Calculators either use the Taylor Series for $\sin / \cos$ or the CORDIC algorithm. A lot of information is available on Taylor Series, so I'll explain CORDIC instead.
The input required is a number in radians $\theta$, which is between $-\pi / 2$ and $\pi / 2$ (from this, we can get all of the other angles).
First, we must create a table of $\arctan 2^{-k}$ for $k=0,1,2,\ldots, N-1$. This is usually precomputed using the Taylor Series and then included with the calculator. Let $t_i = \arctan 2^{-i}$.
Consider the point in the plane $(1, 0)$. Draw the unit circle. Now if we can somehow get the point to make an angle $\theta$ with the $x$-axis, then the $x$ coordinate is the $\cos \theta$ and the $y$-coordinate is the $\sin \theta$.
Now we need to somehow get the point to have angle $\theta$. Let's do that now.
Consider three sequences $\{ x_i, y_i, z_i \}$. $z_i$ will tell us which way to rotate the point (counter-clockwise or clockwise). $x_i$ and $y_i$ are the coordinates of the point after the $i$th rotation.
Let $z_0 = \theta$, $x_0 = 1/A_{40} \approx 0.607252935008881 $, $y_0 = 0$. $A_{40}$ is a constant, and we use $40$ because we have $40$ iterations, which will give us $10$ decimal digits of accuracy. This constant is also precomputed1.
Now let:
$$ z_{i+1} = z_i - d_i t_i $$
$$ x_{i+1} = x_i - y_i d_i 2^{-i} $$
$$ y_i = y_i + x_i d_i 2^{-i} $$
$$ d_i = \text{1 if } z_i \ge 0 \text{ and -1 otherwise}$$
From this, it can be shown that $x_N$ and $y_N$ eventually become $\cos \theta$ and $\sin \theta$, respectively.
1: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$
Here is the code for the first problem:
x1 = ? ; y1 = ? ; // First diagonal point
x2 = ? ; y2 = ? ; // Second diagonal point
xc = (x1 + x2)/2 ; yc = (y1 + y2)/2 ; // Center point
xd = (x1 - x2)/2 ; yd = (y1 - y2)/2 ; // Half-diagonal
x3 = xc - yd ; y3 = yc + xd; // Third corner
x4 = xc + yd ; y4 = yc - xd; // Fourth corner
Best Answer
You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$ at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$ and $|PB|^2=m^2$ that is $$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$ and $$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$ Subtracting (2) from (1) gives a linear equation. Use this to eliminate one variable from (1). This yields a quadratic equation in the other variable. Solving this will give the two possible positions for $P$.