282 Basic Engineering Mathematics

y 155

y (^25)
4
308
Figure 30.12
y 1  5
y (^2) 
4
(^0) 
yR
a
b
30 
150 
Figure 30.13
from which yR=
√
75. 641 = 8. 697
Using the sine rule,
8. 697
sin150◦

4
sinφ
from which sinφ=
4sin150◦
8. 697
= 0. 22996
and φ=sin−^10. 22996
= 13. 29 ◦or 0.232rad
Hence,yR=y 1 +y 2 =5sinωt+4sin(ωt−π/ 6 )
= 8 .697sin(ωt− 0. 232 )
Problem 7. Giveny 1 =2sinωtand
y 2 =3sin(ωt+π/ 4 ), obtain an expression, by
calculation, for the resultant,yR=y 1 +y 2
When timet=0, the position of phasorsy 1 andy 2 are
as shown in Figure 30.14(a). To obtain the resultant,y 1
is drawn horizontally, 2 units long, andy 2 is drawn 3
units long at an angle ofπ/4rad or 45◦and joined to
the end ofy 1 as shown in Figure 30.14(b).
From Figure 30.14(b), and using the cosine rule,
y^2 R= 22 + 32 −[2( 2 )( 3 )cos135◦]
= 4 + 9 −[− 8 .485]= 21. 485
Hence, yR=
√
21. 485 = 4. 6352
Using the sine rule
3
sinφ

4. 6352
   sin135◦
   y 253
   y 152
   /4 or 45 8
   (a)
   y 152
   y 253
   1358
    458
   yR
   (b)
   Figure 30.14
   from which sinφ=
   3sin135◦


5. 
   

   6352
   = 0. 45765
   Hence, φ=sin−^10. 45765
   = 27. 24 ◦or 0.475rad
   Thus, by calculation, yR=^4 .635sin(ωt+^0.^475 )
   Problem 8. Determine
   20sinωt+10sin
   (
   ωt+
   π
   3
   )
   using the cosine and sine rules
   From the phasor diagram of Figure 30.15 and using the
   cosine rule,
   i^2 R= 202 + 102 −[2( 20 )( 10 )cos 120◦]= 700
   i 25 10 A
   i 15 20 A
   iR
   ^1208608
   Figure 30.15
   Hence, iR=
   √
   700 = 26 .46A
   Using the sine rule gives
   10
   sinφ

   
   


6. 46
   sin120◦