Question 195046: I am having a problem solving this word problem. Any help would be appreciated!
A rope is stretched from the ground to the top of an antenna tower. The wire is 30ft long. The height of the tower is 6ft. greater than the distance from the base to the end of the rope. Find the distance from the wire to the tower.
Found 2 solutions by nerdybill, RAY100:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! A rope is stretched from the ground to the top of an antenna tower. The wire is 30ft long. The height of the tower is 6ft. greater than the distance from the base to the end of the rope. Find the distance from the wire to the tower.
.
Draw a diagram of the problem. Doing so, you should recognize that the wire, tower and ground forms a right triangle. You will need to apply Pythagorean theorem.
.
Let x = distance from wire to tower
then
x+6 = height of tower
.
x^2 + (x+6)^2 = 30^2
x^2 + (x+6)(x+6) = 900
x^2 + x^2+12x+36 = 900
2x^2+12x+36 = 900
2x^2+12x-864 = 0
x^2+6x-432 = 0
Factoring the left:
(x+24)(x-18) = 0
x = {-24, 18}
We can toss out the negative solution leaving:
x = 18 feet
Answer by RAY100(1637)  (Show Source):
You can put this solution on YOUR website! Lets view this as a right triangle problem
.
The hypotenuse is 30, the length of the rope
.
one leg is b,,,,the unknown base
.
the other leg is (b+6),,,unknown height, but related to base
,
Using pythagouous, c^2 = a^2 + b^2
.
30 ^2 = b^2 + (b+6)^2
.
900 = b^2 + ( b^2 +12b +36)
.
0 = 2 b^2 +12b -864
.
0 = (2) (b^2 +6b -432)
.
0 = 2 ( b+24) (b-18)
.
b = 18, or -24 ( not realistic)
,
Therefore base is 18, height is 24 (18+6)
.
checking
.
30^2 = 18^2 +(18+6)^2 = 18^2 + 24^2 = 900,,,,ok
|
|
|
