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 .)