Download Section 8.2 – The Law of Sines

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Section 8.2 – The Law of Sines
In this section and the next section, we will be solving triangles that are not
right triangles or what are called oblique triangles. We will derive the laws
of sines (section 8.2) and cosines (section 8.3) to help us do this.
Definition
An oblique triangle is a triangle where none of the angles are right angles.
There are two types of oblique triangles; the first case is when all the
angles are acute and the second case is when the two of the angles are
acute and the third angle is obtuse.
Objective 1:
Derive the Laws of Sines.
To derive the Law of Sines, we will need to consider a case for each type of
oblique triangle. Let ∠A, ∠B, and ∠C be the angles of the oblique triangle
with corresponding sides of length a, b, & c.
Case 1:
Acute Triangle
Case 2:
Obtuse Triangle
B
c
B
a
a
c
A
C
A
b
C
b
Construct h as the height of the triangle and ∠P to be the supplementary
angle of ∠A in the obtuse triangle:
Case 1:
Acute Triangle
Case 2:
B
c
Obtuse Triangle
B
a
h
A
h
C
b
a
c
P
A
C
b
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In both cases, looking at the right triangle formed by ∠C, sides a and h:
sin(C) =
opp
hyp
=
h
a
which implies h = asin(C)
In case 1, looking at the right triangle formed by ∠A, sides c and h:
sin(A) =
opp
hyp
=
h
c
which implies h = csin(A)
In case 2, looking at the right triangle formed by ∠P, sides c and h:
sin(P) =
opp
hyp
=
h
c
which implies h = csin(P)
But, P = 180˚ – A, so
sin(P) = sin(180˚ – A) (use the difference of angles formula)
= sin(180˚)cos(A) – cos(180˚)sin(A)
(but sin(180˚) = 0 and cos(180˚) = – 1)
= 0•cos(A) – (– 1)sin(A)
= sin(A) which implies h = csin(P) = csin(A)
Thus, in both cases, h = csin(A)
This means that asin(C) = h = csin(A).
asin(C) = csin(A)
(divide by ac)
sin(C)
c
=
sin(A)
a
sin(C)
sin(B)
=
.
c
b
sin(C)
for any oblique
c
Similarly, we can show that
Thus,
sin(A)
a
=
sin(B)
b
=
triangle.
The Law of Sines Theorem
For an oblique triangle ABC with opposites of length a, b, and c
respectively,
sin(A)
a
=
sin(B)
b
=
sin(C)
c
There are two cases were we will use the Law of Sines. The first case is
when we know one side and two angles (ASA or SAA). The second case is
when we know two sides and the one angle that is opposite of either the
two known sides (SSA). As we solve for the various sides of an oblique
triangle, it is important to always use the given information if possible rather
than rounded value of something we have already found.
Objective 2:
Solve SAA or ASA Triangles (Case 1)
139
We will solve a couple examples where two angles and a side of the
triangle are given.
Solve the following triangle:
Ex. 1
A = 50˚, C = 20˚, a = 3 m
Solution:
We will begin by sketching a picture
B
of a triangle. Since the sum of the
c
a=3m
angles of a triangle is 180˚, we can
find angle B:
A = 50˚
C = 20˚
B = 180˚ – 50˚ – 20˚ = 110˚
b
Since we know angle A and side a,
(SAA)
we will use those values on one
side of the Laws of Sines:
To find c:
To find b:
sin(A)
a
=
sin(C)
c
sin(A)
a
=
sin(B)
b
Plug in the known values, cross multiply and solve:
sin(50o )
3
=
sin(20o )
c
sin(50˚)c = 3sin(20˚)
c=
o
3sin(20 )
sin(50o )
sin(50o )
3
=
sin(110o )
b
sin(50˚)b = 3sin(110˚)
b=
3sin(110o )
sin(50o )
c ≈ 1.3394… ≈ 1.34 m
b ≈ 3.6800… ≈ 3.68 m
Thus, B = 110˚, c ≈ 1.34 m, and b ≈ 3.68 m
Ex. 2
A = 78˚, B = 41˚, c = 11 in
Solution:
We will begin by sketching a picture
B = 41˚
of a triangle. Since the sum of the
c = 11 in
a
angles of a triangle is 180˚, we can
find angle C:
A = 78˚ C
C = 180˚ – 78˚ – 41˚ = 61˚
b
Since we know angle C and side c,
(ASA)
we will use those values on one
side of the Laws of Sines:
To find a:
To find b:
sin(A)
a
=
sin(C)
c
sin(B)
b
=
sin(C)
c
Plug in the known values, cross multiply and solve:
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sin(78o )
a
=
sin(61o )
11
sin(41o )
b
sin(61˚)a = 11sin(78˚)
a=
=
sin(61o )
11
sin(61˚)b = 11sin(41˚)
11sin(78o )
b=
sin(61o )
11sin(41o )
sin(61o )
a ≈ 12.3020… ≈ 12.30 in
b ≈ 8.2511… ≈ 8.25 in
Hence, C = 61˚, a ≈ 12.3 in, and b ≈ 8.25 in.
