Problem 49 - Algebra and Trigonometry

Question

A major league baseball diamond is actually a square 90 feet on a side. The pitching rubber is located 60.5 feet from home plate on a line joining home plate and second base. (a) How far is it from the pitching rubber to first base? (b) How far is it from the pitching rubber to second base? (c) If a pitcher faces home plate, through what angle does he need to turn to face first base?

Step-by-Step Solution

Answer

(a) 75.4 feet (b) 66.8 feet (c) 36.87 degrees.

Step 1: Understanding the Problem

Visualize the baseball diamond as a square with each side measuring 90 feet. Home plate, first base, second base, and third base are at the corners of the square. The pitching rubber is on a line from home plate to second base, 60.5 feet from home plate.

Step 2: Finding the Distance from the Pitching Rubber to First Base

Use the Pythagorean theorem. The distance from the pitching rubber to first base is the hypotenuse of a right triangle where one leg is the distance from the pitching rubber to home plate (60.5 feet) and the other leg is half the side of the square (90/2 = 45 feet). Formula: \[ \text{Distance} = \text{hypotenuse} = \sqrt{(60.5)^2 + (45)^2} \].

Step 3: Calculating the Hypotenuse

Use the formula to calculate: \[ \text{Distance} = \sqrt{(60.5)^2 + (45)^2} = \sqrt{3660.25 + 2025} = \sqrt{5685.25} \]. Therefore, the distance from the pitching rubber to first base is approximately 75.4 feet.

Step 4: Finding the Distance from the Pitching Rubber to Second Base

The pitching rubber is on the direct line between home plate and second base, which makes it on the same straight path as a diagonal of the square. The diagonal of a square is given by the formula: \[ \text{Diagonal} = a\sqrt{2} \] where \(a = 90 \) feet.

Step 5: Calculating the Distance to Second Base

Subtract the distance from home plate to the pitching rubber from the full diagonal length: \[ \text{Distance to second base} = 90\sqrt{2} - 60.5 \]. Formula: \[ 90\sqrt{2} - 60.5 = 127.279 - 60.5 \approx 66.8 \text{ feet} \].

Step 6: Determine the Angle to First Base

Use the property of right triangles to find the angle. The tangent of the angle is the opposite side (45 feet) over the adjacent side (60.5 feet). Use arctangent (inverse tangent): \[ \theta = \tan^{-1} \left( \frac{45}{60.5} \right) \] which approximately equals 36.87 degrees.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a crucial principle in geometry used to calculate the lengths of sides in a right triangle. It states that for a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The formula is \(\text{c}^2 = \text{a}^2 + \text{b}^2\). Let’s apply this theorem to the problem: We need to find the distance from the pitching rubber to first base. Here, the distance from the pitching rubber to home plate is 60.5 feet, and the distance from the pitching rubber to the baseline (half the side of the square) is 45 feet. By applying the Pythagorean Theorem: \(\text{Distance} = \text{hypotenuse} = \sqrt{(60.5)^2 + (45)^2}\). This gives us approximately 75.4 feet.

Right Triangle

Understanding right triangles is key to solving the exercise. A right triangle is any triangle where one of the angles is exactly 90 degrees. The two sides that form this right angle are known as the legs, and the side opposite the right angle is the hypotenuse. In our baseball diamond problem:

• The leg along the baseline is 45 feet.
• The leg from the pitching rubber to home plate is 60.5 feet.
• The hypotenuse is the distance from the pitching rubber to first base.

To calculate the hypotenuse, we use the Pythagorean Theorem, as mentioned previously.

Trigonometry Applications

Trigonometry helps in solving problems involving angles and distances in right triangles. In our problem, we need to find the angle the pitcher has to turn from facing home plate to face first base. This requires calculating the tangent of the angle using the opposite side (45 feet) and the adjacent side (60.5 feet), thus: \(\theta = \tan^{-1} \left( \frac{45}{60.5} \right)\) which calculates to approximately 36.87 degrees. Additionally, the distance from the pitching rubber to second base can be found by using the diagonal of the square, \(\text{Diagonal} = a\sqrt{2}\), where \(a = 90 feet\). Subtracting the distance from home plate to the pitching rubber gives us: \(\text{Distance to second base} = 90\sqrt{2} - 60.5\). Solving this provides us with approximately 66.8 feet.