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MATH 201 - Week 8
MATH 201 - Week 8
Ferenc Balogh
Concordia University
2008 Winter
Based on the textbook
J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson
MATH 201 - Week 8
Overview
Solving Triangles
MATH 201 - Week 8
Overview
Solving Triangles
Law of Sines (Section 6.4)
The Law of Sines
Solving a Triangle using Law of Sines
The Ambiguous Case
MATH 201 - Week 8
Overview
Solving Triangles
Law of Sines (Section 6.4)
The Law of Sines
Solving a Triangle using Law of Sines
The Ambiguous Case
The Law of Cosines (Section 6.5)
The Law of Cosines
The Area of a Triangle: Heron’s Formula
Navigation
MATH 201 - Week 8
Overview
Solving Triangles
Law of Sines (Section 6.4)
The Law of Sines
Solving a Triangle using Law of Sines
The Ambiguous Case
The Law of Cosines (Section 6.5)
The Law of Cosines
The Area of a Triangle: Heron’s Formula
Navigation
Some applications
MATH 201 - Week 8
Solving Triangles
The data to be known about a triangle:
angles α β γ
sides a b c
Solving a triangle means that we have to find all angles and all
sides from the data provided.
MATH 201 - Week 8
Solving Triangles
The data to be known about a triangle:
angles α β γ
sides a b c
Solving a triangle means that we have to find all angles and all
sides from the data provided.
Cases depending on the given data
ASA or SAA - one side and two angles are known
SSA - two sides and the angle opposite to one of those sides
are known
SAS - two sides and the included angle are known
SSS - all three sides are known
MATH 201 - Week 8
Solving Triangles
data
ASA or SAA
SSA
SAS
SSS
solution
unique
ambiguous
unique
unique
law to use
law of sines
law of sines
law of cosines
law of cosines
MATH 201 - Week 8
The Law of Sines
The Law of Sines
The Law of Sines
The lengths of the sides of a triangle are proportional to the sines
of the corresponding opposite angles:
sin α
sin β
sin γ
=
=
.
a
b
c
MATH 201 - Week 8
The Law of Sines
The Law of Sines
The Law of Sines
The lengths of the sides of a triangle are proportional to the sines
of the corresponding opposite angles:
sin α
sin β
sin γ
=
=
.
a
b
c
Proof. The area of a triangle is expressible as
1
1
1
A = ab sin γ = bc sin α = ca sin β.
2
2
2
Dividing through by 12 abc we get
sin γ
sin α
sin β
=
=
.
c
a
b
MATH 201 - Week 8
The Law of Sines
The Law of Sines
How to solve a triangle in the ASA or SAA case?
1
Find the third angle using
α + β + γ = 180◦ .
2
Find the two other sides using the Law of Sines.
There is no ambiguity, the solution is always uniquely determined.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if c = 12m, α = 23◦ , β = 42◦ .
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if c = 12m, α = 23◦ , β = 42◦ .
Solution. One side and two angles are given (ASA).
Since
α + β + γ = 180◦ ,
the third angle is given by
γ = 180◦ − 23◦ − 42◦ = 115◦ .
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if c = 12m, α = 23◦ , β = 42◦ .
Solution. One side and two angles are given (ASA).
Since
α + β + γ = 180◦ ,
the third angle is given by
γ = 180◦ − 23◦ − 42◦ = 115◦ .
By the Law of Sines, we have
sin β
sin γ
=
.
b
c
Therefore
b=
c sin β
12m · sin 42◦
=
≈ 8.86m.
sin γ
sin 115◦
a=
c sin α
12m · sin 23◦
=
≈ 5.17m.
sin γ
sin 115◦
Similarly,
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if b = 3.2m, γ = 31◦ , β = 74◦ .
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if b = 3.2m, γ = 31◦ , β = 74◦ .
Solution. One side and two angles are given (SAA).
Since
α + β + γ = 180◦ ,
the third angle is given by
α = 180◦ − 31◦ − 74◦ = 75◦ .
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if b = 3.2m, γ = 31◦ , β = 74◦ .
Solution. One side and two angles are given (SAA).
Since
α + β + γ = 180◦ ,
the third angle is given by
α = 180◦ − 31◦ − 74◦ = 75◦ .
By the Law of Sines, we have
sin β
sin γ
=
.
b
c
Therefore
c=
b sin γ
3.2m · sin 31◦
=
≈ 1.71m.
sin β
sin 74◦
a=
b sin α
3.2m · sin 75◦
=
≈ 3.22m.
sin β
sin 74◦
Similarly,
MATH 201 - Week 8
The Law of Sines
The Law of Sines
If two sides and the angle opposite to one of those sides is given
(SSA) (let’s say, a, b and α) then the following cases are possible
depending on a:
there is no solution (no intersection point)
there is exactly one solution (right triangle case)
there are two solutions (two intersection points)
there is only one solution (two intersection points)
MATH 201 - Week 8
The Law of Sines
The Law of Sines
The method of solution for SSA:
1
Try to determine the angle β from
sin β =
2
b
sin α
a
Find the third angle using
α + β + γ = 180◦ ,
and the third side c using the Law of Sines.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 1.1m.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 1.1m.