Objective 3:
Solve SSA Triangles (Case 2).
In solving oblique triangles that fit into second case when we know two
sides and the one angle that is opposite of either the two known sides,
there are several possible solutions.
Consider ∆ABC where the sides a
b
a
and b and ∠A are given. Let h =
h
height of the triangle (dashed line):
A
In examining the right triangle formed
by ∠A, side b and h. Since sin(A) =
opp
hyp
, then sin(A) =
h
b
which implies
that h = bsin(A). The length of side a dictates how many possible triangles
we get for our solution.
No Triangle Case
If a < h = bsin(A), then
a is not long enough to
connect to the base
(side c). Thus, no triangle
can be formed:
b
h a
A
In this case, if we use the
Law of Sines, we would get
sin(B) > 1 which is impossible
since the range of the sine
function is [– 1, 1].
One Right Triangle Case
If a = h = bsin(A), then a is
just barely long enough to
connect to the base (side c)
at a right angle. Thus, one
right triangle can be formed:
b
a=h
A
In this case, if we use the Law
of Sines, we would get
sin(B) = 1 which would tell us that
B is a 90˚ angle.
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Two Triangle Case
If h = bsin(A) < a < b, then
a is long enough to connect
to the base (side c) at two
different points. Thus, there
two triangles that can be formed:
b
a
One Oblique Triangle Case
If a ≥ b then a is long enough
to connect to the base (side c) at
one point, but too long to connect
to the base at another point. Thus,
only one triangle can be formed:
h a
b
A
h
a
A
In this case, when we use the
Law of Sines, we will find an
angle in the first quadrant and
one in the second quadrant that
will satisfy the triangle giving us
two possible answers.
In this case, when we use the
Law of Sines, we will find an angle
only in the first quadrant that will
satisfy the triangle. The angle in
the second quadrant will give us a
sum of angles larger than 180˚.
Solve the following triangle:
Ex. 3
a = 2, c = 3, ∠A = 36˚
Solution:
We will start by making a sketch
of the triangle. Since we know ∠A
and side a, we will use those values
on one side of the Laws of Sines:
To find ∠C:
sin(A)
sin(C)
=
(plug
a
c
sin(36o )
sin(C)
=
2
3
3sin(36o )
sin(C) =
2
3sin(36o )
=
C = sin – 1
2
(
In quadrant I,
C ≈ 61.85˚
)
c=3
B
A = 36˚
a=2
C
b
in the known values, cross multiply and solve)
(find the inverse sine)
sin – 1(0.88167…) = 61.845432…˚
In quadrant II,
C = 180˚ – 61.845432…˚
= 118.154567…˚ ≈ 118.15˚
Both angles are possible, so we have two possible triangles.
