Aspotlight on the ground shines on a wall 12m away. if a man 2m tall walks from the spotlight toward a building at a speed of 16ms, how fast is the length of his shadow on the building decreasing when he is 4m from the building?

we suppose that x be distance of man from spot light is
so 12-x is distance from man to wall
we will draw first triangle ABC, where
a= spotlight (on ground)
b= man (feet)
c= man (head)
we know that AB = x and BC = 2
by extend line AB to the wall at point D and extend line AC to the wall at point E
AD = 12 (distance from spotlight to wall)
DE = s (length of shadow)
Now triangles ABC and ADE are similar
Therefore DE/AD = BC/AB
s / 12 = 2 / x
s = 24 / x
Differentiate both sides with respect to t
ds/dt = -24/x² dx/dt
Man is walking toward building at speed of 16 ms
dx/dt = 16
Man is 4 m from the building/wall
12 - x = 4
x = 8
Find ds/dt when x = 8 and dx/dt = 16
ds/dt = -24/x² dx/dt
ds/dt = -24/(64) * 16
ds/dt = -6

So length of shadow is decreasing at rate of 6 m/s
hope it helps

This is the answer: body 1 + engine 2