Analytical Methods Assignment 2Trigonometric Methods Trigonometry1. A 4.2m long ladder is placed against a perpendicular pylon with its foot 60cm from the pylon. 1.1 Determine how far up the pylon the ladder reaches To solve this equation it is nessasary to find the unknown value, this is achieved by the folllowing equation:

x 2 + 0.6 2 = 4.2 2 x 2 + 0.36 = 17.64 x 2 = 17.28 x = 17.28 x = 4.156m1.2 Calculate how far the top of the ladder rises when the foot of the ladder is moved 20cm towards the pylon To do this I use the exact same method as above except this time I substituye the new values in.

x 2 + 0.4 2 = 4.2 2 x 2 + 0.16 = 17.64 x 2 = 17.48 x = 4.18Therfore the top of the ladder will rise 0.024 meters.

2. From a point on horizontal ground a surveyor measures the angle of elevation of a churchspire as 20. He moves 50m closer to the church and measures the angle of elevation as 25. Calculate the height of the spire. To make the calculations of this equation easier I will represent the church and spire graphically as a series of two trianguls:P P

h R25

20

Q x 50

S

Equation X In triangle PQS

Tan 20 =Hence i.e.

h x + 50

h = tan 20( x + 50)

h = 0.3639( x + 50)

Equation Y In triangle PQR, Hence h

Tan 25 =

h x

= tan 25( x) , i.e. h = 0.4663x

Equating equation x and y gives:

0.3639( x + 50) = 0.4663 x 0.3639 x + (0.3639)(50) = 0.4663 x (0.3639)(50) (0.4663 0.3639) x 18.195 = 0.1024 x x= 18.195 = 177.685 0.1024

From equation y, Height of the building h =

0.4663 x = 0.4663(177.685) = 82.854m

3. Solve the triangle ABC given that:

C = 90, A = 35 and AC = 5mm

B

35 C 5mm A

= 180 90 35 = 55So to find BC: On the calculator Tan 35 x 5mm = 3.501 And to find the hypotenuse (BA):

52 + 3.5012 = BA2 = 37.257 37.257 = 6.103

4. A ship, X, sails at a steady speed of 50km/hr in direction W 30 N (i.e. A bearing of 300 from port). At the same time another ship, Y, leaves port at a steady speed of 40km/hr in a direction N 20 E (i.e. a bearing of 20). Determine their distance apart after 15hrs. The first step would be to find the speed of the ship by multiplying speed by time. Ship 1 travels at 750Km. Ship 2 travels at 600Km.

Using the cosine rule: So:

c2 = a2 + b2

2ab Cos C

c 2 = 750 2 + 600 2

2(750 600) cos C

562500 + 360000 900000 0.1736 = 922500 156240 = 766220 c = 766220c = 875.362 Km to 3 d.p

5. An aero plane is sighted due east from a radar station at an elevation of 50 and a height of 5,000m Later it is sighted at an elevation of 45 and a height of 2500m in a direction of E 70 S. 5.1 If it is descending uniformly, find the angle of decentA B 2500m 5000m

b H3

H3 5000m a

A

H1

A H1 H2

B H2

c

2500m B

50 O 45 O O H3

H1 70 1 O

H2

Work out length AO and OB.

AO =

5000 = 4195.498 Tan50 2500 = 2500 Tan 45

OB =

Then work out the hypotenuse of each triangle H1 and H2.

H 1 = 5000 2 + 4195.498 2 = 42598025 = 6526.716 H 2 = 2500 2 + 2500 2 = 12500000 = 3535.533I will then calculate the first and second positions of the aircraft from there I will then be able to find the angle of desent. 1st position ( ) (4195.498) (5000) 2nd position (2349.232) 2500 sin 70. (855.0504) 2500 cos 7(2500) From this I can then find the difference in the distance between the two different observations using pythagoras theorem.

2349 .232 2 + 855 .0504 4195 .498 2 = 4083 .44 meters

The height difference between the two observations is 2500 meters therefore:

4083 = 0.549rads = 31.5Therfore the angle of desent is 31.5 degrees.

tan = 2500

5.2 Find the speed of the aero plane if the time between observations is 40s Speed = distance(d) time(t)

Speed =

4083.44 = 102.086m / s 40

102.086 x 3600=367509.6 meters covered in one hour

367509.6 = 367.5km / hr 1000

6. Find the magnitude and direction of the resultant of two concurrent and coplanar forces of 30N and 50N when the angle between them is 55. State the direction of the resultant relative to the 50N force.