Solution. The sine of β is given by
sin β =
6m
b
sin α =
sin 53◦ ≈ 4.36
a
1.1m
But we should have sin β ≤ 1 for an angle β!
This means that there is no angle β satisfying the equation above.
Therefore there is no solution in this case.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 30◦ , b = 3.2m, a = 1.6m.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 30◦ , b = 3.2m, a = 1.6m.
Solution. The sine of β is given by
sin β =
3.2m
b
sin α =
sin 30◦ = 1
a
1.6m
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 30◦ , b = 3.2m, a = 1.6m.
Solution. The sine of β is given by
sin β =
3.2m
b
sin α =
sin 30◦ = 1
a
1.6m
There is only one angle 0◦ ≤ β ≤ 180◦ such that sin β = 1:
β = 90◦ .
There is a unique solution in this case.
γ = 180◦ − 90◦ − 30◦ = 60◦ .
The Law of Sines gives
c=
b sin γ
3.2m · sin 60◦
=
≈ 2.77m.
sin β
sin 90◦
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 5m.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 5m.
Solution. The sine of β is given by
sin β =
6m
b
sin α =
sin 53◦ ≈ 0.96
a
5m
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 5m.
Solution. The sine of β is given by
sin β =
6m
b
sin α =
sin 53◦ ≈ 0.96
a
5m
There are two angles 0◦ ≤ β ≤ 180◦ such that sin β = 0.96:
β1 ≈ 73.4◦ and β2 ≈ 180◦ − β1 ≈ 106.6◦ .
Since
γ1 = 180◦ − α − β1 ≈ 53.6◦ and γ2 = 180◦ − α − β2 ≈ 20.4◦ .
are both allowed angles, we have two different solutions in this
case.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
The Law of Sines gives
c1 =
5m · sin 53.4◦
a sin γ1
=
≈ 5.03m
sin α
sin 53◦
a sin γ2
5m · sin 20.4◦
=
≈ 2.18m.
sin α
sin 53◦
There are two different triangles corresponding to this data of SSA
type.
c2 =
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 7m.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 7m.
Solution. The sine of β is given by
sin β =
b
6m
sin α =
sin 53◦ ≈ 0.68
a
7m
MATH 201 - Week 8
The Law of Sines
The Law of Sines
Example. Solve the triangle if α = 53◦ , b = 6m, a = 7m.
Solution. The sine of β is given by
sin β =
b
6m
sin α =
sin 53◦ ≈ 0.68
a
7m
There are two angles 0◦ ≤ β ≤ 180◦ such that sin β = 0.68:
β1 ≈ 43.2◦ and β2 ≈ 180◦ − β1 ≈ 136.8◦ .
Then
γ1 = 180◦ − α − β1 ≈ 83.8◦ and γ2 = 180◦ − α − β2 ≈ −9.8◦ .
γ1 is allowed but γ2 is not: we have a unique solution
corresponding to γ1 in this case.
MATH 201 - Week 8
The Law of Sines
The Law of Sines
The Law of Sines gives
c=
7m · sin 83.8◦
a sin γ1
=
≈ 8.71m
sin α
sin 53◦
There is one triangle corresponding to this data of SSA type.
MATH 201 - Week 8
The Law of Cosines
For a right triangle with sides a, b and c, where γ = 90◦ we have
the Pythagorean Theorem:
c 2 = a2 + b 2 .
The following generalization holds for an oblique triangle:
The Law of Cosines
c 2 = a2 + b 2 − 2ab cos γ.
And we have two other equations corresponding to α and β:
a2 = b 2 + c 2 − 2bc cos α.
b 2 = c 2 + a2 − 2ca cos β.
MATH 201 - Week 8
The Law of Cosines
Example. Solve the triangle if a = 2m, b = 6m, c = 7m.
MATH 201 - Week 8
The Law of Cosines
Example. Solve the triangle if a = 2m, b = 6m, c = 7m.
Solution. This is a problem of the SSS type, we have to find the
angles. The Law of Cosines gives
cos α =
cos β =
cos γ =
27
b 2 + c 2 − a2
=
2bc
28
17
c 2 + a2 − b 2
=
2ca
28
a2 + b 2 − c 2
3
=−
2ab
8
Therefore
27
≈ 15.35◦
28
17
β = cos−1
≈ 52.62◦
28
3
−1
γ = cos
−
≈ 112.02◦
8
α = cos−1
MATH 201 - Week 8
The Law of Cosines
Example. Solve the triangle if a = 2m, b = 6m and γ = 49◦ .
MATH 201 - Week 8
The Law of Cosines
Example. Solve the triangle if a = 2m, b = 6m and γ = 49◦ .
Solution. This is a problem of the SAS type. The Law of Cosines
gives the third side:
c 2 = a2 + b 2 − 2ab cos γ.
