Illuminance on a table directly below a lamp

[Matthewkind]

Homework Statement

Two 640-candle lamps are placed 6.0 ft apart and 8.0 ft above a table. Calculate the illuminance on the table directly below one of the lamps.

Homework Equations

The illuminance equation, to the best of my knowledge is as follows:
E = I/s^2 and E = (I/s^2)(cos x).

The Attempt at a Solution

Alright, so at first I thought that it was just asking for me to find the illuminance on the table with regards to one of these lamps. So that E = (640-candles)/(8.0 ft)^2 = (640-candles)/64.0 ft^2 = 10 lu/ft^2 = 10 ft-c.
However, the answer in the back of the book is 15 ft-c. So I was very puzzled. Then I thought that, hey, maybe right triangles have something to do with this problem. After all, the distance between the lamps and the height of the lamps can form right-triangles if you're only taking the illuminance directly under one lamp into consideration. That hypotenuse would then be 10 ft, since 6^2 + 8^2 = x^2 -> 36 + 64 = x^2 -> 100 = x^2 -> x = 10. So I thought I had to calculate the illuminance like before and then add to that the illuminance of the second lamp with the distance being 10 ft. So that ended up with E2 = I2/(s2)^2 = (640-candles)/100 ft^2 = 6.4 lu/ft^2 = 6.4 ft-c. 10 ft-c + 6.4 ft-c = 16.4 ft-c, but 16.4 ft-c is not 15.0 ft-c, so I am doing something wrong. Can someone please point out my flaw so that I may solve this problem and further understand how to correctly approach this (and others like it)?

 

[Matthewkind]

Ah, I forgot to mention: E is the illuminance, I is the luminous flux, and s is the distance.

 

[Matthewkind]

Argh, I sincerely apologize! I is luminous intensity (obviously!), not luminous flux, which would be F.