Download Mutiple-Choice Questions 1. Which of the following is the solution of

Mutiple-Choice Questions
1.
Which of the following is the solution of the equation tan   1  0 on the interval 0     ?
Answer: C
If tan   1  0 , then tan   1 . Since 0     , the angle is necessarily located in
3
quadrant II. Thus,  
.
4
2.
What is the solution set of the equation cos  2   cos on the interval 0    2 ?
Answer: B
If cos  2   cos , then cos  2   cos  0 . Using the double-angle formula for cosine,
we then have  2 cos 2   1  cos   0 , or 2 cos 2   cos   1  0 .
Here we recognize the quadratic form 2 X 2  X  1  0 , where X  cos .
Therefore, X 
1  1  4(2)(1) 1  3
1

  ,1 .
2(2)
4
2
In the interval 0    2 , cos  1 for   0 and cos   
1
2 4
,
for  
.
2
3 3
 2 4 
, .
Thus, the solution set of the equation is given by 0,
 3 3 
[Check this!]
3.
 x
Which of the following is a zero of the function f ( x)  2sin 2    2 cos x ?
2
Answer: D
By substituting the 4 choices, you can check that only choice D is a zero, i.e.
f  5   0 .
Otherwise, solve the equation f  x   0 to get all the zeros of f :
Using the half-angle formula for sine, we have:
 x
 1  cos x 
f ( x)  2sin 2    2 cos x  2 
  2 cos x  1  cos x
2 
2

Therefore,
f ( x)  0  1  cos x  0  cos x  1
This implies that any odd multiple of  is a zero of f , so D is the answer.
4.
If A  400 , C  450 , b  4 , then what is c ?
Answer: A [ASA case]
Since B  180  40  45  950 , by the Law of Sines we have:
0
sin 450 sin 950
4sin 450

c
 2.84 .
c
4
sin 950
5.
If a  3, b  2, A  500 , then what is the perimeter of ABC?
Answer: A [SSA case]
By the Law of Sines we have:
sin 500 sin B
2sin 500

 sin B 
 B  30.70
3
2
3
0
0
Hence C  180  50  30.7   99.3 and we have:
sin 500 sin 99.30
3sin 99.30

c
 3.86
3
c
sin 500
Therefore, the perimeter of ABC is given by:
a  b  c  3  2  3.86  8.86 .
6.
If a  3, b  7, A  700 , then what can you conclude?
Answer: D [SSA case]
Here the Law of Sines implies that
sin 700 sin B
7sin 700

 sin B 
 2.19  1 … which is impossible since
3
7
3
1  sin   1 for any angle  !
Therefore, there is no triangle ABC that satisfies these conditions.
7.
If b  4, c  6, B  200 , then what is the measure of A?
Answer: C [SSA case]
This is the SSA case where there are two possible triangles ABC (one with angle A
obtuse, the other with angle A acute).
1) If angle A is obtuse, then the Law of Sines yields
sin 200 sin C
6sin 200

 sin C 
 C  30.90
4
6
4
Hence:
A  180  20  30.9   129.10
0
2) If angle A is acute, then C  180  30.9   149.10 and so
0
A  180  20  149.1  10.90 .
0
8.
If a  3, c  2 , and the perimeter of ABC is 8, then what is the measure of A + B?
Answer: D [SSS case]
First, note that ABC is isosceles (a = b = 3). Thus, the measure of A + B is equal to the
measure of 2A.
Applying the Law of Cosines here yields:
cos A 
22  32  32 1
  A  70.530
2(2)(3)
3
Therefore,
A  B  2 A  2(70.530 )  141.10 .
9.
A person in a small boat, offshore from a vertical cliff known to be 100 feet in height, takes a
sighting of the top of the cliff. If the angle of elevation is found to be 25°, how far offshore is
the boat?
Answer: D
Say the distance between the boat and the shore is d . We then have:
tan  250  
10.
100
100
d
 214.45'
d
tan  250 
The dimensions of a triangular lot are 60 meters by 50 meters by 40 meters. If the price of
land is $325.00 per square meter, then how much does the lot cost (rounded to the nearest
dollar)?
Answer: C
First, compute the area of the triangular lot by applying Heron’s formula:
A  75  75  60 75  50  75  40   992.1567m2
Multiplying this area by 325 (the price per square meter) yields a final price of,
approximately, $322,451.
Problem 1
A golfer hits an errand tee shot that lands in the rough. A marker in the center of the fairway is exactly
150 yards from the flag in the green. While standing on this marker and facing the flag, the golfer
turns 110° and paces off 35 yards to his ball (see the illustration below.)
How far is the ball from the flag? Round your answer to the nearest hundredth of a feet.
Let x be the distance from the golf ball to the flag. By the law of cosines, we then have:
x 2  1502  352  2(35)(150) cos1100
 x  165.28
We conclude that the ball is, approximately, 165.28 yards away from the flag.
Problem 2
Consider the complex number z  3  3i .
a)
Write z in trigonometric form.
The modulus of z is given by
r  32 
 
