Problem: A ship sails $10$ miles in a straight line from $A$ to $B$ , turns through an angle between $45^{\circ}$ and $60^{\circ}$ , and then sails another $20$ miles to $C$ . Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$ [asy] unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair B=(0,0), A=(-10,0), C=20*dir(50);  draw(A--B--C); draw(A--C,linetype("4 4"));  dot(A); dot(B); dot(C); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); [/asy]
$\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$ $\textbf{(E)}\ [800,900]$

Solution: Let $C_1$ be the point the ship would reach if it turned $45^\circ$ , and $C_2$ the point it would reach if it turned $60^\circ$ . Obviously, $C_1$ is the furthest possible point from $A$ , and $C_2$ is the closest possible point to $A$
Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);  draw(A--B--C1); draw(A--C1,linetype("4 4")); draw(A--B--C2); draw(A--C2,linetype("4 4"));  draw( arc(A, length(C1-A), 0, 55 ), dotted ); draw( arc(A, length(C2-A), 0, 55 ), dotted ); draw( arc(B, C1, C2) );  dot(A); dot(B); dot(C1); dot(C2); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$C_2$",C2,N); [/asy]
We can find $AC_1^2$ and $AC_2^2$ as follows:
Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$ . The angles in the triangle $BD_1C_1$ are $45^\circ$ $45^\circ$ , and $90^\circ$ , hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$ . Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$ $60^\circ$ , and $90^\circ$ , hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);  draw(A--B--C1); draw(A--C1,linetype("4 4"));  dot(A); dot(B); dot(C1);  pair D1 = (C1.x,0); dot(D1); draw(B--D1--C1);  label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$20/\sqrt 2$",midpoint(B--D1),S); label("$20/\sqrt 2$",midpoint(D1--C1),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$D_1$",D1,S); [/asy]
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60);  draw(A--B--C2); draw(A--C2,linetype("4 4"));  dot(A); dot(B); dot(C2);  pair D2 = (C2.x,0); dot(D2); draw(B--D2--C2);  label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C2),SE); label("$10$",midpoint(B--D2),S); label("$10\sqrt 3$",midpoint(D2--C2),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_2$",C2,NE); label("$D_2$",D2,S); [/asy]
Hence we get:
\[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\]
\[AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\]
Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{700,800}$

Answer: D 700,800