A 75ft diagnol brace on a bridge connects a support of the center of the bridge The horizontal distance that it spans is 15ft longer than the hieght that reaches on the side of the bridge.Find the horizontal and vertical distance spanned by this brace
Answered in previous post.
Let's denote the horizontal distance as x and the vertical distance as y.
According to the given information, the diagonal brace forms a right triangle with the horizontal and vertical distances.
We are given the diagonal brace length as 75 ft, so we can use the Pythagorean theorem to find the relationship between x and y:
x^2 + y^2 = 75^2
We are also given that the horizontal distance is 15 ft longer than the height, so we can write:
x = y + 15
Now we can substitute the value of x in the Pythagorean theorem equation:
(y + 15)^2 + y^2 = 75^2
Simplifying the equation:
y^2 + 30y + 225 + y^2 = 5625
Combining like terms:
2y^2 + 30y + 225 - 5625 = 0
2y^2 + 30y - 5400 = 0
Dividing the equation by 2:
y^2 + 15y - 2700 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. Let's use the quadratic formula:
y = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 15, and c = -2700. Plugging these values into the formula:
y = (-15 ± sqrt(15^2 - 4(1)(-2700))) / (2(1))
Simplifying:
y = (-15 ± sqrt(225 + 10800)) / 2
y = (-15 ± sqrt(11025)) / 2
y = (-15 ± 105) / 2
Now we have two possible values for y:
1. y = (-15 + 105) / 2 = 90 / 2 = 45 ft
2. y = (-15 - 105) / 2 = -120 / 2 = -60 ft (which is not applicable in this context)
Next, we can substitute the value of y into x = y + 15:
x = 45 + 15 = 60 ft
Therefore, the horizontal distance spanned by the brace is 60 ft, and the vertical distance spanned by the brace is 45 ft.
To find the horizontal and vertical distances spanned by the brace on the bridge, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let's denote the horizontal distance as x and the vertical distance as y.
From the given information, we know that the hypotenuse (brace) is 75ft long. Therefore, we have the equation:
x^2 + y^2 = 75^2
We are also given that the horizontal distance x is 15ft longer than the height y. In equation form, this becomes:
x = y + 15
Now we can substitute the value of x from the second equation into the first equation:
(y + 15)^2 + y^2 = 75^2
Expanding the equation:
y^2 + 30y + 225 + y^2 = 5625
Combining like terms:
2y^2 + 30y + 225 = 5625
Moving all terms to one side to form a quadratic equation:
2y^2 + 30y - 5400 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. For simplicity, let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a=2, b=30, and c=-5400:
y = (-30 ± √(30^2 - 4*2*(-5400))) / (2*2)
Simplifying inside the square root:
y = (-30 ± √(900 + 43200)) / 4
y = (-30 ± √44100) / 4
y = (-30 ± 210) / 4
This results in two possible values for y:
y = (-30 + 210) / 4 = 180/4 = 45
y = (-30 - 210) / 4 = -240/4 = -60
Since we are dealing with a physical distance on a bridge, the height cannot be negative. Therefore, we disregard the negative solution.
So, the vertical distance spanned by the brace is 45ft.
To find the horizontal distance, we can substitute this value of y back into the equation x = y + 15:
x = 45 + 15 = 60
Therefore, the horizontal distance spanned by the brace is 60ft, and the vertical distance spanned by the brace is 45ft.