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Suggested solutions V1 MKUMaDNVC06K5 trigonometry NV-College
Make Up Chapter Test: Ch5 trigonometry; MATHEMATICS COURSE D;
Spring 2008: MaDNVC06
Please note that you have to try to solve the problems by yourself before checking your solutionsagainst mine. My solutions are just suggested ones. Usually there are more than one methods ofsolving a given problem.
Warning: Just reading the solutions can never replace your own struggle in solving a givenproblem. By just reading the solutions you may not be able to understand the mathematics of theproblem deep enough and therefore, just going through my solutions may not help you to solve asimilar problem by yourself. Have Fun!
Behzad
InstructionsTest period 8:15-10:45
Tools Formula sheet, ruler and graphic calculator.
The test For most items a single answer is not enough. It is also expected
that you write down what you do
that you explain/motivate your reasoning
that you draw any necessary illustrations.
All together there are 18 problems. 6 problems are multiple choice problems and are compulsory for
all. From the problem 7 to 18 you are given a choice. You may choose from the given alternatives, a
problem that suits you the best. In the box below mark the problems you want me to grade. ONLY
AND ONLY THOSE PROBLEMS MARKED IN THE TABLE BELOW ARE GOING TO BE
GRADED.
After every item is given the maximum mark your solution can receive. [2/3] means that the item can
give 2 g-points (Pass level) and 3 vg-points (Pass with distinction level).
Items marked with give you a possibility to show MVG-quality (Pass with special distinction
quality). This means that you use generalised methods, models and reasoning, thatyou analyse your
results and account for a clear line of thought in a correct mathematical language.
Problems 8 and 10 are larger problems which may take up to 45 minutes to solve completely. It is
important that you try to solve this problem. A description of what your teacher will consider when
evaluating your work is attached to the problem. This problem is heavily graded. Even in G-level
there are a lot of moments of the problem that you can solve.
Mark limits G-Level: The test gives totally at the most 24 points. To pass the test you must have at least 15 points.
VG/MVG-Level: The test gives totally at the most 27 points out of which 15 VG-points and 3
MVG-pints. To pass the test you must have at least 9 points and to get the test character Pass with
distinction (VG) you must have at least 18 points out of which at least 5 points on Pass with
distinction level. Excellence (MVG) requires 20 points out of which at least 10 VG points and
excellent quality presentation of the solutions. Have fun!
Note: Only those problems checked here will be graded!
Problem 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Total
Grade
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In the multiple choice problems below, circle
thethe correct alternative and write clearly
correct answer in the space provided as
Alternative:
Answer the questions 2 to 3 based on the
accompanying diagram of the unit circle O with the radius unitsOBOD 1== . The
line CD is tangent to the unit circle at D
and the line AB is perpendicular to the x
axis.
1. The distance CD represents: [1/0]a. cos
b. cos c. sin
d. tan e.
+
2sin
Answer: Alternative:________________
Answer: Alternative d: CD is tan
2. The distance OA represents: [1/0]a. cos
b. cos c. sin d. tan
e.
+
2sin
Answer: Alternative:________________
Answer: Alternative a: OA is cos
3. The distance AB represents: [1/0]
a. cos b. cos c. sin d. tan
e.
+
2sin
Answer: Alternative:________________
Answer: Alternative c: AB is sin
A
B
D
C
O x
y
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4. If is a positive acute angle and sin=a , an expression for 2cos in terms of a may be written as
a. 212cos a=
b.2
2
1
12cos
a
aa
=
c. 2212cos a= d. a22cos = Answer: Alternative:________________
Suggested solutions: Answer: Alternative c: 2212 a=cos
2sin212cos = 2212cos a=
5. If5
3sin = and tan is positive: [1/0]
a.5
4cos =
b.5
4cos =
c.25
16cos =
d.3
5cos = Answer: Alternative:________________
Suggested solutions: Answer: Alternative a:
5
4=cos
Using the trigonometry unity:
1cossin 22 =+
2sin1cos =
Due to the fact that sin is negative while tan is positive, the angle
lies in the third quadrant and therefore cos must be negative (of
course, the same conclusion may be achieved using
cos
sintan = .)
