Analytical Methods for Engineers: Trigonometric Methods TMA 2 Solution

This document presents a comprehensive solution to a Trigonometric Methods assignment, covering topics such as trigonometric functions, Pythagoras theorem, sinusoidal functions, and complex numbers. It includes detailed step-by-step calculations and explanations for each question, addressing concepts like amplitude, frequency, phase angle, and impedance. The solution also covers trigonometric identities, vector addition, and power calculations in AC circuits. This resource is valuable for students studying analytical methods in engineering, particularly electrical engineering, and those seeking a deeper understanding of trigonometric principles in practical applications. Desklib provides additional resources including solved assignments and past papers for similar topics.

Trigonometric Methods 1
Trigonometric Methods
Student’s Name
Course
Professor’s Name
University
City (State)
Date

Trigonometric Methods 2
Trigonometric Methods
Question 1
Part a
sin Y = WX
WY
sin 58= 32 km
WY
WY = 32
sin58 =37.7338 km
WY ≅ 38 km(¿ the nearest km)
Part b
cosθ= WZ
WY = 27
37.7338 =0.71554
θ=cos−1 0.71554=44.312°

Trigonometric Methods 3
θ ≅ 44 ° (¿ the nearest degree)
Part c
Using Pythagoras theorem,
YZ= √ WY 2−WZ 2= √ 37.73382−272=26.3598 km
YZ ≅ 26 km(¿ the nearest km)
Question 2
i=15 sin(100 πt+ 0.6)
The equation is in the form y¿ asin(bt +c )
a) amplitude ¿ a=15 Amperes
b) period ¿ b=100 π
c) Frequency¿ 1
period = 1
100 π
d) Phase angle¿ c=0.6 radians= (0.6 × 180
π )°=34.3775 °
e) When t=2.5 s ,i=15 sin(100 π × 2.5+ 0.6)
¿ 15 sin ( 250 π + 0.6 )=15 × 0.5646=8.4696 Amperes
f) The current reaches the maximum level when sin ( 100 πt +0.6 )=1 so that,
sin−1 1=(100 πt +0.6)
100 πt =sin−1 1−0.6=1.5706−0.6=0.9708
t= 0.9708
100 π =0.00309 seconds

Trigonometric Methods 4
t=0.00309 ×1000=3.09 milliseconds
Question 3
Part a (i)
cos (270 °−θ)
270 °−θ=180+( 90−θ)
cos ( 270 °−θ ) =cos (180+(90−θ))
But we know that, cos ( 180+∅ ) =−cos ∅ so that, cos ( 270 °−θ ) =−cos (90−θ)
But again, −cos ( 90−θ )=−( sinθ )
Hence, cos ( 270 °−θ )=−sinθ
Part a (ii)
sin (270 °−θ)
270 °−θ=180+( 90−θ)
sin ( 270 °−θ ) =sin (180+(90−θ))
But we know that, sin ( 180+∅ )=−sin∅ so that, sin ( 270 °−θ ) =−sin (90−θ)
But again, −sin ( 90−θ )=−(cosθ )
Therefore, sin ( 270 °−θ )=−cosθ
Part a (iii)
cos (270 °+θ)

Trigonometric Methods 5
270 ° +θ=180+(90+θ)
cos ( 270 ° +θ ) =cos (180+(90+θ))
But we know that, cos ( 180+∅ ) =−cos ∅ so that, cos ( 270 ° +θ ) =−cos (90+θ)
But again, −cos ( 90+θ ) =−(−sinθ)
Hence, cos ( 270 ° +θ ) =sinθ
Part 3b
V 1=3 sin (ωt )
V 2=2 cos (ωt )
V 3=V 1 +V 2=3 sin ( ωt ) +2 cos ( ωt)
but R sin ( ωt +α ) =(Rcosα )sin ( ωt ) +( Rsinα) cos( ωt)
Which implies that,
Rcosα =3 and Rsinα=2
Rsinα
Rcosα = 2
3 =tanα
α =tan−1
( 2
3 )=33.69°
Rcosα =3 , R= 3
cosα = 3
cos 33.69 ° =3.6056
So, V 3=R sin ( ωt +α ) =3.6056 sin (ωt +33.69 °)

Trigonometric Methods 6
Hence, we can see that frequency of V 1 =frequency of V 2=frequency of V 3= 1
ω
Question 4a
Z= Z1 Z2
Z1 +Z2
Z1 =4+ j 10
Z2 =12− j 3
Z= (4+ j 10)(12− j3)
( 4+ j10 )+(12− j 3)= 4 (12− j 3 )− j10(12− j 3)
4 +12+ j 10− j3 =78+ j108
16+ j7
Converting the numerator into polar form we get,
r = √ 782+¿1082
=133.2216 , θ=tan−1
( 108
78 ) =54.1623° ¿
Hence, 78+ j108=¿133.2216¿ 54.1623 °
Converting the denominator into polar form we get,
r = √162+¿72
=17.4642, θ=tan−1
( 7
16 )=23.6294 ° ¿
Hence, 16+ j7=¿17.4642¿ 23.6294 °
78+ j 108
16 + j 7 =¿133.2216¿ 54.1623 ° ÷17.4642¿ 23.6294 °
Z=¿7.6283¿ 30.5329 °
r =7.6283 ,θ=30.532 9

Trigonometric Methods 7
x=rcosθ=7.6283 cos 30.5329=6.5705
y=r sin θ=7.6283 sin 30.5329=3.8754
z=x +i y =6.5705+ j 3.8754
Z=¿7.6283¿ 30.5329 °
Question 4b
Z1 =2+ j 2
Z2 =1+ j 5
Z3 = j 6
Y = 1
Z1
+ 1
Z2
+ 1
Z3
= 1
2+ j 2 + 1
1+ j 5 + 1
j6
1
j 6 = 1 × j
j 6 × j =− j
6
Y = 6 ( 1+ j5 ) +6 ( 2+ j2 ) − j ( 2+ j 2 ) ( 1+ j5 )
6(2+ j2)(1+ j 5)
Y = 6+ j30+ 12+ j12+ j 8+12
6(−8+ j 12) = 30+ j 50
−48+ j 72
30+ j50= √ 302+502 ¿ tan−1 (50/30) ¿ √ 3400 ¿ 59.0362
−48+ j 72= √ 482 +722 ¿ tan−1 (72/−48) ¿ √7488 ¿−56.3099
Y = √3400 ¿ 59.0362 ÷ √7488 ¿−56.3099
Y =¿0.6738 ¿ ( 59.0362+56.3099 )−180

Trigonometric Methods 8
Y =¿0.6738 ¿−64.6539
r =0.6738 ,θ=−64.6539
x=rcosθ=0.6738 cos (−64.6539)=0.2884
y=rsinθ=0.6738 sin ( −64.6539 ) =−0.6089
Y =x +iy=0.2884− j 0.6089
Question 5
Part a
V =40+ j3 5
I =6+3 j

Trigonometric Methods 9
Part b
θV =tan−1
( 35
40 )=tan−1 0.875=41.1859 °
θi =tan−1
( 3
6 )=tan−1 0.5=26.5651 °
Phase difference=θV −θi=41.1859−26.1859=14.6208 °
Part c
Power=¿ V ∨¿ I ∨cos ∅
|V |= √ 4 02+ 352=53.1507
|I |= √32 +62=6.7082
∅ =Phase difference=14.6208 °
Power=|V ||I |cos ∅ =53.1507 ×6.7082 cos 14.6208 °=345 Watts