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: $A_N = \displaystyle\prod_{i=0}^{N-1} \sqrt{1+2^{-2i}}$
OK, I did the Law of Cosines 3 times and came up with 60.647 , 20.404 and 98.949 respectively for angles A, B and C. Remember, the Law of Cosines does not have an ambiguous case, unlike the Law of Sines. I suspect (without further investigating) that his may be the culprit. My advice: Always use the Law of Cosines whenever you can. In this case, when all sides are known, clearly a case for Law of Cosines
Best Answer
The law of cosines applied to right triangles is the Pythagorean theorem, since the cosine of a right angle is $0$. $$ a^2 + b^2 - \underbrace{2ab\cos C}_{\begin{smallmatrix} \text{This is $0$} \\[3pt] \text{if } C\,=\,90^\circ. \end{smallmatrix}} = c^2. $$
Of course, you can also apply the law of cosines to either of the other two angles.
$$\text{sine}=\frac{\text{opposite}}{\text{hypotenuse}};\text{ therefore }\frac{\text{opposite}}{\text{sine}} =\frac{\text{hypotenuse}}{1} = \frac{\text{hypotenuse}}{\sin90^\circ}.$$ Therefore, the law of sines applied to right triangles is valid.