Download Slide 1 - Souderton Math

The Law of Sines
Section 6.1
Mr. Thompson
An oblique triangle is a triangle that has no right angles.
C
a
b
A
c
B
To solve an oblique triangle, you need to know the
measure of at least one side and the measures of any
other two parts of the triangle – two sides, two angles,
or one angle and one side.
2
The following cases are considered when solving oblique triangles.
1. Two angles and any side (AAS or ASA)
A
A
c
c
B
C
2. Two sides and an angle opposite one of them (SSA)
c
C
3. Three sides (SSS)
b
a
c
a
4. Two sides and their included angle (SAS)
c
B
a
3
The first two cases can be solved using the Law of Sines. (The last
two cases can be solved using the Law of Cosines.)
Law of Sines
If ABC is an oblique triangle with sides a, b, and c, then
a  b  c .
sin A sin B sin C
C
C
a
b
h
h
A
c
Acute Triangle
B
a
b
c
A
Obtuse Triangle
B
Example (ASA):
Find the remaining angle and sides of the triangle.
C
The third angle in the triangle is
A = 180 – A – B
= 180 – 10 – 60
= 110.
10
a = 4.5 ft
4.15 ft b
60
110
A
Use the Law of Sines to find side b and c.
0.83 ft
a  b
sin A sin B
a  c
sin A sin C
4.5  b
sin 110 sin 60
4.5  c
sin 110 sin 10
b  4.15 feet
c
c  0.83 feet
B
For the triangle in the figure, C = 102.3°, B = 28.7°, and b =
27.4 feet. Find the remaining angle and sides.
Ex. 2. A pole tilts toward the sun at an 8° angle from the
vertical, and it casts a 22-foot shadow. The angle of
elevation from the tip of the shadow to the top of the
pole is 43°. How tall is the pole?
Remember back to Geometry
-Which cases were enough to
prove two triangles congruent?
-Why was there a problem with
Some of the cases?
Try to come up with counterexamples for these cases
The Ambiguous Case (SSA) Table
Knowing two sides and an angle opposite one of these
sides might not be enough to form a triangle. This
information may allow you to form 0, 1, or 2 triangles!
Example (SSA):
Use the Law of Sines to solve the triangle.
A = 110, a = 125 inches, b = 100 inches
C
21.26
a  b
sin A sin B
125  100
sin 110 sin B
B  48.74
C  180 – 110 – 48.74
= 21.26
a = 125 in
b = 100 in
110
A
48.74
c
B
48.23 in
a  c
sin A sin C
125 
c
sin 110 sin 21.26
c  48.23 inches
Example (SSA):
Use the Law of Sines to solve the triangle.
A = 76, a = 18 inches, b = 20 inches
a  b
sin A sin B
18  20
sin 76 sin B
sin B  1.078
C
b = 20 in
a = 18 in
76
B
There is no angle whose sine is 1.078.
There is no triangle satisfying the given conditions.
A
Show that there is no triangle for which
a = 15, b = 25, and A = 85°.
Example (SSA):
Use the Law of Sines to solve the triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
a  b
sin A sin B
11.4  12.8
sin 58 sin B
B1  72.2
C  180 – 58 – 72.2 = 49.8
C
49.8
b = 12.8 cm
a = 11.4 cm
B1
72.2
58
c
A
10.3 cm
c  b
sin C sin B
c
 12.8
sin 49.8 sin 72.2
c  10.3
Two different triangles can be formed.
Example continues.
Example (SSA) continued:
Use the Law of Sines to solve the second triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
B2  180 – 72.2 = 107.8 
49.8
b = 12.8 cm
a = 11.4 cm
C  180 – 58 – 107.8 = 14.2
c  b
sin C sin B
c
 12.8
sin 14.2 sin 72.2
c  3.3
C
72.2
58
c
A
B1
10.3 cm
C
14.2
b = 12.8 cm
a = 11.4 cm
58
107.8
B2
c
3.3 cm
A
Find two triangles for which a = 12 meters, b = 31 meters, and
A = 20.5°.
My Old House
Art Museum
City Hall
Area of an Oblique Triangle
Area  1 bc sin A  1 ab sin C  1 ac sin B
2
2
2
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
103 in
a
b
A
74
c
58 in
B
The following cases are considered when solving oblique triangles.
1. Two angles and any side (AAS or ASA)
A
A
c
c
B
C
2. Two sides and an angle opposite one of them (SSA)
c
C
3. Three sides (SSS)
b
a
c
a
4. Two sides and their included angle (SAS)
c
B
a
18
The last two cases (SSS and SAS) can be solved using the
Law of Cosines.
(The first two cases can be solved using the Law of Sines.)
Law of Cosines
Standard Form
Alternative Form
a  b  c  2bc cos A
2
2
2
b

c

a
cos A 
2bc
b  a  c  2ac cos B
2
2
2
a

c

b
cos B 
2ac
c  a  b  2ab cos C
2
2
2
a

b

c
cos C 
2ab
2
2
2
2
2
2
2
2
2
19
Example:
Find the three angles of the triangle.
2
2
2
a

b

c
cos C 
2ab
2
2
2
8

6

12

2(8)(6)
 64  36  144
96
 44
96
C  117.3
Law of Sines:
C
117.3
8
6
36.3
A
12
 6
sin 117.3 sin B
26.4
B
12
Find the angle
opposite the longest
side first.
B  26.4
A  180 117.3  26.4  36.3
20
C
Example:
Solve the triangle.
9.9
Law of Cosines:
b 2  a 2  c 2  2ac cos B
A
67.8
6.2
75
37.2
9.5
B
 (6.2)2  (9.5)2  2(6.2)(9.5) cos 75
 38.44  90.25  (117.8)(0.25882)
 98.20
b  9.9
Law of Sines: 9.9
 6.2
sin 75 sin A
A  37.2
C  180  75  37.2  67.8
21
Example 1: Find the three angles of the triangle shown.
22
Example 2: Find the remaining angles and sides of the
triangle shown.
23
Heron’s Area Formula
Given any triangle with sides of lengths a, b, and c, the
area of the triangle is given by
Area  s(s  a)(s  b)(s  c)
where s  a  b  c .
2
10
8
Example:
Find the area of the triangle.
5
s  a  b  c  5  8  10  11.5
2
2
Area  11.5(11.5  5)(11.5  8)(11.5 10)
 19.8 square units
24
Application:
Two ships leave a port at 9 A.M. One travels at a bearing of N
53 W at 12 mph, and the other travels at a bearing of S 67 W
at 16 mph. How far apart will the ships be at noon?
N
At noon, the ships have traveled for 3 hours.
53
36 mi
Angle C = 180 – 53 – 67 = 60
43 mi c
60
c 2  a 2  b 2  2ab cos C
 36  48  2(36)(48) cos 60  1872
2
2
C
48 mi
67
c  43 mi
The ships will be approximately 43 miles apart.
25
Example 5: Find the area of a triangle having sides a = 43 meters,
b = 53 meters, and c = 72 meters.
26