Answer
The pitcher must turn $81{}^\circ $ for a direct throw towards first base.
Work Step by Step
Let $a=60,c=46$.
$AB$ is along the diagonal line joining the home plate and second. Therefore, angle $B=45{}^\circ $
Using the Law of cosines we will find the distance $c$ -- that is, the line joining the pitcher’s mound and first base.
$\begin{align}
& {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\ \cos B \\
& ={{60}^{2}}+{{45}^{2}}-2\cdot 60\cdot 46\ \cos 45{}^\circ \\
& b=\sqrt{1812.77} \\
& b\simeq 43
\end{align}$
Now by using the sine law, we get:
$\begin{align}
& \frac{a}{\sin A}=\frac{b}{\sin B} \\
& \frac{60}{\sin A}=\frac{43}{\sin {{45}^{2}}} \\
& \sin A=\frac{60\cdot \sin 45{}^\circ }{43} \\
& A={{\sin }^{-1}}.9866
\end{align}$
Hence,
$A\simeq 81{}^\circ $
The pitcher must turn $81{}^\circ $ for a direct throw towards first base.
