that is physics ...
but yes, applied math. we need to know the formulas though.
(1)
in general, since there are 2 cables supporting in an equal way.
that means each cable is responsible for 600/2 = 300 pounds to bring and hold up.
a cable or rope at an angle has to handle a combined tension force : horizontally (Fx) and vertically (Fy).
the tension force (Ftens) on the rope is a combination of both.
we know
Fx = Ftens × cos(theta)
Fy = Ftens × sin(theta)
from the problem we do know Fy (the vertical = up/down force), as this is the force needed to lift and keep the 300 pound weight up there.
and that is Fgravity, the force needed to counteract gravity.
Fgravity = mass × g
g being the constant gravitational acceleration of Earth = 9.8 m/s²
forces are described in Newton.
1 N ≈ 0.225 pounds (lifting on Earth)
so, to lift 1 pound requires 1/0.225 ≈ 4.44822 N
to lift 300 pounds requires
4.44822 × 300 ≈ 1334.47 N
that is what Fy is for one of the 2 cables.
the tension on one of the cables is then given by
Fy = Ftens × sin(60)
Ftens = Fy / sin(60) = 1334.47 / sin(60) =
= 1,540.913227... N = 346.41107515867... pounds
≈ 346 pounds per cable.
(2)
3cos(t) + 4 = 2
3cos(t) = -2
cos(t) = -2/3
cosine is negative in the 2nd and 3rd quadrant.
so, for t > pi/2 and t < 3pi/2.
because the given interval is [0, pi), we are only looking at the 2nd quadrant (pi/2, pi).
t = 131.8103149...° = 2.300523983... rad
(3)
well, that are the numbers
1/2
sqrt(2)/2 = 1/sqrt(2)
sqrt(3)/2
1
they are getting bigger and bigger, all positive, so they indicate larger and larger angles
1/2 is :
sin(30° or pi/6 or 150° or 5pi/6)
cos(60° or pi/3 or 300° or 5pi/3)
1/sqrt(2) is :
sin(45° or pi/4 or 135° or 5pi/4)
cos(45° or pi/4 or 315° or 7pi/4)
sqrt(3)/2 is :
sin(60° or pi/3 or 120° or 2pi/3)
cos(30° or pi/6 or 330° or 11pi/6)
1 is :
sin and csc(90° or pi/2)
cos and sec(0° or 0pi or 360° or 2pi)
tan and cot(45° or pi/4 or 225° or 5pi/4)
(4)
the height moves between 4 meters and 70 meters in a circle.
the circumference of the circle is 2pi×r or pi×d, so in our case : 66pi meters.
it takes 5 minutes to move along these 66pi meters.
let's say, when the height is 4 meters (starting position), the angle is 0 and the arc is 0.
after a quarter trip the angle is 90° or 66pi/4, and the height is 4 + 66/2 = 37 meters
and at 70 meters the angle is 180° or 66pi/2.
the function of the height based on the current angle is then for the first half-circle
height(theta) = 4 + (theta/360)×2×66
or
height(theta) = 4 + (theta/(2pi))×2×66
now we need to find the angle theta for which we reach the height of 47 meters :
47 = 4 + (theta/360)×132
43 = (theta/360)×132
theta/360 = 43/132
theta = 360×43/132 = 117.2727272...°
= 2.046795214... rad
so, after starting at the lowest position at 4 meters we reach the height of 47 meters at an angle of about 117°.
then we get and stay above 47 meters until we get to
360 - theta = 242.7272727...°
= 4.236390093... rad
when going down again on the second half-circle of the trip.
that means we are at and above 47 meters for
(360 - theta) - theta = 360 - 2×theta = 125.4545455...°
= 2.18959488... rad
of the whole trip of 360° or 2pi. which takes 5 minutes.
the time we spend there is then
5 × (360 - 2×theta)/360 = 1.742424242... minutes
= 1 minute 44.54545454... seconds