7.2 Cosine Law and Area of Triangle
1.
2.
3.
4.
5.
6.
7.
Determining if the Law of Sines or the Law of Cosines Should
be Used to Begin to Solve an Oblique Triangle
Using the Law of Cosines to Solve the SAS Case
Using the Law of Cosines to Solve the SSS Case
Using the Law of Cosines to Solve Applied Problems Involving
Oblique Triangles
Determining the Area of Oblique Triangles
Using Heron’s Formula to Determine the Area of an SSS Triangle
Solving Applied Problems Involving the Area of Triangles
H.Melikian/1200
Dr .Hayk Melikyan/ Departmen of Mathematics and CS/ melikyan@nccu.edu
1
The Law of Cosines
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 2  b 2  c 2  2bc cos A,
b 2  a 2  c 2  2ac cos B,
c 2  a 2  b 2  2ab cos C
H.Melikian/1200
2
Determining of the Law of Sines or Cosines Should Be
Used to Begin to Solve an Oblique Triangle
Decide whether the Law of Sines or Cosines can be used to
solve each triangle. Do not solve.
a.
b.
C
b = 8.3
68°
83°
45°
A
67°
B
c = 15.2
Neither can be used.
Law of Sines; given the lengths
of two sides and an angle
opposite one of the given sides.
H.Melikian/1200
3
Solving an SAS Oblique Triangle
Step 1: Use the Law of Cosines to determine the length of
the missing side.
Step 2: Determine the measure of the smaller of the
remaining two angles using the Law of Sines or
using the alternate form of the Law of Cosines.
Step 3: Use the fact that the sum of the measures of the
three angles of a triangle is 180 degrees to
determine the measure of the remaining angle.
H.Melikian/1200
4
Solving a SAS Triangle
Solve the given oblique triangle. Round all measurements to
one decimal place.
Angles
C
b = 20
Sides
A = 61
a = 26.8
B=
b = 20
C=
c = 30
61°
A
c = 30
B
a 2  b 2  c 2  2bc cos A
a 2  202  302  2(20)(30)cos61
a  202  302  2(20)(30)cos61
a  26.8
H.Melikian/1200
5
Solving a SAS Triangle-cont
Solve the given oblique triangle. Round all measurements
to one decimal place.
C
b = 20
61°
A
c = 30
Angles
B
Sides
A = 61
a=
B=
b = 20
C=
c = 30
H.Melikian/1200
26.8
C  180  61  40.7
a 2  c2  b2
cos B 
2ac
C  78.3
26.82  302  202

2(26.8)(30)
1218.24

1608
1  1218.24 
B  cos 
  40.7
 1608 
6
Solving an SSS Oblique Triangle
Step 1: Use the alternate form of the Law of Cosines to
determine the length of the largest angle. This is the
angle opposite the longest side.
Step 2: Determine the measure of one of the remaining two
angles using the Law of Sines or the alternate form
of the Law of Cosines.
Step 3: Use the fact that the sum of the measures of the three
angles of a triangle is 180 degrees to determine the
measure of the remaining angle.
H.Melikian/1200
7
Solving a SSS Triangle
Solve oblique triangle ABC if a = 4, b = 3, and
c = 6.
Angles
Sides
A=
a=4
B=
b=3
C=
117.3
H.Melikian/1200
c=6
a 2  b2  c2
cos C 
2ab
42  32  62

2(4)(3)
11

24
1  11 
C  cos     117.3
 24 
8
Solving a SSS Triangle-cont
Solve oblique triangle ABC if a = 4, b = 3, and
c = 6.
Angles
Sides
A=
a=4
B=
b=3
C=
117.3
c=6
sin B sin C

b
c
sin B sin117.3

3
6
3sin117.3
sin B 
6
B  26.4
153.6 degrees
will not work.
C  180  117.3  26.4  36.3
H.Melikian/1200
9
Determining the Distance between Two Airplanes
Two planes take off from different runways at the same time.
One plane flies at an average speed of 350 mph with a bearing
of N 21 E. The other plane flies at an average speed of 420
mph with a bearing of S 84 W. How far apart are the planes
from each other 2 hours after takeoff?
H.Melikian/1200
10
Determining the Distance between Two Airplanes
Two planes take off from different runways at the same time. One plane
flies at an average speed of 350 mph with a bearing of N 21 E. The other
plane flies at an average speed of 420 mph with a bearing of S 84 W. How
far apart are the planes from each other 2 hours after takeoff?
a = 700, b = 840, and D = 117
d 2  a 2  b 2  2ab cos D
d 2  7002  8402  2(700)(840)cos117
a  7002  8402  2(700)(840)cos117
a  1315.1
The planes are about 1315 miles apart after 2 hours.
H.Melikian/1200
11
Area of a Triangle
In any triangle, the area is given by
1
Area  bh
2
where b is the length of the base of the triangle and h is the
length of the altitude drawn to that base (or drawn to an
extension of that base).
H.Melikian/1200
12
Area of a Triangle
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 the area of triangle ABC is given
by
1
Area  bc sin A
2
1
Area  ac sin B
2
1
Area  ab sin C
2
H.Melikian/1200
13
Determining the Area of an Oblique Triangle
Determine the area of the triangle.
C
a = 20 ft
34°
A
c = 22 ft
H.Melikian/1200
B
1
Area  ac sin B
2
1
Area  (20)(22)sin34
2
 123.0 sq ft
14
Determining the Area of an Oblique Triangle
Determine the area of the triangle.
C  180  18  135
C  27
c
a

sin C sin A
c
12

sin 27 sin18
12sin 27
c
sin18
c  17.6
H.Melikian/1200
C
a = 12 cm
135°
18°
A
B
1
Area  ac sin B
2
 (12)(17.6)sin135
 74.7 sq cm
15
Heron’s Formula
Suppose that a triangle has side lengths of a, b, and c. If the
semiperimeter is then the area of the triangle is
1
s  ( a  b  c)
2
Area  s(s  a)(s  b)(s  c)
H.Melikian/1200
16
Determining the Area of an SSS Oblique Triangle
Using Heron’s Formula
A
Determine the area of the triangle.
1
s  ( a  b  c)
2
1
 (28  40  54)
2
 61
Area  s(s  a)(s  b)(s  c)
b = 40
c = 54
C
a = 28
B
 61(61  28)(61  40)(61  54)
 61(33)(21)(7)  543.98
H.Melikian/1200
17