2sin1cos =
54
2516
25925
2591
531sin1cos
2
2 ====
==
54cos =
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6. Which one/ones of the following equations has/have two solution in the interval0 . Why? Explain [NPMaDVT02_modified 1/1]
a.5
4sin =
b. 5
4
cos =
c.25
16tan =
d.5
4sin = Answer: Alternative(s):________________
Suggested solution: Answer: Alternatives a .
5
4sin = has two solutions in the interval 0 . One of the answers
is in the interval
2
0 , and the other is in the interval
2
.
5
4sin = has no solution in the interval 0 .
5
4cos = has only one solution in the interval 0 .
25
16tan = has also only one solution in the interval 0 .
Choose between the problems 7 and 8. Solve only one of them. Note that problem number 7 is
in G level and problem number 8 is in VG-level. Note also that, even if you originally chooseto solve problem 7, if you get extra time over at the end, you may try to solve problem
number 8. If successful, you may then ask me to grade problem 8 instead of problem 7. You
do not need to cross out problem 7. Leave it as it is. If not sure about your solutions of
problem 8, you may ask me to check your solutions of problem 8, and if correct then choose it
instead of 7. Obviously, you will get grade for your solution of problem 7 or 8 but not both
simultaneously.
7. Solve 5.0sin =x . [2/0]
Suggested Solutions:5.0sin =x
( )5.0sin 1=x Answers: ...,3,2,1,036030 =+= nnx
...,3,2,1,036015036030180 =+=+= nnnx
Answers: ...,3,2,1,0360150 =+= nnx
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Suggested solutions V1 MKUMaDNVC06K5 trigonometry NV-College
8. Find all solutions of the equation 15
3sin2 2 =
x in the interval 20 x . [0/4]
Suggested Solutions:
1
5
3sin2 2 =
x
2
1
5
3sin 2 =
x
2
2
2
1
5
3sin ==
x
...,3,2,1,0242
2sin
53 1 =+=
= nnx
...,3,2,1,024
32
42
2sin
53 1 =+=+=
= nnnx
...,3,2,1,024
52
42
2sin
53 1 =+=++=
= nnnx
...,3,2,1,024
7
2422
2
sin531
=+=+=
=
nnnx
Solving the equations above individually result in:
...,3,2,1,0242
2sin
53 1 =+=
= nnx
...,3,2,1,0242
2sin
53 1 =+=
= nnx
...,3,2,1,0220
92
20
542
453 =+=+
+=++= nnnnx
Divide both sides by 3: ...,3,2,1,03
2
20
3=+= n
nx
[0/1]
...,3,2,1,024
32
42
2sin
53 1 =+=+=
= nnnx
nnnx 220
192
20
1542
4
3
53 +=+
+=++= ...,2,1,0
3
2
60
19=+= n
nx
Similarly [0/1]
...,3,2,1,024
52
42
2sin
53 1 =+=++=
= nnnx
nnnx 220
292
20
2542
4
5
53 +=+
+=++= ...,2,1,0
3
2
60
29=+= n
nx
Similarly [0/1]
...,3,2,1,024
72
42
2
2sin
53 1 =+=+=
= nnnx
nnnx 220
392
20
3542
4
7
53 +=+
+=++= ,...2,1,0
3
2
60
39=+= n
nx
Answers:
60
59,
60
49,
60
39,
60
29,
60
19,
60
9 =x [0/1]
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Choose between the problems 9 and 10. Solve only one of them. Note that problem number 9
is in G level and problem number 10 is in MVG-level. Note also that, even if you originally
choose to solve problem 9, if you get extra time over at the end, you may try to solve problem
number 10. If successful, you may then ask me to grade problem 10 instead of problem 9.You do not need to cross out problem 9. Leave it as it is. If not sure about your solutions of
problem 10, you may ask me to check your solutions of problem 10, and if correct then
choose it instead of 9. Obviously, you will get grade for your solution of problem 9 or 10 but
not both simultaneously.
