Two pathways meet at 30° to each other. One pathway has lighting and the other does not. The distance between successive lights on the lighted pathway is 5 metres. Each light has a
range of effective illumination of 6 metres.
What length of the pathway without lights is illuminated by the pathway with lights?

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.

what