a spy uses a telescope to track a rocket launched vertically from a launching pad 6 kmaway. At a certain moment, the angle (θ) between the telescope and the the ground is equal to π3 and is changing at a rate of 0.9 rad/min. What is the rocket’s velocity atthat moment?

Answer:v = 21.6 Km/minStep-by-step explanation:x = 6 Km∅ = π/3 (variable)d∅/dt = 0.9 rad/minLet ∅ be the angle between the spy's line of sight and the ground and y is the distance between the rocket and the ground.Then we have the equationtan ∅ = y / x⇒ (tan ∅)' = (y / x)' = (1/x)*y'⇒ Sec²∅*(d∅/dt) = (1/x)*(dy/dt) = (1/x)*v⇒ v = x*Sec²∅*(d∅/dt)⇒ v = 6 Km*Sec²(π/3)*(0.9 rad/min)⇒ v = 21.6 Km/min