∠APB = π rad - ∠APC (angles on a straight line BPC)
= (π - θ) rad

For triangle APB,

tan ∠ APB = opp/adj

= AB/BP
= 5/x

tan (π - θ) = 5/x

-tanθ = 5/x (since tan (π - θ) = -tanθ)

x = -5/tanθ

dx/dθ = -5(-sec²θ)/tan²θ

= 5sec²θ/tan²θ

= 5(1/cos²θ)/(sin²θ/cos²θ)

= 5/sin²θ

= 5cosec²θ

When x = 5,

5 = -5/tanθ
tanθ = -1
θ = tan-¹ (-1)
θ = 3π/4 (since θ is obtuse)



dx/dt = -3cms-¹
(since x is decreasing as P moves closer to B)

dθ/dt = dx/dt × dθ/dx

= dx/dt ÷ dx/dθ
= -3cms-¹ ÷ (5cosec²θ cm radians-¹)
= -3s-¹ (sin² (3π/4))radians / 5
= -3(½) radians s-¹ / 5
= -3/10 radians s-¹

(Or -0.3 radians s-¹)


∠APC is decreasing

Alternative way to get θ = 3π/4 :

When x = 5, AB = BP so △ APB is isosceles.

Since △APB is a right angled triangle , the base angles would be π/4 radians (or 45°)

( (π - ½π)rad / 2 or (180° - 90°)/2 )

So π - θ = π/4

θ = 3π/4