What is $\tan(8^\circ51'12'')$? I have a copy of a "ten place natural trigonometric tables" by Hans Hof. For fun I tried to check that the numbers are accurate. But I don't seem to be able to get exactly the same numbers as in the table. The difference is just on the last digit. And it isn't just one number that is off, most seem to be off. I tried to check the table against various online calculators.
For example, according to the table,
$$\tan(8^\circ51'12'') \approx 0.155761467\color{red}3.$$
I understand that
$$8^\circ 51'12'' = \bigg (8 + \frac{51}{60} + \frac{12}{3600}\bigg)^\circ.$$
When I use Google to Calculate tan((8 + 51/60 + 12/3600) degrees) I get $$0.155761467\color{red}{19}^\circ$$
If I use another online calculator I get $$0.155761467\color{red}{199}^\circ$$
My table seem to be the one off, but my question is: what is $\tan(8^\circ51'12'')$? Is my table just off on the last digitor maybe Google isn't rounding correctly?
 A: When I tried in Libre Office and in Excel I found the following result: 0.155761467199028
I can't say for sure without seeing your tables, but I expect that the problem is the way that most tables deal with the last decimal place or the last fraction of a degree (as in your case). There is usually a "differences" column that you use to choose how much to add on. To save space, these are accurate for some of the values in the given row, but not all, although they are "good enough" for most purposes.
Sometimes exact values of (say) tan 8.9 and tan 9.0 are known, and a linear interpolation is used for intermediate values.
OK. I have now checked the following in Excel:
$\tan (8^\circ 51'10'') = 0.155751535692957$
$\tan (8^\circ 51'20'') = 0.155801193523325$
A linear interpolation between these gives $\tan (8^\circ 51'12'') = 0.155761467259031$
I'm guessing that the values from your tables are:
$\tan (8^\circ 51'10'') = 0.1557515357$
$\tan (8^\circ 51'20'') = 0.1558011935$
A linear interpolation between these gives $\tan (8^\circ 51'12'') = 0.15576146726$, which is rounded to $0.1557614673$
A: $\tan (8^\circ 51\text{'}12\text{''})$ is probably better converted entirely to seconds. To do this we need to use the formula $ax^2+bx+c$ where $x=60,a=8,b=51,c=12$. It follows that:
$$
(8)60^2 \text{ s}+(51)60\text{ s}+12\text{ s}
$$
$$
28800\text{s}+3060\text{ s}+12\text{ s}=31872\text{ s}
$$
We know that the total amount of angular units must therefore be divided by the number of seconds per decimal degree which follows that the question is more accurately represented as
$
\tan \big(\frac {31872}{3600} \big )
$
. If we make the fraction irreducible the result is:
$
\tan 8\frac {64}{75}
$
or 
$$
\tan \bigg(\frac {664}{75} \bigg )= \tan8.85\dot{3}
$$
Unfortunately, calculators do not often calculate beyond 14 decimal points, yet they display 10 digits. You may find that the "ten place natural trigonometric tables" by Hans Hof predates modern calculators in which case the results in the table were calculated manually by people. People are not infallible and an example is Nasa's human calculators used  to calculate trajectories of spacecraft before the advent of digital calculators. A movie, "Hidden Figures", highlights how important mathematicians were to space exploration regardless of race or gender.
We now live in an age of digital technology that probably surpasses the calculation ability finest mathematical mind of any humans that have ever lived. I say that because $\pi$ has now been calculated to trillions of decimal places which far exceeds any previous human record of memory.
In terms of your question, however, my upgraded Casio fx-9750GII gives an answer:
$$
\tan \bigg(\frac {664}{75} \bigg )\approx 0.15576146712
$$
Whilst this may seem quite inaccurate on your tables, it is worth remembering that that degree of accuracy (see what I did there?), is far above the accuracy required to send the men to the moon and back so you need not be too concerned with the final digits, as interesting as this journey of knowledge was.
I hope this answer meets with your expectations.
