Steve Goddard

Contents

Topic PageTrigonometry 2

Cartesian and Polar Co-ordinates and Radian measure

9

Sinusoidal Functions 12Trigonometric Identities 16

Bibliography 19

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Steve Goddard

Analytical Methods – Assignment 2

Trigonometric Methods

Trigonometry

1. 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

1.2 Calculate how far the top of the ladder rises when the foot of the ladder is moved 20cm towards the pylon

So the top of the ladder rises:

4.18 – 4.156 = 0.024m

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Steve Goddard

2. From a point on horizontal ground a surveyor measures the angle of elevation of a church spire as 20°. He moves 50m closer to the church and measures the angle of elevation as 25°. Calculate the height of the spire.

Equation 1

In triangle PQS

Hence

i.e.

Equation 2

In triangle PQR,

Hence , i.e.

Equating equation 2 and 2 gives:

From equation 2,

Height of the building h =

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20°RS

PP

Q

25°

50x

h

Steve Goddard

3. Solve the triangle ABC given that: and

So to find BC:

Tan 35 x 5mm = 3.501

And to find the hypotenuse (BA):

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35°

θ

5mmA

C

B

Steve Goddard

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.

First of all I worked out how far each ship had traveled by multiplying their speed by time.

Ship 1 travels = 50Km/hr x 15 Hrs = 750Km

Ship 2 travels = 40Km/hr x 15Hrs = 600Km

Using the cosine rule: Cos C

So:

I then square-rooted this to find c

c = 875.362 Km

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C

B

600

750

b

a

c

A

Steve Goddard

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 decent

Calculations

Firstly I worked out length AO and OB.

Next from these Lengths I worked out the hypotenuse of each triangle H1 and H2.

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A

O

B

H1 H2

H3

50

45

5000m

2500m

H1

H2

H3

701

H1

H2

H3

2500m

5000m

O

O

O

B

B

A

A

θ

a

b

c

Steve Goddard

Using the Cosine rule I calculated H3

=39313498.56

Square Root the answer to give H3

I need to work out θ so in order to simplify things these are my essential calculations.

To work out θ I used the SOH-CAH-TOA method and used SOH

23.498 is the angle of decent.

This problem can be solved a number of ways such as firstly solving the length AB using Pythagoras’s theorem and then using TOA to work out the angle.

Eg.

5.2 Find the speed of the aero plane if the time between observations is 40s

Speed = Distance over time but first of all I converted the units:

Page 7 of 19

6270.047

2500m

2500m

BA

θ

b

Steve Goddard

I calculated 40 seconds in hours

Hrs

From this I used the normal speed equation

Or keeping unconverted in seconds

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.

.

Therefore

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

Cartesian and Polar Co-ordinates and Radian Measure

7. Change the following to polar co-ordinates:

Illustrate your answer with a diagram (-2.2, 5.5)

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55°

30N 50N

x

Steve Goddard

From Pythagoras theorem

By trigonometric ratios or 1.19028 rad

Hence

Or rad

Hence the position of 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)

This corresponds to the length OA in the diagram.

This 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

So:

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P

5.5

r

2.2

Y

X0

a

θ

0

B

A

R= 6.4

θ=2.27 rad

Steve Goddard

30 x 0.0174 = 0.523

9.2 90°

90 x 0.0174 = 1.566

9.3

Rad

1.047 x 57.295 = 59.987°

9.4 0.838 rads

0.838 x 57.295 = 48.0132°

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Steve Goddard

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 km/h

Radius of the wheel mm

m

From the equation , from which,

Angular velocity

Rad/sAnd

Where is in rev/s

Meaning angular speed rev/s

Rev/min

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

Since then the time taken to travel 2.7km i.e. 2700m at a constant speed of

30m/s is given by:

Time t =

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

Revolutions

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Steve Goddard

Sinusoidal Functions

11. Solve the following in the range 0° to 360°:

312.47)678.0(1 Cos

The solutions of Cos-1 (-0.678), between 0 & 360 degrees are: 132.69 and 227.31

180 – 47.312 = 132.69 and 180 + 47.312 = 227.31

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

Determine:

12.1 The amplitude, periodic time, frequency and phase angle (in degrees)

Amplitude = 100 V

Angular Velocity,

Hence periodic time, T =

Or

Frequency, f =

Phase Angle = 0.785rad =

= Lagging v = 100 Sin (200πt)

12.2 The voltage when t = 0

12.3 The voltage when t = 5ms

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Steve Goddard

12.4 The time when the voltage first reaches 50v

When v = 50 then 50 =

Hence

Hence when v = 50v

Time, t = seconds

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: = 200

If the fundamental frequency is 200Hz then angular velocity

Rads/s

Hence fundamental voltage of

The third harmonic component has an amplitude equal to 40% of 200 V i.e. 80

The frequency of the third harmonic component is Hz so:

Hence third harmonic voltage is represented by

The fifth harmonic has an amplitude equal to 20% of 200 V i.e. 40

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Resulting complex waveform for V over one cycle of the fundamental waveform

-300

-200

-100

0

100

200

300

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (ms)

Vo

ltag

e

Fundamental Voltage

Third Harmonic

Fifth Harmonic

Total Voltage

Steve Goddard

The frequency of the fifth harmonic component is:

So:

Hence the voltage of the fifth harmonic is shown as

Overall

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

The data used for this graph can be found on the next page.