9. In a triangle ABC, a side is measured to be cmBC 5.77= , and its angles are = 5.33A
and = 5.81C .
a. Find the length of the side AB . [1/0]
b. Find the area of the triangle ABC. [2/0]
Suggested solutions:Data: = 5.33A , = 5.81C , cmBC 5.77=
a. Answer: cm139 .AB
c
C
b
B
a
A sinsinsin==
AB
=
5.81sin
5.77
5.33sin
= 5.81sin5.775.33sinAB
cmcmAB 87.1385.33sin
5.81sin5.77 =
= Answer: cmAB 139 [1/0]
b. Answer: 2cm 4880Area
= 5.33A , = 5.81C ; cmBCa 5.77== , cmABc 139=
( ) ( ) =+=+= 655.815.33180180 CAB [1/0]
We may use the are formula to calculate the area of the triangle:
2
sinBacArea = .
2248806.4881
2
65sin1395.77
2
sincmcm
BacArea =
== [1/0]
Answer: 24880 cmArea
A
B
C5.335.81
cm5.77
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When assessing your work your teacher will consider:
if the method you have used is reasonable if your calculations are correct what conclusions you have made from your investigation
how plain and pronounced your presentation is, and what kind of mathematical knowledge you have shown to possess.
10.In the triangle ABC illustrated blow angle BA = 3 and cmAB 0.12= . If the
relationship between the length of two corresponding sides are 8:5 , i.e. as illustrated in
the figure below if5
8=
AC
BC
a. Find the angles of the triangle analytically. [0/2/]b. If = 27.36B , Find the area of the triangle. [2/0]
The figure is not properly scaled.
xBC 8=
A
C B
xAC 5=
3 cmAB 0.12=
Suggested solutions: Answer: = 27.36B ; = 8.108A ; = 9.34C ;2
5.70 mArea
Data: cmAB 0.12=
, BA=
3 , xBC 8=
, xAC 5=
;
B ; 3
A In order to solve the problem we must first find its angles. We may usethe sine law for this task:
c
C
b
B
a
A sinsinsin==
xx /=
/ 8
3sin
5
sin
Using sincoscossin)sin( +=+ , cossin22sin = and
1cos22cos 2 = 2sincos2cossin)2sin()3sin( +=+=
cossincos21cos2sin)3sin(2 +=
( ) ( )1cos4sincos21cos2sin)3sin( 222 =+=
1cos4sin)3sin(2 = [0/1]
xx /=
/ 8
3sin
5
sin
( )8
1cos4sin
5
sin2
=
8
1cos4
5
12
=
5cos208 2 = 1358cos20 2 =+= 20
13cos
2 = 20
13cos =
=
= 27.36
20
13cos 1
Answer: = 27.36B ; = 8.1083A ; === 9.3427.361804180 C
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==
= 73.14327.36180
20
13cos 1 Not acceptable. > 1803A
[0/1/]
Use again the sine law to calculate the length of each side of thetriangle:
c
C
b
B
a
A sinsinsin==
mACAC
4.129.34sin
27.36sin12
12
9.34sin27.36sin=
=
=
[1/0]
mAC 4.12=
( ) 248.70
2
8.108sin4.1212
2
sinm
AbcArea =
== [1/0] Answer: 25.70 mArea
Choose between the problems 11 and 12. Solve only one of them. Note that problem number11 is in G level and problem number 12 is in VG-level. Note also that, even if you originally
choose to solve problem 11, if you get extra time over at the end, you may try to solve
problem number 12. If successful, you may then ask me to grade problem 12 instead of
problem 11. You do not need to cross out problem 11. Leave it as it is. If not sure about your
solutions of problem 12, you may ask me to check your solutions of problem 12, and if
correct then choose it instead of 11. Obviously, you will get grade for your solution of
problem 11 or 12 but not both simultaneously.
