Download Top of Form Solve the triangle. Round lengths to the nearest tenth

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
B = 54
C = 112
b = 18
A  180  B  C  180  54  112  14
Using the law of sine:
a
b
sin A
sin14

ab
 18
 5.4
sin A sin B
sin B
sin 54
c
b
sin C
sin112

cb
 18
 20.6
sin C sin B
sin B
sin 54
A =12, a = 22.6, c = 7.4
A =14, a = 5.4, c = 20.6
A = 14, a = 7.4, c = 22.6
A = 12, a = 20.6, c = 5.4
Solve the equation on the interval [0, 2]:
cos2x + 2 cos x + 1 = 0
Let u = cox
u 2  2u  1  0
 u  1
2
 0  u  1
cos x  1
x 
/4, 7/4
/2, 3/2

2
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
C = 110
a=5
b = 11
Using the law of cosine:
c 2  a 2  b 2  2ab cos C
c  52  112  2 *5*11cos(110 )  13.6
Using the law of sine:
sin A sin C
a
5

 sin A  sin C 
sin110  20
a
c
c
13.6
B  180  A  C  180  20  110  50
c = 13.6, A = 20, B = 50
c = 19.4, A = 18, B = 52
c = 16.5, A = 22, B = 48
no triangle
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
B = 43
C = 107
B = 14
A  180  B  C  180  43  107  30
Using law of sine:
a
b
sin A
sin 30

ab
 14
 10.3
sin A sin B
sin B
sin 43
c
b
sin C
sin107

cb
 14
 19.6
sin C sin B
sin B
sin 43
A = 30, a = 10.3, c = 19.6
A = 30, a = 2.3, c = 21.6
A = 28 , a = 19.6, c = 10.3
A = 28, a = 21.6, c = 12.3
5 of 25
Two sailboats leave a harbor in the Bahamas at the same time. The first sails at 20 mph in
a direction 330°. The second sails at 34 mph in a direction 220°. Assuming that both
boats maintain speed and heading, after 2 hours, how far apart are the boats?
A
40°
30°
C
B
In the above diagram, AC = b = 20 * 2 = 40 miles.
AB = c = 34 * 2 = 68 miles.
AC is sailing at a direction of 330° , so we get the labeled 30° = 360° - 330°
AB is sailing at a direction of 220°, then we get the labeled 40° = 220° - 180°.
The angle A = 180° - 40° - 30° = 110°.
Using law of cosine
a 2  b 2  c 2  2bc cos A  402  682  2 * 40 * 68cos110
a  89.9 miles
89.9 miles
70.5 miles
58.5 miles
95.9 miles
6 of 25
A vector v has initial point P1 and P2. Write v in terms of ai + bj:
P1 = (-5, 1); P2 = (6, 3)
PP
1 2  P2  P1   x2  x1  i   y2  y1  j  (6  (5))i  (3  1) j  11i  2 j
v = 8i + 5j
v = 11i + 2j
v = 5i + 8j
v = 2i + 11j
7 of 25
Test the equation for symmetry with respect to the give axis, line, or pole:
r = 4 cos ; the polar axis
When
 '  
r '  4cos( ')  4cos( )  4cos   r
So it has symmetry with respect to polar axis.
may or may not have symmetry with respect to polar axis
has symmetry with respect to polar axis
8 of 25
Solve the problem:
5x
cos
3x
+ cos
2
2
5 x 3x
5 x 3x


5x
3x
2 cos 2
2  2cos 2 x cos x
cos  cos  2cos 2
2
2
2
2
2
2 sin 2x sin x
2
2 sin 2x sin x
2 cos 2x
x
2 cos 2x cos
2
9 of 25
Use the given vectors to find the specified scalar:
v = 10i + 5j; Find v  v.
v  v  v  v  10  5  125
2
x
2
y
2
2
100
225
2500
125
10 of 25
Find the absolute value of the complex number:
z = 9 - 5i
z  x 2  y 2  92  (5) 2  106
2^/14
^/106
^/14
2
11 of 25
Solve the equation on the interval [0, 2 ]:
(tan x + ^/�3)(2cos x+1)=0
2 5
tan x  3  0  tan x   3  x 
,
3 3
or
1
2 4
2cos x  1  0  cos x    x 
,
2
3 3
/3, 2/3
1/2, 1
2/3, 4/3, 5/3
/3
12 of 25
Find the product of the complex numbers. Leave answer in polar form.
z1 = 5(cos 20 + i sin 20)
z2 = 4(cos 10 + i sin 10)
z1  520 , z2  410

z1 z2  520
 420   5* 4  20
20(cos 200 + i sin 200)
20(cos 30 + i sin 30)


 10  2030  20 cos30  i sin 30

9(cos 30 + i sin 30)
9(-cos 200 - i sin 200) 20(-cos 200 - i sin 200)
13 of 25
Find the exact value by using a difference identity:
sin 15
sin15  sin(45  30 )  sin 45 cos30  cos 45 sin 30 


2 3 1
2 3
21


2 2
2 2
4
^/2(^/3-1)
4
^/2(^/3-1)
4
^/2(^/3+1)
4
^/2(^/3+1)
4
14 of 25
Use the given information to find the exact value of the expression:
Find cos (2, ), csc  = 5/3,  lies n quadrant II.
1
5
3
csc 
  sin  
sin  3
5
2
18 7
3
cos 2  1  2sin   1  2 *    1 

25 25
5
2
7
25
24
25
7
25
24
25
15 of 25
Find another representation, (r, ), for the point under the given conditions.

