Answer:
12 meters
Step-by-step explanation:
Draw a picture of the two pathways 30° apart. Add a circle representing the illumination of the light on one of the pathways. Draw a line from the center of the circle to the last point where it intersects the pathway without lights.
This forms a triangle. One leg is x, the length of the pathway without lights. Another leg is 5n, where n is an integer. This represents how far the light is from where the pathways meet. The third and final leg is 6, the radius of the illumination.
Use law of cosine to solve:
6² = x² + (5n)² − 2x(5n) cos 30°
36 = x² + 25n² − 5√3 xn
0 = x² − 5√3 xn + 25n² − 36
In order to have a solution, the discriminant must be greater than or equal to 0.
b² − 4ac ≥ 0
(-5√3 n)² − 4(1)(25n² − 36) ≥ 0
75n² − 100n² + 144 ≥ 0
144 − 25n² ≥ 0
144 ≥ 25n²
144/25 ≥ n²
12/5 ≥ n
So n must be an integer less than 12/5, or 2.4. Therefore, the largest value of n is 2. Substituting:
0 = x² − 5√3 x(2) + 25(2)² − 36
0 = x² − 10√3 x + 64
Solve with quadratic formula:
x = [ 10√3 ± √(300 − 4(1)(64)) ] / 2(1)
x = (10√3 ± √44) / 2
x = 5√3 ± √11
x ≈ 5.34 or 11.98
We want the larger value of x. So approximately 12 meters of the pathway without lights is illuminated.
