Download Law of Sines - MATH 160, Precalculus

Law of Sines
MATH 160, Precalculus
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan
Law of Sines
Objectives
In this lesson we will learn to:
use the Law of Sines to solve oblique triangles (AAS, ASA,
and SSA),
find the areas of oblique triangles,
use the the Law of Sines to model and solve real-world
problems.
J. Robert Buchanan
Law of Sines
Oblique Triangles
Oblique triangles are triangles having no right angles.
C
a
b
c
A
J. Robert Buchanan
B
Law of Sines
Solving an Oblique Triangle
Solving a triangle involves determining to the lengths of its
three sides and the measures of its three angles.
J. Robert Buchanan
Law of Sines
Solving an Oblique Triangle
Solving a triangle involves determining to the lengths of its
three sides and the measures of its three angles.
To solve a triangle we must know the length of at least one side
and any two other measures of the triangle:
1
Any side and two angles (AAS or ASA).
2
Two sides and the angle opposite one of them (SSA).
3
Three sides (SSS).
4
Two sides and their included angle.
J. Robert Buchanan
Law of Sines
Solving an Oblique Triangle
Solving a triangle involves determining to the lengths of its
three sides and the measures of its three angles.
To solve a triangle we must know the length of at least one side
and any two other measures of the triangle:
1
Any side and two angles (AAS or ASA).
2
Two sides and the angle opposite one of them (SSA).
3
Three sides (SSS).
4
Two sides and their included angle.
Today we will explore the first two cases.
J. Robert Buchanan
Law of Sines
Law of Sines
Law of Sines
If ABC is a triangle with sides a, b, and c, then
b
c
a
=
=
sin A
sin B
sin C
sin A
sin B
sin C
=
=
.
a
b
c
or
C
C
b
a
h
h
A
c
B
Acute triangle
J. Robert Buchanan
a
b
A
c
Obtuse triangle
Law of Sines
B
Example
Solve the following triangle (not drawn to scale).
C
a
b
40 °
35 °
A
c =10
J. Robert Buchanan
B
Law of Sines
Example
Solve the following triangle (not drawn to scale).
C
a
b
40 °
35 °
A
c =10
B
This is the ASA case.
C = 180◦ − 35◦ − 40◦ = 105◦
c
10 sin 35◦
a =
sin A =
≈ 5.94
sin C
sin 105◦
c
10 sin 40◦
b =
sin B =
≈ 6.65
sin C
sin 105◦
J. Robert Buchanan
Law of Sines
Example
A flagpole at a right angle to the horizontal is located on a slope
that makes an angle of 12◦ with the horizontal. The flagpole’s
shadow is 16 meters long and points directly up the slope. The
angle of elevation from the tip of the shadow to the sun is 20◦ .
Find the height of the flagpole.
J. Robert Buchanan
Law of Sines
Example
A flagpole at a right angle to the horizontal is located on a slope
that makes an angle of 12◦ with the horizontal. The flagpole’s
shadow is 16 meters long and points directly up the slope. The
angle of elevation from the tip of the shadow to the sun is 20◦ .
Find the height of the flagpole.
This is the ASA situation. Let A = 78◦ , B = 32◦ , and c = 16
meters.
J. Robert Buchanan
Law of Sines
Example
A flagpole at a right angle to the horizontal is located on a slope
that makes an angle of 12◦ with the horizontal. The flagpole’s
shadow is 16 meters long and points directly up the slope. The
angle of elevation from the tip of the shadow to the sun is 20◦ .
Find the height of the flagpole.
This is the ASA situation. Let A = 78◦ , B = 32◦ , and c = 16
meters.
C = 180◦ − 78◦ − 32◦ = 70circ
c
16 sin 70◦
b =
sin B =
≈ 9.02
sin C
sin 32◦
J. Robert Buchanan
Law of Sines
meters
SSA: Ambiguous Case
If we know two side lengths and the measure of the angle
opposite one of the sides, three possibles outcomes may
present themselves when solving the oblique triangle.
No such triangle exists.
One such triangle exists.
Two distinct triangles exist.
J. Robert Buchanan
Law of Sines
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 110◦ ,
a = 125,
J. Robert Buchanan
Law of Sines
b = 200.
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 110◦ ,
sin A
a
a = 125,
b = 200.
sin B
b
b sin A
sin B =
a
200 sin 110◦
=
125
= 1.50351 > 1
=
No solution.
J. Robert Buchanan
Law of Sines
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 76◦ ,
a = 34,
J. Robert Buchanan
Law of Sines
b = 21.
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 76◦ ,
sin A
a
a = 34,
b = 21.
sin B
b
b sin A
sin B =
a
21 sin 76◦
=
34
≈ 0.5993
=
B ≈ 36.82◦
C ≈ 67.18◦
a
34 sin 67.18◦
c =
sin C ≈
≈ 32.30
sin A
sin 76◦
J. Robert Buchanan
Law of Sines
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 25◦ ,
a = 9.5,
J. Robert Buchanan
Law of Sines
b = 22.
Example
Use the Law of Sines to solve the triangle, if there is a solution.
If two solutions exist, find them both.
A = 25◦ ,
a = 9.5,
sin A
a
sin B
b = 22.
sin B
b
b sin A
=
a
22 sin 25◦
=
9.5
≈ 0.978695
=
B ≈ 78.15◦ or B ≈ 101.85◦
C ≈ 76.85◦ or C ≈ 53.15◦
9.5 sin 76.85◦
9.5 sin 53.15◦
c≈
≈
21.89
or
c
≈
≈ 17.99
sin 25◦
sin 25◦
J. Robert Buchanan
Law of Sines
Area of an Oblique Triangle
Area
The area of any triangle is one-half the product of the lengths of
two sides times the sine of their included angle.
Area =
1
1
1
bc sin A = ac sin B = ab sin C
2
2
2
C
C
b
a
h
h
A
c
B
Acute triangle
J. Robert Buchanan
a
b
A
c
Obtuse triangle
Law of Sines
B
Example
Find the area of the triangle having the following angle and
sides.
B = 130◦ ,
a = 62,
c = 20.
J. Robert Buchanan
Law of Sines
Example
Find the area of the triangle having the following angle and
sides.
B = 130◦ ,
a = 62,
c = 20.
Area =
1
1
ac sin B = (62)(20) sin 130◦ = 474.948
2
2
J. Robert Buchanan
Law of Sines
Application
The circular arc of a railroad curve has a chord of length 3000
feet corresponding to a central angle of 40◦ .
Find the radius r of the circular arc.
Find the length s of the circular arc.
J. Robert Buchanan
Law of Sines
Application
The circular arc of a railroad curve has a chord of length 3000
feet corresponding to a central angle of 40◦ .
Find the radius r of the circular arc.
r
sin 70◦
=
r
=
3000
sin 40◦
3000 sin 70◦
≈ 4385.71
sin 40◦
Find the length s of the circular arc.
J. Robert Buchanan
Law of Sines
feet
Application
The circular arc of a railroad curve has a chord of length 3000
feet corresponding to a central angle of 40◦ .
Find the radius r of the circular arc.
r
sin 70◦
=
r
=
3000
sin 40◦
3000 sin 70◦
≈ 4385.71
sin 40◦
feet
Find the length s of the circular arc.
The central angle of 40◦ is equivalent to 2π/9 radians
2π
s = r θ = 4385.71
= 3061.8 feet
9
J. Robert Buchanan
Law of Sines
Homework
Read Section 5.6.
Exercises: 1, 5, 9, 13, . . . , 49, 53
J. Robert Buchanan
Law of Sines