MA 15400
Lesson 18
The Law of Sines
Section 8.1
Any triangle without a right angle is called an oblique triangle. As usual, we use α, β, γ for
the angles of the triangle (or the vertices A, B, and C) and a, b, and c, respectively, for the
lengths of the sides opposite those angles.
B
a
β
γ
c
A
α
C
b
The Law of Sines: In any triangle, the ratio of the sine of an angle to the side opposite that
angle is equal to the ratio of the sine of another angle to the side opposite that angle.
The Law of Sines:
If ABC is an oblique triangle labeled in the usual manner, then:
sinα sin β sin γ
=
=
a
b
c
Solve ∆ABC : (This means given 3 pieces of data about a triangle, find the remaining 3
pieces of data.)
α = 41°, γ = 77°, a = 10.5
a
10.5
b
c
α
41.0°
β
γ
77.0°
Always
draw a
triangle.
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MA 15400
Lesson 18
The Law of Sines
γ = 81°, c = 11, b = 12
This triangle does
not exist. The side
opposite the 81°
angle must be
longer than 11.
Draw Triangle:
Review:
Calculate
sin(10°) and sin(170°)
sin(30°) and sin(150°)
Section 8.1
sin(80°) and sin(100°)
sin(1°) and sin(179°)
Notice that the sines of supplementary angles are equivalent.
Find the angle:
sin(x) = 0.3456
sin(x) = 0.9876
There are two angles that make each equation a true statement, one acute and one obtuse.
This may or may not lead to two different triangles.
When you are given one angle and two sides, and inverse sine is used to calculate an
angle, there is a possibility of two triangles existing for the given information. This
happens when there is SSA. Remember that the sum of the angles in a triangle is 180°.
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MA 15400
Lesson 18
The Law of Sines
γ = 53°, a = 140, c = 115
a 140.0 140.0
b
c 115.0 115.0
A
B
C 53.0° 53.0°
Draw ∆1:
Section 8.1
Draw ∆ 2:
sin 53.0 sin A
=
115
140.0
140(sin 53 )
sin A =
115
sin A = 0.972251925
A = 76.47
A = 180 − 76.47 = 103.53
Since 103.5 + 53 is less than 180
degrees, there are two different triangles.
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MA 15400
Lesson 18
The Law of Sines
β = 75°, b = 248, c =195
a
b 248.0
c 195.0
A
B 75.0°
C
Section 8.1
Draw ∆:
sin 75.0 sin C
=
248.0
195.0
195(sin 75 )
sin C =
248
sin C = 0.75949813
C = 49.42
C = 180 − 49.42 = 130.58
However, 130.6 + 75.0 is greater than 180
degrees. There will not be two distinct
triangles here.
Final Answers: The largest side
should always be opposite the
largest angle. The smallest side
should always be opposite the
smallest angle.
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MA 15400
Lesson 18
The Law of Sines
Section 8.1
Fire lookout: A fire is spotted from two fire lookouts stations. If station B reports the
fire at S37°E, and station A, 10 miles due East of station B, reports the fire at S61°W,
how far is the fire from station A? How far is the fire from station B?
From Point B
From Point A
a
4.8 Miles
8.1 Miles
b
Notice: The
angles given
for the
measures are
outside of the
triangle
formed from
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MA 15400
Lesson 18
The Law of Sines
Section 8.1
Finding the Height of an Airplane. An aircraft is spotted by two observers who are
1000 feet apart. As the airplane passes between them and over the line joining them,
observer A sights a 40° angle of elevation to the plane, and observer B sights a 35° angle.
How high is the airplane?
There are 3
triangles in
this picture,
the big one
and two
right
triangles.
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MA 15400
Lesson 18
The Law of Sines
Section 8.1
Finding the Length of a Ski Lift.
a) What is the length of the ski lift?
mountain?
Again, there are
3 triangles, the
big one and two
smaller ones, a
left one and a
right. In the left
triangle, angle
A would be
155° and the
very small
angle up by B is
10°.
b)
What is the height of the
B
Mountain
Ski Lift
C
sin10 sin15
=
Ski Lift
1000
15°
A
25°
D
1000 ft
sin 25 =
Mountain
Ski Lift
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MA 15400
Lesson 18
The Law of Sines
Section 8.1
Surveying
A surveyor notes that the direction from point A to point B is N78°E and
the direction from A to C is N21°E. The distance from A to B is 345 meters and the
distance from B to C is 543 meters. Approximate the distance from A to C.
Note: This
picture may not
be to scale.
57°
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