3
2
 9  3  12  2 3
Since x  r cos and y  r sin  , we have cos  
Thus,   300 and z  2 3  cos 300  i sin 300  .
b)
3
1
and sin   .
2
2
Show that z is a solution of the equation z 6  1728  0 .
Applying De Moivre’s Theorem, we get:
 
z 6  2 3 cos  6  300   i sin  6  300   64  27  cos1800  i sin1800    1728
6
Therefore,
z 6  1728  0

Problem 3
A 10-feet long loading ramp that makes an angle of 18° with the horizontal is to be replaced by one
that makes an angle of 12°with the horizontal.
How long is the new ramp? Round your answer to the nearest hundredth of a feet. Include a diagram
with your work.
Let x be the length AD of the new ramp. Then the problem is illustrated by the diagram above.
ˆ measures  90  18 0  720 .
Looking at the right triangle ABC, we see that angle ABC
ˆ measures 18  12 0  60 and so angle ADC
ˆ
Looking at triangle ACD, we see that angle DAC
measures 180  72  6   1020 .
0
By the Law of Sines, we then have:
sin 720 sin1020
10sin 720

x
 9.72
x
10
sin1020
We conclude that the new ramp measures, approximately, 9.72 feet.
Problem 4
To approximate the area of a lake, Cindy walks around the perimeter of the lake, taking the
measurements shown in the illustration below.
What is the approximate area of the lake? Round your answer to the nearest square meter.
From the illustration above, we see that the polygon ABCDE (an approximate covering of the
lake) can be partitioned into the 3 triangles ABE, BDE, and BCD, whose respective areas we
shall denote by A1, A2, and A3.
We then need to compute each of these 3 areas:
A1 
1
(100)(125) sin 500  4788m 2
2
1
A3  (70)(50) sin1000  1723m 2
2
To compute A2, we first find the lengths BE and BD using the Law of Cosines:
BE  1002  1252  2(100)(125) cos 500  97.75m
BD  702  502  2(70)(50) cos1000  92.82m
Now we use Heron’s formula:
s
So
1
 97.75  92.82  50   120.285 (the semi-perimeter of BDE)
2
A2  s(s  50)(s  97.75)(s  92.82)  2287m 2 .
We conclude that the total area of the lake is given, approximately, by
A1  A2  A3  8798m 2
Bonus Problem 1
Find the complex fifth roots of  i .
 3

 3

 2k   i sin 
 2k  , where k is any integer, we have the following 5
Since i  cos 
 2

 2

complex fifth roots:
 3 
 3 
z0  cos    i sin  
 10 
 10 
 3 2 
 3 2 
z1  cos 


  i sin 

 10 5 
 10 5 
 7 
 7 
 cos 
  i sin 

 10 
 10 
 3 4 
 3 4 
z2  cos 


  i sin 

 10 5 
 10 5 
 11 
 11 
 cos 
  i sin 

 10 
 10 
(k 0)
( k  1)
(k  2)
 3 6 
 3 6 
z3  cos 


  i sin 

 10 5 
 10 5 
 15 
 15 
 cos 
  i sin 

 10 
 10 
 3 
 3 
 cos    i sin    i !
 2 
 2 
( k  3)
 3 8 
 3 8 
z4  cos 


  i sin 

 10 5 
 10 5 
 19 
 19 
 cos 
  i sin 

 10 
 10 
(k  4)
Bonus Problem 2
Solve the trigonometric equation sin   1  cos  on the interval 0    2 .
Using the sum formula for sine, we have:
sin   1  sin   cos1  cos  sin1
So the equation is equivalent to
cos1 sin   (sin1  1)  cos   0
Here we recognize a trigonometric equation linear in sine and cosine of the form
a sin   b cos  c , where a  cos1, b  sin1 1, and c  0 .
Using the general solution given in Example 7 on pages 499-500, we get:
cos  
a
a b
2
2

cos1
cos 1  (sin1  1)
2
2
   0.2853 or   16.350
,
sin  
b
a b
2
2

sin1  1
cos 1  (sin1  1) 2
[located in quadrant IV]
2
So
  sin 1  0    2k , where k is any integer.
This implies that there are two solutions:
  16.350 or   196.350