.Figure 1 Figure 2

30 N30 cos 5555

30N

55

50N Sin 55 30N 50 NResultant55

Origin Figure 3 24.57

71.56

67.21

Unknown

From the above drawings I was able to calculate the magnitude and direction of the resultant from the two forces, the calculations are as follows: Resultant = -30sin55=-24.57 = 50+30cos 55= 67.21 From this I can the find the magnitude by:

Magnitude = 67.212 + 24.57 2 = 71.56Therefore the Magnitude is 71.56N Using the above information I was then able to draw figure 3 leaving only the unknown internal angle to be found which would give us the direction.

24.57 = 0.3434 71.56 = 0.35 = 20.8 Sin =To conclude the magnitude is 71.56N and the direction is 20.8 degrees.

.

Therefore

40.978 30 Sin55 SinR SinR = 30 Sin55 = 0.599 40.978

R = 36.848Resultant of 40.978 N at 36.848 degrees from the 50 N force.

Cartesian and Polar Co-ordinates and Radian Measure7. Change the following to polar co-ordinates: Illustrate your answer with a diagramP Y

(-2.2, 5.5)

5. 5

r

a 0

2.2

X

From Pythagoras theorem

r = 2.2 2 + 5.5 2 = 5.92368

By trigonometric ratios Hence = 180

a = tan 1

5.5 = 68.19859 or 1.19028 rad 2.2

68.19859 = 111.80141

Or = 1.19028 = 1.95131 rad Point P in polar co-ordinates form is (5.923, 111.801) or (5.923, 1.951) 8. Change the following to Cartesian co-ordinates: Illustrate your answer with a diagram (6.4, 2.27 rad)

=2.27 rad A

0

R= 6.4

B

x = r cos = 6.4 cos 2.27 = 4.1197This corresponds to the length OA in the diagram.

y = r sin = 6.4 sin 2.27 = 4.898This corresponds to the length AB in the diagram. Thus (-4.119, 4.898) in Cartesian coordinates corresponds to the polar coordinates (6.4, 2.27 rad) 9. Convert the following between degrees and radians as appropriate: 9.1 30 1 = 0.0174 Therefore: 30 x 0.0174 = 0.523 9.2 90 90 x 0.0174 = 1.566 9.3

rads 3

= 1.047 Rad 31.047 x 57.295 = 59.987 9.4 0.838 rads 0.838 x 57.295 = 48.0132

10. A train is traveling at 108km/h and has wheels of diameter 80cm. Determine: 10.1 The angular velocity of the wheels in both rad/s and rpm Linear velocity v=108 km/h

= 108

km m 1 h 1000 = 30m / s h km 3600 s=0.4m

Radius of the wheel 800mm 2

v = r , so,Angular velocity =75Rad/s And

v 30 = = r 0.4

= 2 n Where n is in rev/sangular speed

n=

75 rev/s = 2 2 = 60 75 Rev/min 2

= 716.19 Rev/min

10.2 The number of revolutions made by one of the wheels if the speed remains constant for 2.7km and there is no slipping

v=

s then the time taken to travel 2.7km i.e. 2700m at a constant speed of 30m/s is given by: t s 2700m = = 90 s v 30m / s

Time t =

Since the wheel is rotating at 716.19 rev/min, then in 90/60 minutes it makes:

716.19 rev / min

90 min = 1074.285 Revolutions 60

Sinusoidal Functions 11. Solve the following in the range 0 to 360: Cos 1 ( 0.678) = 0180 47.312 = 132.69 and 180 + 47.312 = 227.31

C1(.067)=431The solution of Cos-1 (-0.678), between 0 & 360 degrees are: 132.69 and 227.31 o 8 72 s

12. The voltage, v, in an alternating current circuit at any time, t, (seconds) is given by:

v = 100Sin 2000+Determine:

volts 4

12.1 The amplitude, periodic time, frequency and phase angle (in degrees) Amplitude = 100 V Angular Velocity,

= 200 2 2 1 = = 200 100= 0.01s Or 10ms

Hence periodic time, T =

Frequency, f =

1 1 = = 100 Hz T 0.01 180 rad = 0.785rad = 0.785 4

Phase Angle =

= 45 leading v = 100 Sin (200t) 12.2 The voltage when t = 0

v = 100 Sin(200 0 + 0.785) = 100Sin(628.32 + 45) = 663.1V

12.3 The voltage when t = 5ms

5 v = 100 Sin 200 3 + 0.785 10 = 100Sin(2.356) = 100 Sin 135.436 = 70.17V12.4 The time when the voltage first reaches 50v When v = 50 then 50 =