11.Solve
2
2cos =x has two solutions in the interval 20 x , one of them is
4
=x .
What is the other solution? [1/0]
Suggested solutions: Answer:4
7=x
Data:2
2cos =x ,
4
=x , 20 x
4
7
4
8
44
8
42
=
===x [1/0] Answer:
4
7=x
12.Find exact value of ( )+cos if ( ) 33.0cos = . [0/1]Suggested solutions: Answer: ( ) 33.0cos =+ Data: ( ) 33.0cos =
( ) 33.0coscos ==+ [0/1] Answer: ( ) 33.0cos =+
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Choose between the problems 13 and 14. Solve only one of them. Note that problem number
13 is in G level and problem number 14 is in VG-level. Note also that, even if you originally
choose to solve problem 13, if you get extra time over at the end, you may try to solve
problem number 14. If successful, you may then ask me to grade problem 14 instead of
problem 13. You do not need to cross out problem 13. Leave it as it is. If not sure about your
solutions of problem 14, you may ask me to check your solutions of problem 14, and ifcorrect then choose it instead of 13. Obviously, you will get grade for your solution of
problem 13 or 14 but not both simultaneously.
13.If 3.0sin = find the exact values of:a. cos [1/0]
b. 2sin [2/0]c. tan [1/0]
Suggested solutions: 91.0cos =a ; 9106.02sin = ;91
913tan =
1cossin 22 =+ aa aa22 sin1cos = aa
2sin1cos =
( ) 91.009.013.01sin1cos 22 ==== aa 91.0cos =a
cossin22sin = 91.03.02cossin22sin == 9106.02sin =
91
913
91
3
91.0
3.0
cos
sintan ==
==
91
913tan =
14.If the point ( )4,3 is on the terminal side of the angleA find the exact values of:a. cos [0/1]
b. 2sin [0/2]c. tan [1/0]
Suggested solutions: Answers: 6.0cos = ; 96.02sin = ;3
4
cos
sintan ==
The radius of the unit circle may be calculated using the informationabout the coordinates of the point ( )4,3A
( ) 52516943 222 ==+=+= rr Using the properties of the unit circle thatsinus is the y-component and cosine is the x-coordinates of the point onthe unit circle, we may conclude that:
a. Answer: 6.03 =5
cos = [0/1]
b.2555
42cossin22sin
==
243= Answer: 96.0= [0/2]
25
242sin =
c. Answer:3
4=
cos
sintan =
[1/0]
Of course, the value of the tangent could directly be calculated from theinformation regarding the coordinates of the point on the unit circle.:tangent is simply the ratio of the y-component of the point and that of its
x-component:
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Choose between the problems 15 and 16. Solve only one of them. Note that problem number
15 is in G level and problem number 16 is in MVG-level. Note also that, even if you
originally choose to solve problem 15, if you get extra time over at the end, you may try to
solve problem number 16. If successful, you may then ask me to grade problem 16 instead of
problem 15. You do not need to cross out problem 15. Leave it as it is. If not sure about yoursolutions of problem 16, you may ask me to check your solutions of problem 16, and if
correct then choose it instead of 15. Obviously, you will get grade for your solution of
problem 15 or 16 but not both simultaneously.
15.The point ( )ba, lies on a unit circleas illustrated below. With the help of the
coordinates of the point
P
( )baP , :
a. Find ( )+180sin . [2/0]
b. Find ( )90 . [0/2]sin
c. Find ( )+ 90cos . [0/2]
[NPMaDVT97 modified]
Suggested solutions:First Method:
a. As illustrated below, if we add
180 to the point ( )ba, moves to a new location with thecoordinates:P
( )ba , . [1/0]A This claim may be proven by just making a normal from the pointP and from the pointA to the x-axis. The created triangles ACO and
PDO are exactly equal. This is due to the fact that they have identical
hypotenuse unitsAOPO 1== (which is the radius of the unit circle), as
well as an angle the angle == AOCPOD .
