1. If a segment of length r is at an angle A to a base, and the top of the segment is at distance r sin A from the base, using modern sines. We’re given that A = 90° and that the distance is 200. This makes r = 200 as well.

2. We can divide 141, the given value for sin 45, by the "total sine", 200, and get 0.705, which is correct to two decimal places. In fact, all the values in the table are 200 times their usual values.

3. It is still true that cos A = sin (90°- A). So, for example, cos 40° = sin (90°- 40°) = sin 50° = 153.

4. For this calculation, let’s use capital letters for the functions given in the table, and lower case letters for the modern functions. Then SIN x/COS x = (200 sin x)/(200 cos x) = tan x = (TAN x)/200, and this is not equal to TAN x, so the statement is not true. A student at the time would have learned to write TOT SIN for the radius of the circle involved, and then learned trigonometric identities like SIN x/COS x = TAN x/TOT SIN. That student would have learned it in ratio form, though, so the identity would have looked like SIN x : COS x :: TAN x : TOT SIN.

Similarly, 1/COS x = 1/(200 cos x) = (sec x)/200 = (SEC x)/(200^{2}) = (SEC x)/((TOT SIN)^{2}), and this is not equal to SEC x. Again, the proposed statement is not true.

5. You might start with a/b = TAN A/TOT SIN. This makes 168/200 = a/25, so a = 25•168/200 = 21. Modern methods give a value of 20.9775.

6. You might start with a/c = SIN A/TOT SIN. This gives a/40 = 68/200, so a = 68•40/200 = 13 3/5. Modern methods give 13.6808

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Trigonometry exercises

Note to teachers

The other eight problems

Conclusions