Hello, jasonk!

1. A major League baseball diamond square 90 feet on a side.

The pitching rubber is located 60.5 feet from home plate.

(A) How far is it from the pitching rubber to first base?

(B) If a pitcher faces home plate, through what angle does he need to turn to face first base? Did you make a sketch? Code:

*
* *
* *
* *
* *
* *
* *
* *
* o F
* P * *
* o θ *
* | *
* 60.5| * 90
* | *
* |45°*
* | *
o
H

Draw line segment $\displaystyle PF.$

The pitcher is at $\displaystyle P.$

Home plate is at $\displaystyle H\!:\;PH = 60.5$

First base is at $\displaystyle F\!:\;HF = 90$

Let $\displaystyle \angle FPH = \theta$

Note that $\displaystyle \angle PHF = 45^o.$

(A) Law of Cosines: .$\displaystyle PF^2 \:=\:60.5^2 + 90^2 - 2(60.5)(90)\cos45^o \:=\:4059.857183$

. . .Therefore: .$\displaystyle PF \:\approx\:63.7$ ft

(B) Law of Cosines: .$\displaystyle \cos\theta \:=\:\frac{60.5^2 + 63.7^2 - 90^2}{2(60.5)(63.7)} \:=\:-0.049568613$

. . .Therefore: .$\displaystyle \theta \;\approx\;92.8^o$

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And I completely agree with your reasoning and work on #2.

Good work!