Download 162 Precalculus 2 Review 4

162
Precalculus 2
Review 4
1. Solve the right triangle with an angle 37 and opposite side 14.
Solution
On the picture above angle A is 37º, angle C is 90º, and side BC is 14. First we can
find the third angle B  90  A  90  37  53 . Next we notice that
BC
BC
14
BC

 18.58 . Finally, sin A 
whence AC 
AC
tan A tan 37
AB
BC
14

 23.26 .
whence AB 
sin A sin 37
tan A 
2. Solve the right triangle with the hypotenuse 50 and one side 20.
Solution
x
On the picture above A  90 , AB  20, BC  50 . Then sin C 
AB 20

 0.4 and
BC 50
C  arcsin 0.4  23.58 whence B  90  arcsin 0.4  66.42 . We have many ways to find
AC. For example, by the Pythagoras’ theorem AC 2  BC 2  AB 2  502  202  2100 and
AB  2100  10 21  45.83 .
3. From an airplane flying 7500 ft above level ground, one can see two towns
directly to the east. The angles of depression to the towns are 5 10 and 7730.
What is the distance between the towns in miles?
Solution; look at the picture below
We see two right triangles with horizontal sides X and Y. The distance between the
towns is D  X  Y . From the smaller triangle we see that
7500
7500
 tan 77 30  tan 77.5  4.5107 and Y 
 1663 ft . Similarly from the larger
Y
4.5107
7500
7500
 tan 5 10  tan 5.17  0.0905 and X 
 82892 ft .
triangle we obtain
X
0.0905
Therefore D  82892  1663  81229 ft . Recalling that there are 5280 feet in one mile
81229
 15.4 mi .
we get the distance in miles D 
5280
4.
Solve the triangle ABC if B  30 , C  45 , and AC  6 .
Solution; the picture of the triangle is shown below. First notice
that A  180  B  C  180  30  40  105 . Next we apply the law of Sines:
sin A sin B sin C


BC
AC
AB
whence
sin105 sin 30 sin 45


.
BC
6
AB
From the last equation we obtain that BC 
that AB 
6sin105
 11.59 and
sin 30
6sin 45
 8.49 .
sin 30
5.
Solve the triangle ABC if A  35 , BC  6, and AC  8.
Solution; first of all let us see whether the problem has no solution, one solution,
or two solutions. Because AC sin A  8sin 35  4.6 we have the inequality
AC sin A  BC  AC , which means that the problem has two solutions.
Next, applying the law of Sines we see that
sin35 sin B sin C
.


6
8
AB
From the proportion above we can find that
sin35 sin B sin C


6
8
AB
sin B 
8sin 35
 0.7648 .
6
In our first solution we take
B  arcsin(0.7648)  49.89 .
Therefore
C  180  35  49.89  95.11 .
And finally,
AB 
6sin C 6sin 95.11

 10.42 .
sin 35
sin 35
The triangle corresponding to our first solution is shown below.
To get the second solution we take
B  180  arcsin(0.7648)  130.11 .
Then
C  180  35  130.11  14.89 .
And
AB 
6sin C 6sin14.89

 2.69 .
sin 35
sin 35
The triangle corresponding to the second solution is shown below.
6.
A pole leans away from the sun at an angle of 100 from the vertical. When
the angle of elevation of the sun is 450, the pole casts a shadow 52 ft long on level
ground. How long is the pole?
Solution; the picture below illustrates the problem. In the triangle ABC angle C is
45º, angle A is 90º-10º=80º, and side AC is 52 ft. The length of the pole equals to
side AB. Angle C is equal to 180º-80º-45º=55º. By the law of Sines we have
sin 45 sin 55
52sin 45
whence AB 

 44.9 ft .
AB
52
sin 55
7.
Solve the triangle ABC if AB  12, AC  10, and A  30
Solution;
We apply the law of Cosines.
BC 2  AB 2  AC 2  2 AB  BC cos A  122  102  2 10 12 cos 30  36.15 .
Whence
BC  36.15  6.01 .
Next we apply the law of Cosines again to find angle B,
cos B 
AB 2  BC 2  AC 2 122  6.012  102

 .5555 .
2 AB  BC
2 12  6.01
Therefore
B  arccos.5555  56.26 .
Finally, C  180  A  B  180  30  56.26  93.74 .
8.
Solve the triangle ABC if BC  16, AC  13, and AB  19.
Solution; by the law of Cosines
cos A 
AB 2  AC 2  BC 2 192  132  162

 0.5547 .
2 AB  AC
2 19 13
Whence
A  arccos(0.5547)  56.31 .
Similarly
cos B 
AB 2  BC 2  AC 2 192  162  132

