since the angle is obtuse and the opposite side is greater than the adjacent side, one triangle can be formed.
125
110sin
100
Bsin 7.48Bm
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c. A = 70°, a = 3.2, b = 3
d. A = 76, a = 18, b = 20C
AB
b = 20 a = 18
76
a = 3.2
C
AB
b = 3
c70
Using the Law of Sines Mixed Review Find mB
Sketch and classify
Sketch and classify
C
BA
3.23
c70
C
BA
1820
c76
SSA acute
SSA acute
2.3
70sin
3
Bsin 8.61Bm
18
76sin
20
Bsin
No such triangle exists.
180 - 61.8 = 118.2
Since118.2 + 70 > 180 only one triangle exists.
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e. A = 58, a = 11.4, b = 12.8
Using the Law of Sines Mixed Review Find mB
Sketch and classify C
BA
11.412.8
c58
SSA acute
4.11
58sin
8.12
Bsin 2.72Bm
180 - 72.2 = 107.8
C
AB
b = 12.8
c
a = 11.4
58
a = 11.4
C
AB
b = 12.8
c58
Since107.8 + 58 < 180 two triangles exist.
2.72Bm 8.107Bm
Holt Geometry
f. B = 60°, C = 10, a = 4.5
g. A = 65, B = 80, a = 17
Using the Law of Sines Mixed Review Find the length of AB
Sketch and classify
Sketch and classify
ASA acute
AAS acute
AB
10sin
5.4
110sin
8.AB
AB
35sin
17
65sin
ASA just use the Law of Sines
C
BA
b
c
60
10
a = 4.5
C
BA
4.510
60
The missing angle is 110
AAS just use the Law of Sines
C
BA
17
c65 80
The missing angle is 35
8.10AB
Holt Geometry
AssignmentGeometry:
Ambiguous Case 1
Law of Sines
Holt Geometry
Area of an Oblique Triangle
C
BA
b
c
aFind the area of the triangle.A = 74, b = 103 inches, c = 58 inches
Example:
74
103 in
58 in 1Area = sin
2bc A
103 51= ( )( )8 sin2
74
2871 square inches
Holt Geometry
AssignmentGeometry:
10-4 Area