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
◦
).