25. In adding these with the phasor method (as opposed to, say, trig identities), we may set t = 0 (see

Sample Problem 36-3) and add them as vectors:

y

h

=

10 cos 0

◦

+ 15 cos 30

◦

+ 5.0 cos(

−45

◦

) = 26.5

y

v

=

10 sin 0

◦

+ 15 sin 30

◦

+ 5.0 sin(

−45

◦

) = 4.0

so that

y

R

=

y

2

h

+ y

2

v

= 26.8

β

=

tan

−1



y

v

y

h



= 8.5

◦

.

Thus, y = y

1

+ y

2

+ y

3

= y

R

sin(ωt + β) = 26.8 sin(ωt + 8.5

◦

).