( 6,
), r > 0 and 2 <  < 4
3

3
 2 
( 6, -
5
)
7

7
and
, then
are coterminal. so
3
3
3
 7 
 6,
 is the same as
 3 
 
 6, 
 3
3
2
( 6, -
)
3
4
( 6, -
)
3
7
( 6,
)
3
16 of 25
Find the solutions of the equation:
2 cos  + 1 = 0
2cos  1  cos   
1
2
 is then in the 2nd and 3rd quadrants
 1  2
With 0 to 2 , when in the 2nd quadrant,   cos 1    
 2 3
2 4
 1

When in the 3rd quadrant,   2  cos 1     2 
3
3
 2
Then the coterminal angles are:   2n , where n is integer.
 = 2/3 + n  -or-  = 4 /3 + n 
 = 2 /3 + 2n  -or-  = 4 /3 + 2n 
 = /2 + 2n  -or-  = 3/2 + 2n 
 = 3/2 + n
17 of 25
Use the given information to find the exact value of the expression.
Find cos (2), sin  = 20/29,  lies in quadrant I.
2
41
 20 
Using the double angle formula: cos 2  1  2sin 2   1  2 *   
 29  841
540
841
-41
841
39
841
41
841
18 of 25
Use a half-angle formula to find the exact value of the expression:

2
sin 22.5
 22.5    45
22.5 is in the first quadrant, then its sine is > 0.
sin

2

1  cos
1  cos 45


2
2
1
2
2  1 2 2
2
2
1
-
^/2+^/2
2
1
^/2+^-/2
2
1
-
^/2-^/2
2
1
^/2+^/2
2
19 of 25
Express the product as a sum or difference:
sin 6x cos 2x
1
sin( A  B)  sin( A  B)
2
1
1
sin 6 x cos 2 x  sin(6 x  2 x)  sin(6 x  2 x)    sin8 x  sin 4 x 
2
2
sin A cos B 
1/2(sin 8x + sin 4x)
sin (cos 12x2)
1/2(sin 8x + cos 4x)
1/2(cos 8x - cos 4x)
20 of 25
Complete the identity:
sin
cos
+
cos
sin
sin  cos sin 2   cos 2 
1
1



 sec csc
cos sin 
cos sin 
cos sin 
sin  tan 
sec  csc 
1 + cot 
-2 tan2
21 of 25
Solve the equation on the interval [0, 2]:
^/3
sin 4x =
2
Let u = 4x
sin u 
 3
3

 u  sin 1 
  2n   2n
2
3
 2 
or
 3

2
u    sin 1 
 2n
  2n     2n 
2
3
3


When n = 0, u 
When n = 1,

3
or
2
u 

x 
or
3
4 12
6
u

3
 2 
7
u 7
x 
3
4 12
or
u
2
8
u 2
 2 
x 
3
3
4
3
When n = 2
u

3
 4 
13
u 13
x 
3
4 12
or
u
2
14
u 7
 4 
x 
3
3
4
6
When n = 3
u

3
 3* 2 
19
u 19
x 
3
4 12
or
u
2
20
u 5
 3* 2 
x 
3
3
4 3
/4, 7/4
/12, /6, 2/3, 7/12, 7/6, 13/12, 5/3, 19/12
0
0, /4, 
22 of 25
Complete the identity:
cos ( + ) cos ( - ) = ?
cos(   )  cos  cos   sin  sin 
cos(   )  cos  cos   sin  sin 
So
cos(   )cos(   )   cos  cos   sin  sin   cos  cos   sin  sin  
 cos 2  cos 2   sin 2  sin 2 
 1  sin 2   cos 2   sin 2  sin 2 
 cos 2   sin 2  cos 2   sin 2  sin 2 
 cos 2   sin 2   cos 2   sin 2  
 cos 2   sin 2 
cos(2) cos(2) + sin(2) sin(2)
2 - sin2  - sin2
cos2 - sin2
cos2 - 2 sin 2 sin2
23 of 25
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
B = 18
C = 113
b = 44
A= 180C = 180
Using the law of sine:
a
b
sin A
sin 49

ab
 44
 107.5
sin A sin B
sin B
sin18

c
b
sin C
sin113

cb
 44
 131.1
sin C sin B
sin B
sin18
A = 49, a = 107.5, c = 131.1
A = 47, a = 131.1, c = 107.5
A = 47, a =133.1, c = 109.5
A = 49, a =109.5, c = 133.1
24 of 25
Two sides and an angle (SSA) of a triangle are given. Determine whether the given
measurements produce one triangle, two triangles, or no trianagle at all. Solve each
triangle that results. Round lengths to the nearest tenth and angle measure to the nearest
degree.
A = 30
a = 20
b = 40
Using law of sine:
sin B sin A
b
40
1

 sin B  sin A  sin 30  2 *  1  B  90
b
a
a
20
2
Then C = 180 = 180
c
b
sin C
sin 60
3

cb
 40
 40 *
 34.6
sin C sin B
sin B
sin 90
2
B = 60, C = 90, c = 34.6
B = 60, C = 60, c = 34.6
B = 90, C = 60, c = 34.6
no triangle
25 of 25
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
B = 29
C = 112
b = 35
A= 180C = 180
Using the law of sine:
a
b
sin A
sin 39

ab
 35
 45.4
sin A sin B
sin B
sin 29
c
b
sin C
sin112

cb
 35
 66.9
sin C sin B
sin B
sin 29
A = 37, a = 68.9, c = 47.4
A = 37, a = 66.9, c = 45.4
A = 39, a = 45.4, c = 66.9
A = 39, a = 47.4, c = 68.9