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A major League baseball diamond has four bases forming a square whose sides measure 90 feet each. The pitcher’s mound is 60.5 feet from home plate on a line joining home plate and second base. Find the distance from the pitcher’s mound to first base. Round to the nearest tenth of a foot.

This problem aims to familiarize us with trigonometric laws. The concepts required to solve this problem are related to the law of cosines, or more commonly known as the cosine rule, and the significance of postulates.

The Law of cosines represents the connection between the lengths of sides of a triangle with reference to the cosine of its angle. We can also define it as the method to find the unknown side of a triangle if the length and the angle between any of the two adjacent sides are known. It is presented as:

\[c^2 = a^2 + b^2 – 2ab cos\gamma \]

Where $a$, $b$, and $c$ are given as the sides of a triangle and the angle between $a$ and $b$ is represented as $\gamma$.

To know the length of any side of a triangle, we can use the following formulas as per the given information:

\[ a^2 = b^2 + c^2 – 2bc cos \alpha \]

\[ b^2 = a^2 + c^2 – 2ac cos \beta \]

\[ c^2 = b^2 + a^2 – 2ba cos \gamma \]

Similarly, if the sides of a triangle are known, we can find the angles using:

\[ cos\alpha = \dfrac{[b^2 + c^2 – a^2]}{2bc} \]

\[ cos\beta = \dfrac{[a^2 + c^2 – b^2]}{2ac} \]

\[ cos\gamma = \dfrac{[b^2 + a^2 – c^2]}{2ab} \]

Expert Answer

As per the statement, we are given the lengths of all the four bases forming a square with each side measuring about $90$ feet (one side of a triangle), whereas the length of the pitcher mound from the home plate is $60.5$ feet, which makes up our second side to construct a triangle. The angle between them is $45^{\circ}$.

So we have the lengths of $2$ adjacent sides of a triangle and the angle between them.

Let’s say $B$ and $C$ be the sides of the triangle that are given, and $\alpha$ is the angle between them, then we have to find the length of the side $A$ using the formula:

\[ A^2 = B^2 + C^2 – 2BC cos \alpha \]

Substituting the values in the above equation:

\[ A^2 = 60.5^2 + 90^2 – 2\times 60.5 \times 90 cos 45 \]

\[ A^2 = 3660.25 + 8100 – 10890 \times 0.7071 \]

Further simplifying:

\[ A^2 = 11750.25 – 7700.319 \]

\[ A^2 = 4049.9 \]

Taking square root on both sides:

\[ A = 63.7 \space feet\]

This is the distance from the pitcher mound to the first base plate.

Numerical Answer

The distance from the pitcher mound to the first base plate is $63.7 \space feet$.

Example

Consider a triangle $\bigtriangleup ABC$ having sides $a=10cm$, $b=7cm$ and $c=5cm$. Find the angle $cos\alpha$.

Finding the angle $\alpha$ using the cosine law:

\[ a^2=b^2 + c^2 – 2bc cos \alpha\]

Rearranging the formula:

\[ cos\alpha=\dfrac{(b^2 + c^2 – a^2)}{2bc}\]

Now plug in the values:

\[cos\alpha = \dfrac{(7^2 + 5^2 – 10^2)}{2\times 7\times 5} \]

\[ cos\alpha = \dfrac{(49+25-100)}{70} \]

\[ cos\alpha = -0.37 \]

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