A street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

The correct option is ΔABC ≅ ΔFED by SAS

Step-by-step explanation:

Looking at the given diagram,

Side AC = Side DF = 3.2

Side AB = Side FE = 4 units

Angle A = angle F = 72 degrees

Since two sides and one angle are equal, then triangle FED = triangle ABC by SAS(side angle side)

This makes the ratios possible

AB/FE = AC/DF

The HL given means hypotenuse leg. It will not hold because the triangles are not right angle triangles.