A 24 meters pole is held by a three guy wire into vertical position. Two of the guy wires are of equal length. The third wire is 5 meters longer than the other two and is attached to the ground 11 meters farther from the foot of the pole than the two equal wires. Find the length of the wire.

Let the shorter wire be x m long and let that wire be y m from the base of the pole

so x^2= 576 + y^2 ----> x^2 - y^2 = 576

let the longer be x+5 and we have
(x+5)^2 = 24^2 + (y+11)^2
x^2 + 10x + 25 = 576 + y^2 + 22y + 121
x^2 - y^2 = 22y - 10x + 672

22y - 10x + 672 = 576
22y - 10x = - 96
y = (10x-96)/10

sub this into
x^2 - y^2 = 576
x^2 - (100x^2 - 1920x + 9216)/100 = 576
times 100
100x^2 - 100x^2 + 1920x - 9216 = 57600
1920x = 66816
x = 34.8 , then y = 25.2

check:
x^2 - y^2 = 576
LS = 34.8^2 - 25.2^2 = 576 = RS

also checks out for the 2nd equation, I will let you verify that.

A wire extends from the top of a 12-foot light pole to the ground. The wire forms an angle of 53°with the ground. Find the length of the wire to the nearest tenth.

Let's denote the length of the two equal guy wires as x meters.

According to the problem, the third wire is 5 meters longer than the other two, so its length is x + 5 meters.

Also, we know that the third wire is attached to the ground 11 meters farther from the foot of the pole than the two equal wires. This means that the distance from the foot of the pole to the attachment point of the third wire is x + 11 meters.

Since the three guy wires are used to hold the pole into a vertical position, we can create a right-angled triangle using the pole as the hypotenuse and the guy wires as the two legs.

Using the Pythagorean theorem, we can relate the lengths of the guy wires and the height of the pole:

x^2 + x^2 = (x + 11)^2

Simplifying the equation:

2x^2 = x^2 + 22x + 121

Rearranging the terms:

x^2 - 22x - 121 = 0

Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula.

However, after attempting to factor the quadratic equation, we find that it cannot be factored easily. So, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -22, and c = -121. Substituting these values into the quadratic formula:

x = (-(-22) ± √((-22)^2 - 4(1)(-121))) / (2(1))

x = (22 ± √(484 + 484)) / 2

x = (22 ± √(968)) / 2

x = (22 ± √(4 * 242)) / 2

x = (22 ± 2√242) / 2

Simplifying:

x = 11 ± √242

Since the length of the guy wire cannot be negative, we ignore the negative solution:

x = 11 + √242

Therefore, the length of the two equal guy wires is 11 + √242 meters.

To find the length of the third wire, we add 5 to the length of the equal guy wires:

Length of the third wire = x + 5 = (11 + √242) + 5 = 16 + √242 meters.

To find the length of the wire, we can break down the problem into smaller parts and use basic geometry concepts.

Let's denote the length of the two equal wires as "x". We are given that the third wire is 5 meters longer than the other two, so its length would be "x + 5".

We are also given that the third wire is attached to the ground 11 meters farther from the foot of the pole than the two equal wires. This means that the distance from the foot of the pole to where the third wire is attached is "x + 11".

Now, we can set up a equation using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

In this case, the pole acts as the hypotenuse, so the equation becomes:

x^2 + x^2 = (x + 5)^2 + (x + 11)^2

Simplifying the equation, we get:

2x^2 = x^2 + 10x + 25 + x^2 + 22x + 121

Combine like terms:

2x^2 = 2x^2 + 32x + 146

Subtract 2x^2 from both sides:

0 = 32x + 146

Subtract 146 from both sides:

-146 = 32x

Divide both sides by 32:

x = -146/32

x ≈ -4.5625

Since we are dealing with lengths, the value of "x" cannot be negative. Therefore, the length of the wire cannot be negative, and we conclude that there is no valid solution to this problem.