( +180sin ) is the y-coordinate of thepoint ( baA , ). Therefore
( ) b=+180sin [1/0]
Second Method:Of course, we may as well use thetrigonometric relationship
( ) ( ) b==+ sin180sin
( )baP ,
x
y
( )baP ,
x
y
+180
( )baA ,
C DO
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b. First Method: We may use the trigonometric relationship:( ) = a= cos90sin [0/2]
Second method:As illustrated below, the new point on thecircle ( abE , ) is created by just switchingthe x and y-coordinates of the point
( baP , ) moves to a new location with thecoordinates: ( abE , ) .[0/1]This claim may be proven by just making anormal from the point P and from thepointEto the x-axis. The created trianglesECO and PDO are exactly equal. This is due
to the fact that they have identicalhypotenuse unitsEOPO 1== (which is the
radius of the unit circle), as well as an angle the angle
== CEOPOD .
Therefore, the corresponding sides must be the same: bDPOC == , and
aODEC == .
But is the sinus of the angleOC 90 .
Therefore: ( ) aOC== 90sin [0/1] Answer: ( ) a= 90sin
c. As illustrated below, if we add 90 to thepoint ( )ba, moves to a new locationwith the coordinates:
P
( )ab, .[0/1]
B
This claim may be proven by just makinga normal from the point P and from thepointB to the x-axis. The created trianglesBCO and PDO are exactly equal. This is
due to the fact that they have identicalhypotenuse unitsBOPO 1== (which is the
radius of the unit circle), as well as anangle the angle == BOCPOD .
Therefore, the corresponding sides must be the same: bDPOC == . But
is the cosine of the angleOC + 90 . Considering the sign of the cosine
which is in the negative direction:( ) b=+ 90cos . [0/1]
Second Method: Of course, we may as well use the trigonometricrelationship: ( ) b==+ sin90cos .Third Method:
( )+ 90cos is the x-coordinate of the point ( )abB , . Therefore( ) b=+ 90cos
( )baP ,
x
y
+ 90
( )abB ,
C DO
( )baP ,
x
y
CDO
( )abE ,
90
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When assessing your work your teacher will consider:
if the method you have used is reasonable if your calculations are correct what conclusions you have made from your investigation
how plain and pronounced your presentation is, and what kind of mathematical knowledge you have shown to possess.
16.In the triangle ABC illustrated blow the side a is opposite to the angle A , the side b isopposite to the angle B , and the side c is opposite to the angle C .Show that:
( )
+=
2sin4 2
22 Baccab [0/4/]
a
bc
A
CB
Suggested solutions:
Use the cosine law: Abccba cos2222 +=
Baccab cos2222 += Add and subtract to the equation above:ac2
Bacacaccab cos222222 ++= [0/1]
Use ( ) accaca 2222 += and factorize :ac2( ) ( )Baccab cos1222 += [0/1]
Use2
cos1
2sin
2 = 2
cos1
2sin
2 BB = [0/1]
Multiply it by 2 and switch the sides:2
sin2cos1 2B
B =
Therefore: ( ) ( )Baccab cos1222 += ( )
+=
2sin22 2
22 Baccab [0/1/]
( ) 2sin4222 B
accab+=
QED.
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Choose between the problems 17 and 18. Solve only one of them. Note that problem number
17 is in G level and problem number 18 is in MVG-level. Note also that, even if you
originally choose to solve problem 17, if you get extra time over at the end, you may try to
solve problem number 18. If successful, you may then ask me to grade problem 18 instead of
problem 17. You do not need to cross out problem 17. Leave it as it is. If not sure about your
solutions of problem 18, you may ask me to check your solutions of problem 18, and ifcorrect then choose it instead of 17. Obviously, you will get grade for your solution of
problem 17 or 18 but not both simultaneously.
