1. Law of Sines 10.4 1

2. Calculator Review
1. 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°

3. 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.

4. 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°

5. 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.

7. Using the Law of Sines AAS
AAS, not the ambiguous case, no extra care required.
Find the length of FG. Round lengths to the nearest tenth and angles to the nearest degree.
Law of Sines
Substitution.
FG sin 39° = 40 sin 32°
Cross Multiply.
Divide both sides by sin 39.

8. 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.
Law of Sines
Substitution.
NP sin 39° = 22 sin 88°
Cross Multiply.
Divide both sides by sin 39°.

9. 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.
mA = 69°
69
mA + 67° + 44° = 180°
Prop of ∆.
Subtraction.
Law of Sines
Substitution.
AC sin 69° = 18 sin 67°
Cross Multiply
Divide both sides
by sin 69°.

10. Assignment
Geometry:
10-4 Law of Sines

11. Using the Law of Sines SSA
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
Find mL.
K = 51°, k = 10, and l = 6.
Law of Sines
Substitute the given values.
Cross Products Property
10 sin L = 6 sin 125°
Use the inverse sine function to find mL.

12. Using the Law of Sines SSA
SSA, the ambiguous case, extra care required.
Using the Law of Sines SSA
Find mB with
A = 51°, a = 3.5, and b = 5.
Law of Sines
Substitute the given values.
3.5 sin B = 5 sin 51°
Cross Products Property
Using the inverse function on the calculator indicates no such triangle exists.

13. Using the Law of Sines SSA
EXAMPLE 3
See if 127.7+ 40<180
If so, then 2 triangles
exist, as in this case.
Find mB with
A = 40°, a = 13,
and b = 16.
SSA, the ambiguous case, extra care required.
Law of Sines
Substitute the given values.
13 sin B = 16 sin 40°
Cross Products Property
Use the inverse function on the calculator, then check to see if another triangle exists.
180  = 127.7

14. Using the Law of Sines SSA
See if sin 144 + 51 < 180.
If not, then only one triangle exists, as in this case.
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.
Use the inverse function on the calculator, then check to see if another triangle exists.
180 - 36 = 144

15. Using the Law of Sines Mixed Review Find mB
a. A = 105°, b = 13, a = 6 C B A
b = 13 c
a = 6
105
Since the angle is obtuse and the opposite side is less than the adjacent side, a triangle cannot be formed.
Sketch and classify C B A 6 13 c
105
SSA obtuse C B A
b = 100 c
a = 125
110
Since the angle is obtuse and the opposite side is greater than the adjacent side, one triangle can be formed.
b. A = 110°, b = 100, a = 125
Sketch and classify C B A
125
100 c
110
SSA obtuse

16. Using the Law of Sines Mixed Review Find mB
c. A = 70°, a = 3.2, b = 3
Sketch and classify
a = 3.2 C A B
b = 3 c
70 C B A
3.2 3 c
70
180  = 118.2
Since
118.2 + 70 > 180 only one triangle exists.
SSA acute
d. A = 76, a = 18, b = 20 C A B
b = 20
a = 18
76
Sketch and classify C B A 18 20 c
76
No such triangle exists.
SSA acute

17. Using the Law of Sines Mixed Review Find mB
e. A = 58, a = 11.4, b = 12.8
Sketch and classify
180  = 107.8 C B A
11.4
12.8 c
58 C A B
b = 12.8 c
a = 11.4
58
Since
107.8 + 58 < 180 two triangles exist.
SSA acute

18. 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 B A b c
60
10
a = 4.5 C B A
4.5
10
60
ASA acute
g. A = 65, B = 80, a = 17
AAS just use the Law of Sines
Sketch and classify
The missing angle is 35 C B A 17 c
65
80
AAS acute

19. Assignment
Geometry:
Ambiguous Case 1
Law of Sines

20. Area of an Oblique TriangleC B A b c a
Find the area of the triangle.
A = 74, b = 103 inches, c = 58 inches
Example:
103 in
74
58 in

21. Assignment
Geometry:
10-4 Area