There are 2 way to solve this.
one using Pythagoras theorem and 2nd using trigonometry
so lets solve it by both
using Pythagoras theorem we know
base^2 + perpendicular^2 = hypotanes^2
6^2 + x^2 = 12^2
36 + x^2 = 144
x^2 = 144- 36 = 108
x = √(108) = √( 2×2×3×3×3)
= (2×3) √ (3) = 6 √3
so answer is option 2
bow lets use trigonometry
we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3
So the bigger number are the smallest
And
The small number are biggest
So it would be 1/6,2/3,3/4,5/8,7/12
Hey just to let you now it came out blury
I'd suggest you sketch this situation. If you did, you'd likely see right away that the shortest distance back to the starting point would be the hypotenuse of the right triangle whose legs are of lengths 5 miles and 12 miles.
Apply the Pythagorean Theorem to solve this.
