Chapter9PartyPlanner
Name: __________________________ Date: _____________
1. Given A  53 , B  78 , and a  6.1 , use the Law of Sines to solve the triangle for the
value of b. Round answer to two decimal places.
C
b
A
a
c
B
b
a

sin  B  sin  A 
b
6.1

sin  78  sin  53
b
6.1sin  78 
sin  53
b  7.47
2. Given C  127 , B  34 , and c  15 , use the Law of Sines to solve the triangle for the
value of a. Round answer to two decimal places.
A  B  C  180
A  180  B  C
A  180  34  127
A  19
a
c

sin  A  sin  C 
15
a

sin 19  sin 127 
a
15sin 19 
sin 127 
a  6.11
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Chapter 9 Party Planner
3. Given A  17 , b  9 , and a  7 , use the Law of Sines to solve the triangle (if possible)
for the value of c. If two solutions exist, find both. Round answer to two decimal places.
Pay attention to the directions and the words “If two solutions exist…” .This is a warning that you are looking at SSA, which we learned may result in two different triangles, one triangle or even NO triangles. Start with a picture: IMPORTANT: Determine the height of the triangle. h  9sin 17 
h  2.63
Since mA  90 and 2.63  7  9 we are looking for TWO possible triangles. A sketch of each is advised. CASE 1: sin  B  sin  A 

b
a
sin  B  sin 17 

9
7
9sin 17 
sin  B  
7
 9sin 17  
B  sin 1 

7


B  22.08
C  180  17  22.08
C  140.92
7
c

sin 17  sin 140.92 
c
7 sin 140.92 
sin 17 
c  15.09
(Continued) Page 2
Chapter 9 Party Planner
CASE 2: In case 2, the measure of angle B is the supplement to 22.08 degrees (linear pairs). B  180  22.08
B  157.92
C  180  17  157.92
C  5.08
c
7

sin 17  sin  5.08 
c
7 sin  5.08 
sin 17 
c  2.12
Therefore, the two solutions are c  2.12,15.09 . NOTE: Back in step 1, if we had found the length of the side opposite the given angle equaled the height, we would have only one triangle to solve. Additionally, if the height was larger than the opposite side, then there is no triangle that could be formed. Page 3
Chapter 9 Party Planner
4. A straight road makes an angle, A, of 20 with the horizontal. When the angle of
elevation, B, of the sun is 59 , a vertical pole beside the road casts a shadow 6 feet
long parallel to the road. Approximate the length of the pole. Round answer to two
decimal places.
The red markings were added to the picture. The Law of Sines can now be used to find the height of the pole. pole
6

sin  39  sin  31
pole 
6sin  39 
sin  31
pole  7.33 feet
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Chapter 9 Party Planner
5. After a severe storm, three sisters, April, May, and June, stood on their front porch and
noticed that the tree in their front yard was leaning 3 from vertical toward the house.
From the porch, which is 108 feet away from the base of the tree, they noticed that the
angle of elevation to the top of the tree was 22 . Approximate the height (length) of the
tree. Round answer to two decimal places.
Note: the phrasing of the question should be changed from height of the tree to length of the tree. As before, start with a sketch: 108
tree

sin  71 sin  22 
tree 
108sin  22 
sin  71
tree  42.79 feet
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Chapter 9 Party Planner
6. Given a  4 , b  13 , and c  11 , use the Law of Cosines to solve the triangle for the
value of A. Round answer to two decimal places.
C
a
b
Figure not drawn to scale
A
c
B
a 2  b 2  c 2  2bc cos  A 
2bc cos  A   a 2  b 2  c 2
2bc cos  A   b 2  c 2  a 2
b2  c 2  a 2
2bc
 b2  c2  a 2 
A  cos 1 

2bc


cos  A  
 132  112  42 
A  cos 1 

 2 13 11 


A  cos 1 132  112  42  /  2 13 11  Ti-83s need parenthesis
A  16.66
7. Given C  109 , a  8 , and b  7 , use the Law of Cosines to solve the triangle for the
value of c. Round answer to two decimal places.
c 2  a 2  b 2  2bc cos  C 
c  a 2  b 2  2bc cos  C 
c  8  7  2  8  7  cos 109 
2
2
c  12.23
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Chapter 9 Party Planner
8. In the figure below, a  8 , b  12 , and d  14 . Use this information to solve the
parallelogram for  . The diagonals of the parallelogram are represented by c and d.
Round answer to two decimal places.
c

a
d

figure not drawn to scale
b
NOTE: The diagonal’s entire length is 14 units. Also,  and  are the measures of the angles formed by the sides of the parallelogram. Also recall consecutive angles of a parallelogram are supplementary. Consider the following picture with important information highlighted: Use the law of cosines to find the measure of  then subtract  from 180 to find  , (again consecutive angles of a parallelogram are supplementary.)  82  122  142 
  cos 

 2  8 12  
  86.42
  180  
  93.58
1
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Chapter 9 Party Planner
9. A vertical pole 28 feet tall stands on a hillside that makes an angle of 14 with the
horizontal. Determine the approximate length of cable that would be needed to reach
from the top of the pole to a point 58 feet downhill from the base of the pole. Round
answer to two decimal places.
Since the hillside makes and angle of 14 degrees with horizontal, start your drawing with a hill that is inclined 14 degrees. Add the rest of the parts described in the problem to your picture. (green and blue below) Next throw in the red perpendicular lines at B that give you a line parallel to horizontal through B. By alternate interior angles we see a portion of angle B is 14 degrees and the other portion is 90 degrees. Therefore: B = 104 degrees. We can now use the law of cosines to find the length of the wire. b  582  282  2  58  28  cos 104 
b  70.24 '
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Chapter 9 Party Planner
10. A triangular parcel of land has sides of lengths 250, 520, and 650 feet. Approximate
the area of the land. Round answer to nearest foot.
The area of a triangle formula is 1
Area  bc sin  A  2
Note the SAS format of the area formula’s requisite information. We will need the measure of an angle between two of the sides. Therefore, we start with the law of cosines.  2502  6502  5202 
A  cos 1 

 2  250  650   A  48.68
We can now use our area formula: 1
Area  bc sin  A 
2
1
Area   250  650  sin  48.68  2
Area  61, 021 ft 2
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Chapter 9 Party Planner
11. Determine the quadrant in which the angle lies. (The angle measure is given in radians.)
–3
7
Once again, a picture will help. Notice the relationship between the given fraction and the quartile cutoff at 
 3.5

2
 7

 . Clearly, the angle is to be found in quadrant IV. 
12. Determine the area of a triangle having the following measurements. Round your
answer to two decimal places.
A  130, b  8, and c  14
1
Area  bc sin  A 
2
1
Area   8 14sin 130 
2
1
Area   8 14sin 130 
2
Area  42.90 units 2
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Chapter 9 Party Planner
Answer Key
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
b  7.47
a  6.11
c  2.12 and 15.09
7.33 feet
42.79 feet
16.66
12.23
93.58
70.24 feet
61,018 ft2
IV
42.90 sq. units
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