Therefore
p
p
c = a2 + b 2 − 2ab cos γ = 22 + 62 − 2 · 2 · 6 cos 49◦ m ≈ 4.925m
cos α =
cos β =
b 2 + c 2 − a2
≈ 0.952
2bc
c 2 + a2 − b 2
=≈ −0.396
2ca
Therefore
α = cos−1 0.95 ≈ 17.85◦
β = cos−1 (−0.4) ≈ 113.15◦
MATH 201 - Week 8
The Law of Cosines
Heron’s Formula
If we know the sides a, b and c of a triangle, then it is uniquely
determined and therefore its area is expressible in terms of the
lengths of the sides:
Heron’s Formula
The area of a triangle of sides a, b, c is given by
p
A = s(s − a)(s − b)(s − c),
where
a+b+c
2
is the semiperimeter of the triangle.
s=
MATH 201 - Week 8
The Law of Cosines
Heron’s Formula
Example. Suppose that a triangle has sides a = 5cm, b = 4cm,
c = 7cm. Calculate the area of the triangle.
MATH 201 - Week 8
The Law of Cosines
Heron’s Formula
Example. Suppose that a triangle has sides a = 5cm, b = 4cm,
c = 7cm. Calculate the area of the triangle.
Solution. The semiperimeter is given by
s=
a+b+c
5cm + 4cm + 7cm
=
= 8cm.
2
2
Using Heron’s formula, we get the area:
p
p
s(s − a)(s − b)(s − c) = 8(8 − 5)(8 − 4)(8 − 7)cm2
A =
√
√
=
8 · 3 · 4 · 1cm2 = 96cm2 ≈ 9.798cm2
MATH 201 - Week 8
The Law of Cosines
Navigation
Navigation
In navigation, a direction is usually given as a bearing, for example:
N 30◦ W
MATH 201 - Week 8
The Law of Cosines
Navigation
Navigation
In navigation, a direction is usually given as a bearing, for example:
N 30◦ W
The first letter is either N or S indicating north or south.
The last letter is either E or W , either east or west.
The acute angle between the letters indicates the direction
measured from N/S to E /W .
The goal is to avoid the use of negative angles and angles
exceeding 90◦ .
MATH 201 - Week 8
The Law of Cosines
Navigation
Example. An airplane takes off from airport A heading to
N 52◦ E . After flying 300 kms, it makes a course correction above
the point B and heads to the new direction N 11◦ W . Flying 120
kms more, it lands at point C .
Find the distance between the points A and C .
Find the bearing from A to C .
Solution. First we realize that we have to solve a triangle from
data of the type SAS because c = 300km, a = 120km and the
angle β is given by
β = 90◦ − 11◦ + 90◦ − 52◦ = 117◦ .
MATH 201 - Week 8
The Law of Cosines
Navigation
The Law of Cosines gives the third side:
b 2 = c 2 + a2 − 2ca cos β.
Therefore
p
c 2 + a2 − 2ca cos β
p
=
3002 + 1202 − 2 · 300 · 120 cos 117◦ km ≈ 370.253km
b =
a2 + b 2 − c 2
≈ 0.957
2ab
b 2 + c 2 − a2
=≈ 0.692
cos α =
2bc
α = cos−1 0.957 ≈ 16.78◦
cos γ =
β = cos−1 0.692 ≈ 46.22◦
The bearing from A to C is approximately
N 5.78◦ E
MATH 201 - Week 8
The Law of Cosines
Applications
Example. An Unidentified Flying Object is observed from
observatories A and B simultaneously. The angles of elevations are
α = 43◦ and β = 29◦ respectively.
How far is the object from A and B if we know that the distance
between A and B is c = 134km?
Solution. We have to solve a triangle from data of the form ASA.
γ = 180◦ − 43◦ − 29◦ = 108◦ .
MATH 201 - Week 8
The Law of Cosines
Applications
Example. An Unidentified Flying Object is observed from
observatories A and B simultaneously. The angles of elevations are
α = 43◦ and β = 29◦ respectively.
How far is the object from A and B if we know that the distance
between A and B is c = 134km?
Solution. We have to solve a triangle from data of the form ASA.
γ = 180◦ − 43◦ − 29◦ = 108◦ .
By the Law of Sines, we have
sin γ
sin β
=
.
b
c
Therefore
c sin β
134km · sin 29◦
b=
=
≈ 68.31km.
sin γ
sin 108◦
Similarly,
c sin α
134km · sin 43◦
a=
=
≈ 96.09km.
sin γ
sin 108◦
MATH 201 - Week 8
The Law of Cosines
Applications
Example. What is the area of the triangle-shaped area where
Norman Bethune’s statue stands near Metro Guy-Concordia?
(The sides are approximately 77.3m, 18.7m and 80.6m.)
Solution.We can use Heron’s Formula to calculate the
approximate area.
s=
77.3m + 18.7m + 80.6cm
a+b+c
=
= 88.3m.
2
2
Using Heron’s formula, we get the area:
p
s(s − a)(s − b)(s − c)
A =
p
=
88.3(88.3 − 77.3)(88.3 − 18.7)(88.3 − 80.6)m2
√
=
88.3 · 10.0 · 69.6 · 7.7m2 = 721.5m2