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?

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.

Let the pitching rubber be C and the first base be B.

Now, $a^2=b^2+c^2-2ac\cos A$

$=60.5^2+{90}^2-2\times 60.5\times 90\cos 45°$

$a\approx 63.7$

So, the first base is $63.7$feet far from the pitching rubber.

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.

Let the pitching rubber be C and the second base be D.

Now, $b^2=c^2+d^2-2cd\cos B$

$={90}^2+63.7^2-2\times 90\times 63.7\cos 45°$

$b\approx 63.6$

So, the second base is $63.6$feet far from the pitching rubber.

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.

If a pitcher faces home plate, the angle he need to turn to face first base is -

$c^2=a^2+b^2-2ab\cos C$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$C\approx {\cos }^{-1}\frac{63.7^2+60.5^2-{90}^2}{2\times 63.7\times 60.5}$

$C\approx 92.8°$