100 Sin( 200t 0.785)

50 = Sin( 200t 0.785) 100

( 200 t 0.785) = arcHence

sin

50 100 = 30 or 0.522rad 200t = 0.522 + 0.785 = 1.307

Hence when v = 50v Time, t =

1.307 = 2.0801 103 seconds 200

13. A complex voltage waveform, v, is comprised of a 141.4v rms fundamental voltage at a frequency of 200 Hz, a 40% third harmonic component leading the fundamental voltage at zero time by and a 20% fifth harmonic component lagging the fundamental by 13.1 Write down an expression for the voltage, v Voltage = 141.4 V (rms). So the maximum value or amplitude is: If the fundamental frequency is 200Hz then angular velocity

rads 4

rads . 3

2 (141.4) = 200

= 2 f = 2 200 = 400 Rads/sHence fundamental voltage of

V = 200 sin(400 t )

The third harmonic component has an amplitude equal to 40% of 200 V i.e. 80 The frequency of the third harmonic component is 3 200 = 600 Hz so:

= 2 600 = 1200

Hence third harmonic voltage is represented by

The fifth harmonic has an amplitude equal to 20% of 200 V i.e. 40 The frequency of the fifth harmonic component is: So: = 2 1000 = 2000 Hence the voltage of the fifth harmonic is shown as Overall

V = 80 sin(1200 t + ) 4 5 200 = 1000 Hz ) 3

V = 40 sin( 2000 t

V = 200 sin( 400t ) + 80 sin 1200 t +

+ 40 sin 2000t 4

3

13.2 Plot the resulting complex waveform for v over one cycle of the fundamental waveform.

Re300The following formulas were used in m.s spreadsheet and the data optained follows on the next three pages. A1 = Time B1 = Fundamental Voltage = 200*SIN(400*PI()*A3*0.001) C1 = Third Harmonic = 80*SIN(1200*PI()*A3*0.001+PI()/4) D1 = Fifth Harmonic = 40*SIN(2000*PI()*A3*0.001-PI()/3) E1 = Total Voltage = B3+C3+D3

200

Time (ms) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3

Fundamental voltage 0 25.06664671 49.73797743 73.62491054 96.35073482 117.5570505 136.9094212 154.1026486 168.8655851 180.9654105 190.2113033 196.4574501 199.6053457 199.6053457 196.4574501 190.2113033 180.9654105 168.8655851 154.1026486 136.9094212 117.5570505 96.35073482 73.62491054 49.73797743

Third Harmonic 56.56854249 73.42037005 79.96052483 75.27046152 60.00888557 36.31923998 7.528665065 22.31928848 49.03256429 68.85936216 79.01506725 78.07334096 66.16644594 44.96667023 17.45145931 12.5147572 40.72331326 63.21240099 76.82349485 79.64495717 71.28052194 52.90494923 27.09903362 2.512860726

Fifth Harmonic 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2

Total Voltage 21.927526 34 82.217551 04 138.01496 99 178.621165 1 196.14049 62 188.51730 66 160.70755 2 123.46689 24 90.107227 79 72.325172 52 76.555219 86 102.114643 5 141.75536 74 184.36446 85 218.78686 66 237.36707 66 237.95818 95 223.76151 85 201.20035 04 176.77350 25 154.19655 62 132.98621 83 109.040411 8 76.950909 73

2.4 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4 4.1 4.2 4.3 4.4 4.5 4.6

25.06664671 -6.43149E-14 -25.06664671 -49.73797743 -73.62491054 -96.35073482 -117.5570505 -136.9094212 -154.1026486 -168.8655851 -180.9654105 -190.2113033 -196.4574501 -199.6053457 -199.6053457 -196.4574501 -190.2113033 -180.9654105 -168.8655851 -154.1026486 -136.9094212 -117.5570505 -96.35073482