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36˚
61.85˚
Triangle 1: C ≈ 61.85˚
B = 180˚ – 36˚ – 61.845432…˚
= 82.154567… ≈ 82.15˚
Use the Laws of Sines to find b:
sin(A)
sin(B)
=
a
b
o
sin(36 )
sin(82.154567...o )
=
2
b
sin(36˚)b = 2sin(82.154567…˚)
b=
2sin(82.154567...o )
sin(36o )
= 3.370754… ≈ 3.37
Hence, C = 61.85˚, B = 82.15˚
and b ≈ 3.37.
Ex. 4
118.15˚
36˚
Triangle 2: C ≈ 118.15˚
B = 180˚ – 36˚ – 118.154567…˚
= 25.845432…˚ ≈ 25.85˚
Use the Laws of Sines to find b:
sin(A)
sin(B)
=
a
b
o
sin(36 )
sin(25.845432...o )
=
2
b
sin(36˚)b = 2sin(25.845432…˚)
b=
2sin(25.845432...o )
sin(36o )
= 1.483347… ≈ 1.48
Hence, C = 118.15˚, B = 25.85˚
and b ≈ 1.48.
a = 9 m, b = 5 m, ∠A = 81˚
Solution:
We will start by making a sketch
of the triangle. Since we know ∠A
and side a, we will use those values
on one side of the Laws of Sines:
To find ∠B:
c
B
A = 81˚
C
b = 5m
sin(A)
sin(B)
=
(plug in the known values, cross multiply
a
b
sin(81o )
sin(B)
=
9
5
5sin(81o )
(find the inverse sine)
sin(B) =
9
5sin(81o )
= sin – 1(0.548715…) = 33.278952…˚
B = sin – 1
9
(
In quadrant I,
B ≈ 33.28˚
a=9m
and solve)
)
In quadrant II,
B = 180˚ – 33.278952…˚
= 146.721047…˚ ≈ 146.72˚
The second angle, 146.72˚ is not possible since the sum of 146.72˚
and 81˚ exceeds 180˚.
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Triangle 1: B ≈ 33.28˚
C = 180˚ – 81˚ – 33.278952…˚
= 65.721047…˚ ≈ 65.72˚
Use the Laws of Sines to find c:
sin(A)
sin(C)
=
a
c
o
sin(81 )
sin(65.721047...o )
=
9
c
sin(81˚)c = 9sin(65.721047…˚)
c=
9sin(65.721047...o )
sin(81o )
= 8.306253… ≈ 8.31 m
Hence, B = 33.28˚, C = 65.72˚ and c ≈ 8.31 m.
Ex. 5
a = 3 ft, c = 1.6 ft, ∠C = 54˚
Solution:
We will start by making a sketch
of the triangle. Since we know ∠C
and side c, we will use those values
on one side of the Laws of Sines:
To find ∠A:
sin(A)
a
sin(A)
3
c = 1.6 ft
A
B
a = 3 ft
54˚ = C
b
sin(C)
(plug in the known values, cross multiply and
c
sin(54o )
=
1.6
3sin(54o )
(find the inverse sine)
sin(A) =
1.6
3sin(54o )
= sin – 1(1.516906…) which is undefined.
A = sin – 1
1.6
=
(
solve)
)
Thus, there is no triangle with such measurements.
Objective 4:
Solve Applications.
Solve the following:
Ex. 5
An astronomer on Earth measures the angle between the Sun
and Venus to be 22.4˚. If the Earth is approximately 93 million miles
from the Sun and Venus is approximately 67 million miles from the
sun, how far away is Venus from the Earth?
Solution:
We will first need to draw a picture of the situation:
144
Venus
x
V
67 million miles
22.4˚
93 million miles S
Earth
Sun
Since we know 22.4˚ and the opposite side (67 million miles), we will
plug in those values on one side of the Law of Sines and plug in 93
million miles in on the other to find the angle on top:
sin(22.4o )
67
sin(V)
93
93sin(22.4o )
sin(V) =
67
93sin(22.4o )
V = sin – 1
67
=
(
(cross multiply and solve)
(find the inverse sine)
) = sin
–1
(0.528948…) = 31.934432…˚
In quadrant I,
V ≈ 31.93˚
In quadrant II,
V = 180˚ – 31.934432…˚
= 148.06556…˚ ≈ 148.07˚
Both angles are possible, so we have two possible triangles.