17.Area of a triangle of sides a , b and c may be calculated using Herons formula:
( )( )( )cbaA = llll
where22
cbap ++==l is half of the triangles perimeter.
Heron was a Greek Mathematician and inventor who advanced
developments in hydraulics, pneumatics, and automated devices. Some
inventions credited to him are disputed by historians and may have beeninvented by others. One thing is certain by the 1'st century AD
technology had advanced to the point that if it had advanced just a little
more (maybe if they had spent just a little more time perfecting the
steam engine for instance) the industrial revolution may have started then.
A triangles sides are cma 57= , cmb 31= and cmc 63= .
a. Find the area of the triangle using sine and cosine laws. [3/0]
b. Find the area of the triangle using Herons formula. [2/0]
Suggested solutions:Data: cma 57= , cmb 31= and cmc 63= .
Use the cosine law to calculate one of the angles, for example theangle A
Abccba cos2222 += 222cos2 acbAbc += bc
acbA
2cos
222 +=
( )( )63312576331
cos222 +
=A 43.0cos =A [1/0]
( ) == 5.6443.0cos 1A [1/0] Answer: = 5.64A
2
sinArea
Abc=
( )( )2
5.64sin6331Area = [1/0] Answer: 2880Area cm
Using Herons formula:
( )( )( )cbaA = llll
5.752
316357
22=
++=
++==
cbapl [1/0]
( )( )( ) 24.881315.75635.75575.755.75 cmA == [1/0]
Answer: 2880 cmArea
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18.In the triangle ABC illustrated blow, h is the height of the triangle from the angle A tothe opposite side to it, the side a . Show that:
( )CBCBa
h+
=
sin
sinsin [0/4/]
Suggested solutions:
( )CB
CBah
+
=
sin
sinsin
In the right triangle ADB :c
hB =sin ,
c
BDB =cos . [0/1]
In the right triangle :ADCb
hC=sin ,
b
CDC=cos
Using sincoscossin)sin( +=+
b
h
c
BD
b
DC
c
hCBCBCB +=+=+ sincoscossin)sin( [0/1]
( ) DChBDhhha
b
h
c
BD
b
DC
c
h b
h
c
ha
bc
bc
b
h
c
BD
b
DC
c
h b
h
c
ha
CB
CBaRHS +
=+
=+
=+
= sin
sinsin [0/1]
( ) ( )h
a
ha
DCBDh
hha
CB
CBaRHS =
/
/=
+/
/=
+
=
sin
sinsin [0/1/] QED
Second method (Martins method):
In the triangle ADB :c
hB =sin Bch sin=
In the triangle , sinus law requires:ABCc
CbB
aA sinsinsin ==
CaAc sinsin = A
Cac
sin
sin=
( )( ) ( )CBCBa
CB
BCa
A
BCaB
A
CaBch
+
=
+
=
=
==
sin
sinsin
180sin
sinsin
sin
sinsinsin
sin
sinsin
Answer:( )CB
CBah
+
=
sin
sinsin
a
bc
A
CB
h
D
8/3/2019 Suggested+Solutions+V1+MKUMaDNVC06K5+Trigonometry
15/15
Suggested solutions V1 MKUMaDNVC06K5 trigonometry NV-College
Third method (Mikaels method):
In the triangle :ADCb
hC=sin Cbh sin=
Replace Cbh sin= in the equation( )CB
CBah
+
=
sin
sinsin:
( )CBCBaCb
+= sinsinsinsin ( )CB
Bab+= sin
sin ( )ABab= 180sin
sin ABab
sinsin=
In the equations above we used ( ) ( ) AACB sin180sinsin ==+ .Multiply both sides by :SinA
BaAb sinsin =
Divide both sides by ba
b
B
a
A sinsin=
This is the Sinus law in the triangle .ABC
Therefore, starting from ( )CBCBa
h +
= sin
sinsin
we were able to get a well-
known trigonometric law named sinus law, proving that( )CB
CBah
+
=
sin
sinsin.
a
bc
A
CB
h
D
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