we make a drawing to find the relationship between the angle and the sides
given
dh/dt=4m/sec
we get
tan(x)=h/40
where x is the angle and h is the height
take the derivitive of both sides
1/cos²(x)(x)dx/dt=[tex] \frac{h'40-0}{40^2} [/tex]
1/cos²(x)dx/dt=[tex] \frac{h(dh/dt)40}{1600} [/tex]
1/cos²(x)dx/dt=[tex] \frac{h(dh/dt)}{40} [/tex]
solve for dx/dt
times both sides by cos²(x)
dx/dt=[tex] \frac{h(dh/dt)cos²(x)}{40} [/tex]
remember that dh/dt=4
also, solve for the angle at that time
h=40 and base=40 so x=45 degrees (pi/4 radians)
input that
dx/dt=[tex] \frac{4cos²(x)}{40} [/tex]
dx/dt=[tex] \frac{cos²(x)}{10} [/tex]
dx/dt=0.05
A is the answer