Triangle 1: V ≈ 31.93˚
Triangle 2: V ≈ 148.07˚
S = 180˚ – 22.4˚ – 31.934432…˚ S = 180˚ – 22.4˚ – 148.06556…˚
= 125.665567… ≈ 125.67˚
= 9.534432…˚ ≈ 9.53˚
Use the Laws of Sines to find ?: Use the Laws of Sines to find ?:
sin(22.4o )
sin(125.665567...o )
=
67
x
67sin(125.665567...o )
x=
sin(22.4o )
sin(22.4o )
sin(9.534432...o )
=
67
x
o
67sin(9.534432... )
x=
sin(22.4o )
= 142.84259… ≈ 142.84
= 29.122965… ≈ 29.12
Thus, Venus is either 142.84 million miles or 29.12 million miles from
the Earth.
Ex. 6
Lighthouse #1 is located 150 miles directly north of Lighthouse
#2. Each lighthouse receives a distress signal from a ship out at sea.
If the signal indicates that the ship has a bearing of S48˚W from
lighthouse #1 and a bearing of N32˚W from lighthouse #2, how far is
the ship from each lighthouse?
Solution:
145
We will begin by draw a picture
of the situation. We will first need
to find angle B:
C = 180˚ – 48˚ – 32˚ = 100˚
Since we know angle C and
the opposite side (150 miles) ,
we can plug those values in
on one side of the Laws of Sines:
a
48˚
C
150
miles
b
To find a:
sin(A)
a
=
32˚
To find b:
sin(C)
c
sin(B)
b
=
sin(C)
c
Plug in the known values, cross multiply and solve:
sin(32o )
sin(100o )
=
a
150
o
150sin(32 )
a=
sin(48o )
sin(100o )
=
b
150
o
150sin(48 )
b=
sin(100o )
sin(100o )
a ≈ 80.7141… ≈ 80.71 miles
b ≈ 113.19135… ≈ 113.19 miles
The ship is ≈ 80.71 miles from lighthouse #1 and ≈ 113.19 miles from
lighthouse #2.
Prove the following:
Ex. 7
Mollweide's Theorem: For any triangle,
a+b
c
=
1
2
cos[ (A −B)]
1
2
sin( C)
Solution:
We will begin by using the Law of Sines to build the left side:
a
:
c
sin(A)
sin(C)
=
a
c
a
sin(A)
=
c
sin(C)
a
b
a+b
add c + c = c :
a+b
sin(A)
sin(B)
=
+
c
sin(C)
sin(C)
b
:
c
sin(B)
sin(C)
=
b
c
b
sin(B)
=
c
sin(C)
Solve for
Solve for
Now,
sin(A) + sin(B)
sin(C)
α +β
α −β
Use the formula (sin(α) + sin(β) = 2sin
cos
2
2
1
numerator and rewrite C as 2( 2 C) in the denominator.
=
(
) (
)) on the
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A +B
=
A −B
2 sin( 2 ) cos( 2 )
1
(use the double angle formula for sine)
sin( 2( 2 C ))
A +B
=
A −B
2 sin( 2 ) cos( 2 )
1
1
(reduce)
2 sin( 2 C ) cos( 2 C )
A +B
=
A −B
sin( 2 ) cos( 2 )
1
1
(since A + B + C = 180˚, then A + B = 180˚ – C)
sin( 2 C ) cos( 2 C )
=
=
=
=
180o −C
A −B
) cos( 2 )
2
(simplify)
1
1
sin( 2 C ) cos( 2 C )
1
A −B
sin( 90 0 − 2 C ) cos( 2 )
(but, sin(90˚
1
1
sin( 2 C ) cos( 2 C )
1
A −B
cos( 2 C ) cos( 2 )
(reduce)
1
1
sin( 2 C ) cos( 2 C )
1
A−B
cos[ (A − B)]
cos( 2 )
2
=
.
1
1
C)
sin(
sin( 2 C )
2
sin(
– θ) = cos(θ))