Transcript ppt

7.1 The Sines Law
1. Determining if the Law of Sines Can Be Used to Solve an
Oblique Triangle
2. Using the Law of Sines to Solve the SAA Case or ASA Case
3. Using the Law of Sines to Solve the SSA Case
4. Using the Law of Sines to Solve Applied
5. Problems Involving Oblique Triangles
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Dr .Hayk Melikyan/ Departmen of Mathematics and CS/ [email protected]
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The Law of Sines
If A, B, and C are the measures of the angles of any triangle
and if a, b, and c are the lengths of the sides opposite the
corresponding angles, then
a
b
c
sin A sin B sin C


or


sin A sin B sin C
a
b
c
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The Law of Sines
The following information is needed to use the Law of Sines.
1. The measure of an angle must be known.
2. The length of the side opposite the known angle must be
known.
3. At least one more side or one more angle must be known.
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Determining of the Law of Sines Can Be Used to Solve an
Oblique Triangle
Decide whether or not the Law of Sines can be used to solve
each triangle.
A
a.
b.
c.
A
47°
A
99°
b
b = 40
c = 54
c
b = 15.4
c = 14
37°
50°
B
C
B
C
a
a
C
a = 28
B
No; We are not given
the measure of any
angle.
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Yes; we are given an
angle and the measure
of its opposite side and
an additional angle.
Yes; we are given an
angle and the measure
of its opposite side and
an additional angle.
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Text Example (ASA)
Solve triangle ABC if A  50º, C  33.5º, and b  76.

Solution
We begin by drawing a picture of triangle ABC and labeling it
with the given information. The figure shows the triangle that we must solve.
We begin by finding B.
C
A + B + C  180º
33.5º
b 76
a
50º + B + 33.5º  180º
83.5º + B  180º
50º
A
c
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B
B  96.5º
The sum of the measurements of a
triangle’s interior angles is 180º.
A = 50º and C = 33.5º.
Add.
Subtract 83.5º from both sides.
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Text Example cont.
Solve triangle ABC if A  50º, C  33.5º, and b  76.
Keep in mind that we must be given one of the three ratios to apply the
Law of Sines. In this example, we are given that b  76 and we found that B  96.5º.
Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law
of Sines to find a and c.
Solution
C
Find a:
Find c:
This is the known ratio.
33.5º
b 76
a
50º
A
c
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B
a
b

sin A sin B
a
76

sin 50
sin 96.5
76sin 50
a
 59
sin 96.5
c
b

sin C sin B
c
76

sin 33.5 sin 96.5
76sin 33.5
c
 42
sin 96.5
The solution is B  96.5º, a  59, and c  42.
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Solving a SAA Triangle Using the Law of Sines
Solve the given oblique triangle. Round lengths to one
decimal place.
Angles
A
99°
b
c = 14
50°
C
B
Sides
A = 99
a=?
B = ??
b=?
C = 50
c = 14
a
A + B + C  180
99 + B + 50  180
149 + B  180
B  31
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Solving a SAA Triangle Using the Law of Sines
Solve the given oblique triangle. Round lengths to one
decimal place.
Find a:
Find b:
A
c = 14
99°
b
50°
C
B
a
Angles
Sides
A = 99
a=?
B = 31
b=?
C = 50
c = 14
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a
c

sin A sin C
a
14

sin 99 sin 50
14sin 99
a
sin 50
a  18.1
b
c

sin B sin C
b
14

sin 31 sin 50
14sin 31
b
sin 50
b  9.4
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Solving a ASA Triangle Using the Law of Sines
Solve oblique triangle ABC if B = 42, A = 57, and c = 18.6 cm.
C  180  57  42  81
A
c = 18.6
57°
b
42°
B
a
C
Angles
Sides
A = 57
a=?
B = 42
b=?
C=
c = 18.6
81
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a
c

sin A sin C
a
18.6

sin 57 sin81
18.6sin 57
a
sin81
a  15.8
b
c

sin B sin C
b
18.6

sin 42 sin81
18.6sin 42
a
sin81
a  12.6
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The Ambiguous Case (SSA)
a is less than h
and not long
enough to form a
triangle.
a
b
h = b sin A
A
a = h and is just
the right length
to form a right
triangle.
b
a
h = b sin A
A
No Triangle
One Right Triangle
a is greater than h
and a is less than
b. Two distinct
triangles are
formed.
A
b
a
a h = b sin A
a is greater than h
and a is greater
than b. One
triangle is formed.
b
a
A
Two Triangles
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One Triangle
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Solving a SSA Triangle Using the Law of Sines ( Quiz No)
Solve the triangle; if possible.
Angles
Sides
A=?
a = 20
B=?
b=
C = 50
c = 10
sin A sin C

