Basic Engineering Mathematics, Fifth Edition

Methods of adding alternating waveforms 283

from which sinφ=

10sin120◦

26. 46

= 0. 327296

and φ=sin−^10. 327296 = 19. 10 ◦

= 19. 10 ×

π

180

= 0 .333rad

Hence, by cosine and sine rules,
iR=i 1 +i 2 = 26 .46sin(ωt+ 0. 333 )A

Now try the following Practice Exercise

PracticeExercise 120 Resultant phasors by

the sine and cosine rules (answers on

page 352)

1. Determine, using the cosine and sine rules, a
   sinusoidal expression for
   y=2sinA+4cosA.


2. Givenv 1 =10sinωtvolts and
   v 2 =14sin(ωt+π/ 3 )volts, use the cosine
   and sine rules to determine sinusoidal expres-
   sions for (a)v 1 +v 2 (b)v 1 −v 2
   In problems 3 to 5, express the given expressions
   in the formAsin(ωt±α)by using the cosine and
   sine rules.


3. 12sinωt+5cosωt


4. 7sinωt+5sin

(

ωt+

π

4

)

5. 6sinωt+3sin

(

ωt−

π

6

)

30.5 Determining resultant phasors

by horizontal and vertical

components

If a right-angled triangle is constructed as shown in
Figure 30.16, 0a is called the horizontal component of
Fandabis called the vertical component ofF.

(^0) H a
V
b
F

Figure 30.16
From trigonometry(see Chapter 21 and remember SOH
CAH TOA),
cosθ=
0 a
0 b
,from which 0a= 0 bcosθ=Fcosθ
i.e. the horizontal component of F,H=Fcosθ,
and
sinθ=
ab
0 b
,from whichab= 0 bsinθ=Fsinθ
i.e.the vertical component ofF,V=Fsinθ.
Determiningresultant phasorsbyhorizontal andvertical
components is demonstrated in the following worked
problems.
Problem 9. Two alternating voltages are given by
v 1 =15sinωtvolts andv 2 =25sin(ωt−π/ 6 )
volts. Determine a sinusoidal expression for the
resultantvR=v 1 +v 2 by finding horizontal and
vertical components
The relative positions ofv 1 andv 2 at timet=0are
shown in Figure 30.17(a) and the phasor diagram is
shown in Figure 30.17(b).
v 15 15 V
(a)
v 25 25 V
/6 or 30 8
(b)
(^0) 
vR
v 2
v 1
1508 308
Figure 30.17
The horizontal component ofvR,
H=15cos0◦+25cos(− 30 ◦)= 36 .65V
The vertical component ofvR,
V=15sin0◦+25sin(− 30 ◦)=− 12 .50V