SOLUTION: In a 10 foot wide alley, two ladders lean against opposite walls. One Ladder reaches 30 feet up on the wall, while the other reaches 20 ffet up on the opposite wall. How high abo

Algebra ->  Equations -> SOLUTION: In a 10 foot wide alley, two ladders lean against opposite walls. One Ladder reaches 30 feet up on the wall, while the other reaches 20 ffet up on the opposite wall. How high abo      Log On


   

Question 106360: In a 10 foot wide alley, two ladders lean against opposite walls. One Ladder reaches 30 feet up on the wall, while the other reaches 20 ffet up on the opposite wall. How high above the ground will the two ladders cross?
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
I assume that the ladders bases are in the corner of the side opposite which they are leaning.
L=longest ladder, s=shortest
cos(L)=10/30=1/3
cos^-1(1/3)=70.53 deg
.
cos(s)=10/20=1/2
cos^-1(1/2)=60 deg
.
Now we have a triangle with a base of 10 and we know the three angles 70.5,60 and 49.47
a/sin(A)=b/sin(B)
10/sin(49.47)=b/sin(60)
10*sin(60)/sin(49.47)=11.39 ft (one of the sides)
.
Now if you draw a line from the angle of the ladders crossing perpendicular to the base we have the height (h) of the triangle.
It is opposite the 70.53 deg angle and the 60 deg angle. It forms two right triangles. The 11.39 side is the hypotenuse of one of the right triangles.
sin(70.53)=opp/11.39
opp=sin(70.53)*11.39=10.74 ft Answer
.
Ed