Question

A wire is stretched from the ground to the top of an antenna tower.The wire is 17 feet long. The height of the tower is 7 feet greater than the distance d from the tower's base to the end of the wire. Find the distance d and the height of the tower

Answer 1

Therefore,

The distance d is 8 feet,

and the height of the tower is 15 feet.

Explanation:

Consider a diagram shown below such that

Let,

AC = length of wire = 17 feet

BC = d = distance from the tower's base to the end of the wire.

The height of the tower is 7 feet greater than the distance d

AB = height of tower = 7 +d

To Find:

AB = ? ( height of tower)

BC = d =?

Solution:

In Right Angle Triangle ABC by Pythagoras theorem we have

`(\textrm{Hypotenuse})^(2) = (\textrm{Shorter leg})^(2)+(\textrm{Longer leg})^(2)`

`AC^(2)=BC^(2)+AB^(2)`

Substituting the values we get

`17^(2)=d^(2)+(7+d)^(2)`

Using (A+B)²=A²+2AB+B² we get

`17^(2)=d^(2)+49+14d+d^(2)\\2d^(2)+14d-240=0`

Dividing through out by 2 we get

`d^(2)+7d-120=0`

Which is a quadratic equation, so on factorizing we get

`(d-8)(d+15)=0\\d-8=0\ or\ d+15=0\\d=8\ or\ d=-15`

d cannot be negative therefore ,

`d =8\ feet`

Now substitute d in AB we get

`AB=7+8=15\ feet`

Therefore,

The distance d is 8 feet,

and the height of the tower is 15 feet.