TRIGONOMETRICAL PRoposiITIONS. 71
Now add the logarithm of the half sum, namely 1819061,
| either | or |
| to the logarithm 1218382, and the sum will be 3037443; from this subtract the logarithm 1900221 and there will remain 1137222. | to the logarithm of the complement of the half difference, namely 11307, and the sum will be 1830368; from this subtract 693147 and there will remain 1137221. |
Halve the latter and you have the logarithm 568611, to which corresponds the arc 34° 30’, and double this arc is the base required, namely 69°.
Conversely, given the three sides, to find any angle. The solution of this problem ts given in my work on. Logarithms, Book II. chap. vi. sect. 8, but partly by logarithms and partly by prosthapharesis of arcs.
It ts to be observed that in the preceding and following problems there ts no need to discriminate between the different cases, since the form and magnitude of the several parts appear in the course of the calculation.
Another direct converse of the preceding problem follows.—
[Given the sides and the base, to find the vertical angle.]
HAlve the given base, namely 69°, and you have 34° 30’, the logarithm of which is 568611. Double the latter and you have 1137222; corresponding to this is the arc 18° 42’, which note as the second found.
As before, take for the first found the arc 0° 44’, corresponding to the logarithm 4357140.
The complements of the two arcs are 89° 16’ and 71° 18’; their half sum is 80° 17’, and its logarithm
