91Ó°ÊÓ

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

$=60.5^2+{90}^2-2×60.5×90\cos 45°$

$a≈63.7$

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

$={90}^2+63.7^2-2×90×63.7\cos 45°$

$b≈63.6$

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 -

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

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

$C≈{\cos }^{-1}\frac{63.7^2+60.5^2-{90}^2}{2×63.7×60.5}$

$C≈92.8°$