A golfer putts a ball to a point 22.0 ft. east of his position. The ball passed south of the hole. In order to sink the ball into the hole from his new location, the golfer will have to putt the ball 7.0 ft. on a bearing of 302T. How many feet away from the hole was he before he made his first putt?
23 feet
To find the distance away from the hole before the first putt was made, we can use the Pythagorean theorem.
Let's call the distance away from the hole before the first putt as "x" ft.
From the given information, we know that the first putt was made 22.0 ft east. So, the horizontal distance from the hole after the first putt is x + 22.0 ft.
The ball passed south of the hole, so the vertical displacement from the hole is given by x + 22.0 ft.
Now, using the Pythagorean theorem, we can write:
(x + 22.0)^2 + (x + 22.0)^2 = 7.0^2
Simplifying the equation:
2(x + 22.0)^2 = 7.0^2
Expanding and rearranging the equation:
2(x^2 + 44.0x + 484.0) = 49.0
Distributing:
2x^2 + 88.0x + 968.0 = 49.0
Simplifying further:
2x^2 + 88.0x + 919.0 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 88.0, and c = 919.0.
Calculating the values:
x = (-88.0 ± √(88.0^2 - 4 * 2 * 919.0)) / (2 * 2)
Simplifying further:
x = (-88.0 ± √(7744.0 - 3676.0)) / 4
x = (-88.0 ± √4068.0) / 4
Calculating the square root:
x = (-88.0 ± 63.78) / 4
Using both positive and negative values:
x ≈ (-88.0 + 63.78) / 4 ≈ -5.06
x ≈ (-88.0 - 63.78) / 4 ≈ -37.70
Since distance cannot be negative, we can disregard the negative value.
Therefore, the golfer was approximately 5.06 ft away from the hole before he made his first putt.
To solve this problem, we need to use trigonometry and vector addition.
Let's start by drawing a diagram to visualize the situation. We can represent the starting point of the golfer as point A and the hole as point H. The first putt took the golfer from point A to a new point B, 22.0 ft east of his original position.
A -------- B
\ |
\ |
\302T |7.0 ft
\ |
\ |
H
Note that 302T represents a bearing of 302 degrees counterclockwise from the positive x-axis.
Now, let's break down the problem into components using trigonometric functions. We can consider the displacement of the first putt in terms of its horizontal and vertical components.
The horizontal component of the first putt is given by 7.0 ft * cos(302°). The cosine function gives us the ratio of the adjacent side to the hypotenuse in a right triangle.
The vertical component of the first putt is given by 7.0 ft * sin(302°). The sine function gives us the ratio of the opposite side to the hypotenuse in a right triangle.
Next, let's find the coordinates of point B. We know that it is 22.0 ft to the east of point A. Since there is no north or south displacement mentioned, we can assume the vertical position of B is the same as that of A.
To find the coordinates of point H, we need to add the horizontal and vertical components of the first putt to point B.
Finally, we can calculate the distance between points A and H using the distance formula in two-dimensional space. The distance formula is based on the Pythagorean theorem and can be expressed as follows:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
So, by plugging in the coordinates of points A and H, we can calculate the distance between them.
Let's go ahead and do the calculations:
Horizontal component = 7.0 ft * cos(302°)
= 7.0 ft * (-0.6691)
= -4.6837 ft (rounded to four decimal places)
Vertical component = 7.0 ft * sin(302°)
= 7.0 ft * (-0.7431)
= -5.2017 ft (rounded to four decimal places)
Coordinates of point B = (22.0 ft, 0 ft)
Coordinates of point H = (22.0 ft - 4.6837 ft, 0 ft - 5.2017 ft)
= (17.3163 ft, -5.2017 ft)
Distance between points A and H = sqrt((17.3163 ft - 0 ft)^2 + (-5.2017 ft - 0 ft)^2)
= sqrt(300.4517 ft^2 + 27.0304 ft^2)
= sqrt(327.4821 ft^2)
= 18.11 ft (rounded to two decimal places)
Therefore, the golfer was approximately 18.11 feet away from the hole before he made his first putt.