31.77183125 56.56854249 73.42037005 79.96052483 75.27046152 60.00888557 36.31923998 7.528665065 22.31928848 49.03256429 68.85936216 79.01506725 78.07334096 66.16644594 44.96667023 17.45145931 -12.5147572 40.72331326 63.21240099 76.82349485 79.64495717 71.28052194 -

39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657 2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5 16.2694657

33.075691 28 21.927526 3 -82.217551 -138.01497 178.621165 196.14049 6 188.51730 7 160.70755 2 123.46689 2 90.107227 8 72.325172 5 76.555219 9 102.114643 141.75536 7 184.36446 8 218.78686 7 237.36707 7 237.95818 9 223.76151 8 -201.20035 176.77350 3 154.19655 6 -

52.90494923 4.7 4.8 4.9 5 -73.62491054 -49.73797743 -25.06664671 1.2863E-13 27.09903362 2.512860726 31.77183125 56.56854249

2 8.31646763 3 29.7257930 2 39.7808758 1 34.6410161 5

132.98621 8 109.04041 2 76.950909 7 33.075691 3 21.927526 34

Trigonometric Identities14. Show that: If

1 Sin 2 = (1 Cos 2) and explain where this would be useful 2

Cos 2 = Cos 2 Sin 2

And Cos 2 + Sin 2 = 1 Then:

Cos 2 = 1 Sin 2 Cos 2 = (1 Sin 2) Sin 2 Cos 2 = 1+ 2 Sin 2 (1 Cos 2) = Sin 2 2This equation is useful when calculating RMS values. 15. Solve the following in the range:

0 to 360 5Sin 2 x + 3Sin x = 4

Rearranging the equation to make the answer equal to zero would give me:

5Sin 2 x + 3Sin x 4 = 0Now that the equation is in this format I can easily apply the quadratic formula: So: a = 5, b = 3 and c = -4

3 32 4(5 4) 3 9.433 Sinx = = 25 10+ Sinx = 0.6433 - Sinx = -1.2433 The values in the range

0 to 360 are:

X = 40038 and 139962

16. Solve the following for values:

0 2

5Sin

= 8Cos 6

Sin - 6 Sin( A - B ) = SinACosB - SinBCosA

Sin = SinCos CosSin 6 6 6 Cos

3 = 6 2 Sin = 0.5 6 3 1 8 Sin = Sin Cos = Cos 6 2 2 5 3 8 1 sin = cos + cos 2 5 2 3 8 1 16 5 21 sin = ( + ) cos = ( + ) cos = cos 2 5 2 10 10 10 3 21 sin = cos 2 10 3 sin = 2 cos 3 21 42 tan = tan 2 10 10 3

3.42 = tan 30 = 1.18rads also tan = tan ( + ) hence1.18 + = 4.321rads = 247.59 Therefore 1 = 1.18rads = 67.59 2 = 4.321rads = 247.59

17. Express the equation: 17.1 In the form R Sin (t+) and hence

5.5Cos t + 7.8Sin t

7.8Sin(t ) + 5.5Cos (t ) = RSin(t + ) = R[ Sin(tCos + Cos (t ) Sin)] = ( RCos) Sin(t ) + ( RSin)Cos (t )Equating coefficients:

7.8 R 5.5 Sin = R Cos = 7.82 + 5.52 = 9.544 so, 7.8Sin(t ) + 5.5Cos (t ) = 9.544 Sin(t + ) RSin = 5.5 RCos = 7.8 5.5 7.8

Tan =

= 0.614 radians 7.8Sin(t ) + 5.5Cos (t ) = 9.544 Sin(t + 0.614)17.2 Solve, for values

0 2

5.5Cos t + 7.8 Sin t = 4.5

9.544 Sin (t + 0.614 ) = 4.5 4 .5 si (t + 0.614 ) = = 0.471495 9.544 a, sin 0.471495 = 0.490985 = t + 0.614 t = 012301Which is not in range of:

0 2 so, add 2 for one solution. t = 2 0.12301 = 6.16rad

Also

sin(0.490985)=o.471495 then

sin( 0.490985) = 0.471495 0.490985 = t + 0.614 t = 0.490985 0.614 = 2.036608 hence t = 6.16rads t = 2.04rads

Bibliography

1. en.wikipedia.org/wiki/Waveform John Bird Higher Engineering Mathematics Barry college module1 Maths Barry college module 2 Physics Yeovil college Moodle. Www.freestudy.com.