A kite 100 ft above the ground moves horizontally at a speed of 6 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? rad/s g

0.015 radians per second.

Step-by-step explanation:

They tell us that at the moment the speed would be 6 ft / s, that is, dx / dt = 6 and those who ask us is dθ / dt.

Which we can calculate in the following way:

θ = arc sin 100/200 = pi / 6

Then we have the following equation of the attached image:

x / 100 = cot θ

we derive and we are left:

(1/100) * dx / dt = - (csc ^ 2) * θ * dθ / dt

dθ / dt = 0.01 * dx / dt / (- csc ^ 2 θ)

dθ / dt = 0.01 * 6 / (- csc ^ 2 pi / 6)

dθ / dt = 0.06 / (-2) ^ 2

dθ / dt = -0.015

So there is a decreasing at 0.015 radians per second.

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it