math question?

Question

a roller coaster is held up by two steel beams. there is acable that runs from the top of one to the bottom of the other, as well as another that runs from the top of the other to the bottom of the first one.(think bowtie) one beam is 40 feet high and the other is 30 feet high. how high above ground do the two cables cross? use similar triangles to solve.

Answer 1

Draw it ... say the 40 foot high beam is at the left, then A feet to the right of that is under where the two cables cross, then B feet to the right of that is where the 30 foot beam stands. (So the two beams are A + B feet across.).

Let the crossing height be H.

Legs of similar right triangles with right angle at the left:
40 : (A + B)
H : B

Legs of similar right triangles with right angle at the right:
30 : (A + B)
H : A

So:

40 / (A + B) = H / B or H = 40 B / (A + B)
30 / (A + B) = H / A or H = 30 A / (A + B)

From which which we 40 B = 30 A or A/B = 40 / 30 = 4/3 -- the crossover point separates the two beams in the same ratio as the beams on the respective sides of the crossover point.

A / (A + B) = 4 / 7
B / (A + B) = 3 / 7

H = 120 / 7 feet (17 and 1/7 feet)

(The actual values of A, B, and A + B do not matter, just the ratios.)

Dan