53
8. SOLVING OBLIQUE TRIANGLES: THE LAW OF COSINES When two sides and the included angle (SAS) or three sides (SSS) of a triangle are given, we cannot apply the law of sines to solve the triangle. In such cases, the law of cosines may be applied. Theorem 8.1: The Law of Cosines
To prove the theorem, we place triangle ABC in a coordinate plane with vertices labeled counterclockwise and so that one side lies on the positive x axis and one vertex is at O. Suppose that A is at ( 0 , 0 ) . Then B = ( c , 0 ) and C = ( b cos , b sin ).
Thus, 2
CB = 22 ) sin ( ) cos ( bcb = 2a .
222222 sin cos 2 cos abcbcb .
So cos 2 222 bccba .
b a
A B
C
c
In the general triangle ABC , the square of the length of any side is equal to the sum of the squares of the lengths of the other two sides minus twice the product of those side lengths times the cosine of the angle between them. cos2222 abbac
cos2222 accab
cos2222 bccba
b a
A B = ( c , 0 )
C = ( b cos , b sin )
c x
y
54
o60
14
A
C
B 10
6
A
C
B 7
5
Now rotate the triangle so that B is at the origin and C is on the positive x axis. An analogous argument now gives
cos 2 222 cacab . When C is at the origin, we find
2 2 2 2 cosc a b ab .
Example 8.1 ---------------------------- ------------------------------------------------------------ SAS case: Solve the triangle ABC if = o60 , b = 14, c = 10. Since cos2222 bccba )60cos()10)(14(2)10()14( 22 o
= 196 + 100 - 140
1562 a a = 12.49. It is geometrically evident that is acute and by the law of sines
49.12
37
2
3
12.49
14sin
14
sin
49.12
60sin o
arcsin o10.7649.12
37
.
Then = oo 10.7660180o = o90.43 .
___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Example 8.2 ---------------------------- ------------------------------------------------------------ SSS case: Solve the triangle ABC if a = 5, b = 6, c = 7.
cos2222 bccba cos)7)(6(2493625
7
5
84
60coscos848525
7
5arccos = o42.44
Note: there is no other angle for which:
o0 < < o180 and cos =7
5.
55
b
h = b sin
c
a
Then by the law of sines
oo
42.44sin5
6sin
6
sin
5
42.44sin
arcsin o13.5742.44sin5
6 o
. Clearly, must be acute.
Then = oo 13.5742.44180o = o45.78 . ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The proof follows from the law of cosines expressed in the form:
2 bc cos = 222 acb
Note that A = 2
1 ch = 2
1 bc sin 2222 sin 41 cbA .
Now we may obtain the desired formula by algebraic manipulation.
) cos1 ( sin 2222222 4
1
4
1 cbcbA
) cos1 ( ) 2 ( ) cos1 ( ) 2 ( 16
1 bcbc
) 2 ( ) 2 ( 222222
16
1acbbcacbbc
Theorem 8.2: Heron’s Area Formula The area of a triangle with sides a , b and c
and semiperimeter s = 2
cba has area A
given by A = )( )( )( csbsass .
56
2)(2 22)(
16
1cbaacb
2
)(
2
)(
2
)(
2
)( cbacbaacbacb
ccba
bcba
acbacba
2
2
2
2
2A = )( )( )( csbsass . Section 8 Problems--------------- ------- ----------------------------------------------------------- In problems 1 to 5 use the law of cosines to find the specified part of the triangle ABC. Round off angles to the nearest hundredth of a degree and side lengths to four significant digits.
1. Find c if a = 3, b = 10, = o60 .
2. Find a if b = 3.2, c = 2.4, = o117 .
3. Find if a = 200, b = 50, c = 177.
4. Find a if b = 68, c = 14 and = o5.24 .
5. Find if a = 2, b = 3 and c = 4. 6. Find the length of side AB in the quadrilateral shown in the figure. In problems 7 through 9 use Heron’s Formula to find the area of the triangle.
7. Find the area of a right triangle with sides 3, 4, and 5.
8. Find the area of the triangle with sides 31, 42, and 53.
9. Find the area of the triangle with sides 5.9, 6.7, and 10.3.
10. Use the answer you obtained in problem 8 to find the length h of the shortest altitude of the triangle with sides 31, 42, and 53. 11. A highway cuts a corner from a parcel of land. Find the number of acres in the triangular lot ABC. (Note: 1 acre = 43,560 2ft .)