Trigonometric Methods 1

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−WZ2= √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