SOLUTION: I am not sure but I think I need to use this theorem to solve this problem. The problem is:
A guy wire of length 50 feet is attached to the ground and to the top of an antenna. Th
Algebra ->
Pythagorean-theorem
-> SOLUTION: I am not sure but I think I need to use this theorem to solve this problem. The problem is:
A guy wire of length 50 feet is attached to the ground and to the top of an antenna. Th
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Question 77982: I am not sure but I think I need to use this theorem to solve this problem. The problem is:
A guy wire of length 50 feet is attached to the ground and to the top of an antenna. The height of the antenna is 10 feet longer than the distance from the base of the antenna to the point where the guy wire is attached to the ground. What is the height of the antenna?
You can put this solution on YOUR website! A guy wire of length 50 feet is attached to the ground and to the top of an antenna. The height of the antenna is 10 feet longer than the distance from the base of the antenna to the point where the guy wire is attached to the ground. What is the height of the antenna?
:
The 50 feet will be the hypotenuse of the right triangle formed with the tower
and the ground,
:
Let x = the height of the antenna
Then (x-10) = distance from the base to the tie point on the ground
:
Using pythag. a^2 + b^2 = c^2
:
x^2 + (x-10)^2 = 50^2
:
x^2 + x^2 - 20x + 100 = 2500; FOILed (x-10)(x-10)
:
2x^2 - 20x + 100 - 2500 = 0
:
2x^2 - 20x - 2400 = 0; a quadratic equation
:
x^2 - 10x - 1200 = 0; simplified, divided eq by 2
:
Factors to:
(x - 40)(x + 30) = 0
:
x = 40; it's the positive solution we want here
:
40' is the height of the antenna,
30' is the distance from the base to the tie point on the ground
:
:
Check:
40^2 + 30^2 =
1600 + 900 = 2500 which is 50^2