Spreadsheet Formula

A1 = TimeB1 = 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

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Steve Goddard

Time (ms)

Fundamental voltage

Third Harmonic

Fifth Harmonic

Total Voltage

0 0 56.56854249 -34.64101615 21.927526340.1 25.06664671 73.42037005 -16.26946572 82.217551040.2 49.73797743 79.96052483 8.316467633 138.01496990.3 73.62491054 75.27046152 29.72579302 178.62116510.4 96.35073482 60.00888557 39.78087581 196.14049620.5 117.5570505 36.31923998 34.64101615 188.51730660.6 136.9094212 7.528665065 16.26946572 160.7075520.7 154.1026486 -22.31928848 -8.316467633 123.46689240.8 168.8655851 -49.03256429 -29.72579302 90.107227790.9 180.9654105 -68.85936216 -39.78087581 72.32517252

1 190.2113033 -79.01506725 -34.64101615 76.555219861.1 196.4574501 -78.07334096 -16.26946572 102.11464351.2 199.6053457 -66.16644594 8.316467633 141.75536741.3 199.6053457 -44.96667023 29.72579302 184.36446851.4 196.4574501 -17.45145931 39.78087581 218.78686661.5 190.2113033 12.5147572 34.64101615 237.36707661.6 180.9654105 40.72331326 16.26946572 237.95818951.7 168.8655851 63.21240099 -8.316467633 223.76151851.8 154.1026486 76.82349485 -29.72579302 201.20035041.9 136.9094212 79.64495717 -39.78087581 176.7735025

2 117.5570505 71.28052194 -34.64101615 154.19655622.1 96.35073482 52.90494923 -16.26946572 132.98621832.2 73.62491054 27.09903362 8.316467633 109.04041182.3 49.73797743 -2.512860726 29.72579302 76.950909732.4 25.06664671 -31.77183125 39.78087581 33.075691282.5 -6.43149E-14 -56.56854249 34.64101615 -21.92752632.6 -25.06664671 -73.42037005 16.26946572 -82.2175512.7 -49.73797743 -79.96052483 -8.316467633 -138.014972.8 -73.62491054 -75.27046152 -29.72579302 -178.6211652.9 -96.35073482 -60.00888557 -39.78087581 -196.140496

3 -117.5570505 -36.31923998 -34.64101615 -188.5173073.1 -136.9094212 -7.528665065 -16.26946572 -160.7075523.2 -154.1026486 22.31928848 8.316467633 -123.4668923.3 -168.8655851 49.03256429 29.72579302 -90.10722783.4 -180.9654105 68.85936216 39.78087581 -72.32517253.5 -190.2113033 79.01506725 34.64101615 -76.55521993.6 -196.4574501 78.07334096 16.26946572 -102.1146433.7 -199.6053457 66.16644594 -8.316467633 -141.7553673.8 -199.6053457 44.96667023 -29.72579302 -184.3644683.9 -196.4574501 17.45145931 -39.78087581 -218.786867

4 -190.2113033 -12.5147572 -34.64101615 -237.3670774.1 -180.9654105 -40.72331326 -16.26946572 -237.9581894.2 -168.8655851 -63.21240099 8.316467633 -223.7615184.3 -154.1026486 -76.82349485 29.72579302 -201.200354.4 -136.9094212 -79.64495717 39.78087581 -176.7735034.5 -117.5570505 -71.28052194 34.64101615 -154.1965564.6 -96.35073482 -52.90494923 16.26946572 -132.9862184.7 -73.62491054 -27.09903362 -8.316467633 -109.0404124.8 -49.73797743 2.512860726 -29.72579302 -76.95090974.9 -25.06664671 31.77183125 -39.78087581 -33.0756913

5 1.2863E-13 56.56854249 -34.64101615 21.92752634

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Steve Goddard

Trigonometric Identities

14. Show that: and explain where this would be

useful

If

And

Then:

This equation is useful with integrating when calculating RMS values.

15. Solve the following in the range:

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

Now that the equation is in this format I can easily apply the quadratic formula:

So: a = 5, b = 3 and c = -4

+ Sinx = 0.6433- Sinx = -1.2433

Therefore the values in the range are:

X = 40°038’ and 139°962

16. Solve the following for values:

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Steve Goddard

Or 1.043 radians

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Steve Goddard

17. Express the equation:

17.1 In the form R Sin (t+α) and hence

Equating co-efficients:

17.2 Solve, for values

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Steve Goddard

Bibliography

www.google.com

John Bird – Higher Engineering Mathematics

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