Solving triangles, please help!?

Question

A light, on top of a post, is 14 feet above the ground. At night a 5 foot woman is standing 18 feet from the post. The woman's shadow is 10 feet long. Her friend is standing 16 feet away from the post.

Starting 18 feet from the post, the woman walks directly away from the post for 4 seconds at a rate of 3 feet per second. Find the average speed(average speed = total distance/total time) on the tip of her shadow during that 4 second interval

http://i37.tinypic.com/2us8y7s.jpg

How do i solve this...

Answer 1

At the beginning, the tip of her shadow is (18 + 10) = 28 feet from the post. If from this position the woman then walks away from the post for 4 sec at 3 ft/sec, she will be (18+12) = 30 feet from the post. Let the length of her shadow at this point be x. The right triangle determined by her 5-ft height and this later shadow is similar to the right triangle determined by the 14 ft post and the the line segment from the base of the post to the tip of the later shadow. This line segment has a length of 30+x. Therefore, by similar triangles,

5/14 = x/(30+x)

150 + 5x = 14x

150 = 9x

50/3 = x

So in that four seconds, the tip of her shadow moved from 28 feet from the post to (30 + 50/3) feet from the post.

total distance traveled/total time = (30 + 50/3 - 28)/4 = 14/3 ft/sec

Answer 2

(@ 18' from post) (14 -5)= 9

her shadow ...10' front of her ...

4 sec @ 3' / sec = 12'

(30' from post) 30-5 = 25

tan^-1 (9/ 25) = .36 = 19.7988

5/.36 = 13.888 ' (shadow

(13.888 - 10.0 ) final shadow - original shadow

3.888 + 12(3@4sec) = 15.8888'

15.888 / 4 sec = 3.9722' / sec

Answer 3

let shadow's tip total distance be d.

x / 5 = (18 + 3*4 + x) / 14

x / 5 = (x + 30) / 14

14x = 5x + 150

9x = 150

x = 150 / 9 = 50 / 3

d = (x + 30) - (10 + 18)

= 50/3 + 2

= 56/3

average speed

= d / (4 seconds)

= 56 / 3*4

= 14 / 3

= 4.667 feet per second

Answer 4

you should probably go back and read the problem again.  your drawing is wrong.  the 6' friend (where did you get that information?) at 16 feet from the post should cast exactly the same shadow as a 5' person, 18' from the post.  both the triangles are 30-60-90.