a
c
sin A sin 50

20
10
20sin 50
sin A 
10
sin A  1.5
Since there is no angle A for which sin A > 1, there can
be no triangle with the given measurements.
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Solving a SSA Triangle Using the Law of Sines (One Triangle)
Solve the triangle; if possible.
Angles
Sides
A = 40
a = 30
B=?
b = 20
C=?
c=?
Two possible angles:
sin A sin B

a
b
sin 40 sin B

30
20
20sin 40
sin A 
30
sin B  0.4285
25.4 or 180  25.4  154.6
154.6 is too large when combined with the given angle.
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Solving a SSA Triangle Using the Law of Sines (One Triangle)-cont
Solve the triangle; if possible.
Angles
Sides
A = 40
a = 30
B = 25.4
b = 20
C = 114.6
c=?
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C  180  40  25.4  114.6
sin A sin C

a
c
sin 40 sin114.6

30
c
30sin114.6
c
sin 40
c  42.3
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Solving a SSA Triangle Using the Law of Sines (Two Triangles)
Solve the triangle; if possible.
Angles
Sides
A = 35
a = 60
B=?
b = 80
C=?
c=?
sin A sin B

a
b
sin 35 sin B

60
80
80sin 35
sin B 
60
sin B  0.7648
Two possible angles:
49.9 or 180  49.9  130.1
There are two possible triangles.
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Solving a SSA Triangle Using the Law of Sines (Two Triangles)
Solve the triangle; if possible.
Angles
Sides
Angles
Sides
A = 35
a = 60
A = 35
a = 60
B = 49.9
b = 80
B = 130.1
b = 80
C=
c = 104.2 C = 14.9
95.1
C  180  35  49.9  95.1
c=
sin A sin C

a
c
sin 35 sin 95.1

60
c
60sin 95.1
c
sin 35
c  104.2
or
C  180  35  130.1  14.9
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Solving a SSA Triangle Using the Law of Sines (Two Triangles)
Solve the triangle; if possible.
Angles
Sides
Angles
Sides
A = 35
a = 60
A = 35
a = 60
B = 49.9
b = 80
B = 130.1
b = 80
C=
95.1
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c = 104.2 C = 14.9
c=
26.9
sin A sin C

a
c
sin 35 sin14.9

60
c
60sin14.9
c
sin 35
c  26.9
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Solving an Applied Problem
Martin wants to measure the distance across a river. He has
made a sketch. Find the distance across the river, a.
C  180  23  126  31
23°
348 feet
31
a
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126°
a
b

sin A sin B
a
348

sin 23 sin 31
348sin 23
a
sin 31
a  264 ft
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Determining the Distance a Ship is from Port
A ship set sail from port at a bearing of N 53E and sailed 63 km to point
B. The ship then turned and sailed an additional 69 km to point C.
Determine the distance from port to point C if the ship’s final bearing is
N 74E.
Draw a diagram.
Find angle A.
A  74  53  21
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Determining the Distance a Ship is from Port-cont
A ship set sail from port at a bearing of N 53E and sailed 63 km to point
B. The ship then turned and sailed an additional 69 km to point C.
Determine the distance from port to point C if the ship’s final bearing is N
74E.
Angles
Sides
A = 21
a = 69
B=
b=
C=
c = 63
Two possible angles:
19.1 or 180  19.1  160.9
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sin C sin A

c
a
sin C sin 21

63
69
63sin 21
sin A 
69
sin A  0.3272
160.9 will not work
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Determining the Distance a Ship is from Port-cont
A ship set sail from port at a bearing of N 53E and sailed 63 km to point B.
The ship then turned and sailed an additional 69 km to point C. Determine
the distance from port to point C if the ship’s final bearing is N 74E.
Angles
Sides
A = 21
a = 69
B = 139.9
b=
C = 19.1
c = 63
b
a

sin B sin A
b
69

sin139.9 sin 21
69sin139.9
b
sin 21
b  124.0
The ship is 124 miles from port to point C.
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