Transcript Slide 1

Homework, Page 494
Solve the triangle.
B
1.
8
131
13
A
C
b  a  c  2ac cos131  b  13  8  2 13 8  cos131
2
2
2
2
2
b  19.221
sin A sin131
1  13sin131 

 A  sin 
  30.693
13
19.221
 19.221 
C  180  131  30.693   18.307
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 1
Homework, Page 494
Solve the triangle.
5. A  55, b  12, c  7
A  55, b  12, c  7  a  b 2  c 2  2bc cos55
a  122  7 2  2 12 7  cos55  9.831
sin C sin 55
1  7sin 55 

 C  sin 
  35.680
c
a
 9.831 
B  180   55  35.680   89.320
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 2
Homework, Page 494
Solve the triangle.
9.
a  1, b  5, c  4
a  1, b  5, c  4  a  c  b
No triangle, only a straight line.
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Slide 6- 3
Homework, Page 494
Solve the triangle.
13. A  42, a  7, b  10
sin A sin B
A  42, a  7, b  10 

a
b
1  10sin 42 
B  sin 
  B1  72.921
7


C1  180   42  72.921   65.079
B2  180  72.921  107.079
C2  180   42  107.079   30.921
7 sin 65.079
7 sin 30.921
c1 
 9.487  c2 
 7.679
sin 42
sin 42
Slide 6- 4
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Homework, Page 494
Find the area of the triangle.
17. A  47, b  32 ft , c  19 ft
A  47, b  32 ft , c  19 ft
1
1
Area  bc sin A   32 19sin 47   222.332 ft 2
2
2
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 5
Homework, Page 494
Decide if a triangle can be formed from the three sides. If so, use
Heron’s formula to find the area of the triangle.
21. a  4, b  5, c  8
a  4, b  5, c  8  4  5  8  triangle formed
458
s
 8.5
2
A  8.5  8.5  4  8.5  5  8.5  8   8.182
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 6
Homework, Page 494
Decide if a triangle can be formed from the three sides. If so, use
Heron’s formula to find the area of the triangle.
25. a  19.3, b  22.5, c  31
a  19.3, b  22.5, c  31  19.3  22.5  31
triangle formed
19.3  22.5  31
s
 36.4
2
A  36.4  36.4  19.3 36.4  22.5  36.4  31
 216.149
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Slide 6- 7
Homework, Page 494
29. Find the radian measure of the largest angle in the
triangle with sides 4, 5, and 6.
a  4, b  5, c  6
c 2  a 2  b 2  2ab cos C  2ab cos C  a 2  b 2  c 2
2
2
2


a 2  b2  c2
a

b

c
1
cos C 
 C  cos 

2ab
2
ab


2
2
2


4

5

6
1
C  cos 
  1.445 rad
 245 
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Slide 6- 8
Homework, Page 494
33. Find the area of a regular hexagon circumscribed
about a circle of radius 12 in..
30 30
12
x
1
x
A  bh  h  12  tan 30   x  12 tan 30  6.928
2
12
1

b  2 x  13.856  A  6  13.856 12   498.830 in 2
2

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Slide 6- 9
Homework, Page 494
37. In softball, adjacent bases are 60 ft apart. The
distance from the center of the front edge of the
pitcher’s rubber to the far corner of home plate is 40 ft.
(a) find the distance from the center of the
pitchers rubber to the far corner of first base.
c 2  a 2  b2  2ab cos C  c  a 2  b 2  2ab cos C
c  402  602  2 40 60cos 45  42.495 ft
(b) Find the distance from the center of the
pitcher’s rubber to the far corner of second base.
d  602  602  40  44.853 ft
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Homework, Page 494
37. (c) Find B in ABC.
b 2  a 2  c 2  2ac cos B  2ac cos B  a 2  c 2  b 2
2
2
2


a 2  c2  b2
a

c

b
1
cos B 
 B  cos 

2ac
2ac


2
2
2


40

42.495

60
1
B  cos 
  93.274
2 40 42.495


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Slide 6- 11
Homework, Page 494
41. A player waiting for a kick-off stands at his 5 yd
line, 65 yards from the ball. The ball travels 73 yd at an
angle of 8º to the right of the receiver. Find the distance
the receiver runs to catch the ball
b 2  a 2  c 2  2ac cos B  b  a 2  c 2  2ac cos B
b  652  732  2 65 73cos8  12.504 yd
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 12
Homework, Page 494
45. If ABC is any triangle with sides and angles labeled
in the usual way, then b 2  c 2  2bc sin A.
True. If b 2  c 2  2bc sin A, then a 2  0, which is not
possible if a, b, and c are real numbers.
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Slide 6- 13
Homework, Page 494
49. Two boats start at the same point and speed away on
courses that form a 110º angle. If one boat travels at 24
mph and the other at 32 mph, how far apart are the boats
after 30 min?
a 2  b 2  c 2  2bc cos A
(a) 21 miles
2
2
a

b

c
 2bc cos A
(b) 22 miles
2
2
(c) 23 miles
 24   32 
 24 32 
a        2
cos110

(d) 24 miles
 2   2 
 4 
(e) 25 miles
 23.051
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Slide 6- 14
Homework, Page 494
53. Two ships leave the same port at 8:00 AM and travel
at a constant rate of speed. Each ship keeps a log of its
distance from the port and distance from the other ship.
Time
To port
To B
Time
To port
To A
9:00
15.1
8.7
9:00
12.4
8.7
10:00
30.2
17.3
11:00
37.2
26
(a) How fast is each ship traveling?
spd A  15.1 kts; spd B  12.4 kts
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Slide 6- 15
Homework, Page 494
53. (b) What is the angle of intersection of the courses
of
the2 two2ships?
2
c  a  b  2ab cos C  2ab cos C  a 2  b 2  c 2
2
2
2


a 2  b2  c2
a

b

c
1
cos C 
 C  cos 

2ab
2ab


2
2
2


15.1

12.4

8.7
1
C  cos 
  35.180
2 15.1 12.4


(c) How far apart are the ships at 12:00 noon, if
they maintain the same courses and speeds?
By similar triangles, distance  4*8.7  34.8 nmi
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 16