Suggested+Solutions+V1+MKUMaDNVC06K5+Trigonometry

8/3/2019 Suggested+Solutions+V1+MKUMaDNVC06K5+Trigonometry

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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

[email protected] for sale. Free to use for educational purposes. 1/15


2/15



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 a: OA is cos

3. The distance AB represents: [1/0]

a. cos b. cos c. sin d. tan

e.

+

2sin


Answer: Alternative c: AB is sin

A

B

D

C

O x

y


3/15



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 =


4/15



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


5/15


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]



6/15



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


7/15



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


8/15



==

= 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 =+


9/15



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


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


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:


10/15




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


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


11/15



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


12/15



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.


13/15




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


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


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


14/15



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


15/15


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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