 0.7368
2 AB  BC
2 19 16
and
B  arccos(0.7368)  42.54 .
Finally,
C  180  A  B  180  56.31  42.54  81.15
9. A hill is inclined 100 to the horizontal. A 26-ft pole stands on the top of the hill.
How long a rope will it take to reach from the top of the pole to a point 32 ft
downhill from the base of the pole?
Solution; look at the picture below. The line OC represents the slope of the hill, so
the angle ‫ﮮ‬COD equals to 10º. The side AC is 32 ft and the side BC is 26 ft. Te
length of the rope is equal to the side AB.
The angle ‫ﮮ‬OCD equals to 90º - ‫ﮮ‬COD = 90º - 10º=80º. Therefore the angle
º
º
º
‫ﮮ‬ACB equals to 180 -80 =100 . By the law of Cosines we have
AB 2  AC 2  BC 2  2 AC  BC cos100  322  262  2  32  26cos100  1988.95 .
Therefore
AB  1988.95  44.60 ft .
Review of extra credit problems (will be discussed only if time allows).
10. Convert to polar notation and then multiply (4  4i 3)(2 3  2i) .
Solution; to convert a complex number x  yi to its polar form r (cos   i sin  ) we
use the formulas
r  x2  y 2 ,
b

arctan a if a  0,

arctan b   if a  0,

a



   if a  0, b  0,
2
 
 2 if a  0, b  0,

undefined if a  b  0.

In particular, for z  x  yi  4  4i 3 we have r  42  (4 3)2  64  8 ,
and   arctan
4 3



 arctan 3  , whence 4  4i 3  8(cos  i sin ) . Similarly,
4
3
3
3
2 3  2i  (2 3) 2  22 (cos(arctan
2
2 3
)  i sin(arctan
2
2 3
))  4(cos


 i sin ) .
6
6
Next we use the formula
r1 (cos 1  i sin 1 )  r2 (cos 2  i sin 2 )  r1r2 (cos(1  2 )  i sin(1  2 ))
to get
  
  
(4  4i 3)(2 3  2i)  32(cos     i sin   ) 
3 6
3 6
 32(cos
11.


 i sin )  32i.
2
2
Convert to polar notation and then divide
Solution; first we notice that
3  i  ( 3)2  12 (cos(arctan
and that
2 3  2i  (2 3) 2  (2) 2 (cos(arctan
3 i
.
2 3  2i
1
1


)  i sin(arctan
))  2(cos  i sin )
6
6
3
3
2
2


)  i sin(arctan
))  4(cos( )  i sin( )) .
6
6
2 3
2 3
Next we apply the formula
r1 (cos 1  i sin 1 ) r1
 (cos(1   2 )  i sin(1   2 ))
r2 (cos  2  i sin  2 ) r2
to get
3 i
1


1
  
   1
 (cos     i sin   )  (cos  i sin )  i .
2
2
2
2 3  2i 2
3 6
3 6 2
12. Find ( 3  i)5 .
Solution; first we convert ( 3  i) to the polar form


( 3  i )  2(cos( )  i sin( )) and then we use de Moivre’s formula
6
6
[r (cos   i sin  )]n  r n (cos n  i sin n ) .
Therefore
 5
( 3  i )5  25 (cos  
 6

 5
  i sin  

 6
 32(

 5
)  32(cos 

 6
3 1
 i )  16( 3  i ).
2 2

 5
  i sin 

 6

) 

13. Solve x 6  64 .
Solution; first we convert the right part to the polar form 64  64(cos   i sin  ) .
Next recall how we solve the equation xn  r (cos  i sin  ) . The formula is
  2m
  2m

x  n r  cos
 i sin
n
n


 , m  0,1,..., n  1.

In our case we have
  2m
  2m

x  6 64  cos
 i sin
6
6


 , m  0,1,...,5.

We get the following six solutions of our equation.



x0  2  cos  i sin   3  i ,
6
6

  2
  2 




x1  2  cos
 i sin
  2  cos  i sin   2i ,
6
6 
2
2


  4
  4 
5
5 


x2  2  cos
 i sin
 i sin
  2  cos
   3 i,
6
6 
6
6 


  6
  6 
7
7 


x3  2  cos
 i sin
 i sin
  2  cos
   3 i,
6
6 
6
6 


  8
  4 
3
3 


x4  2  cos
 i sin
 i sin
  2  cos
  2i ,
6
6 
2
2 


  10
  10 
11
11 


x5  2  cos
 i sin
 i sin
  2  cos
  3 i .
6
6 
6
6 


Notice that we have three pairs of conjugate complex solutions.