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:**

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

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

\[ cos\alpha = -0.37 \]