68 TRIGONOMETRICAL PROPOSITIONS.
three times applied. Therefore substitute A for D and D for A, and the problem will be as follows:—
Given A D & the angle A with the angle D, to find the side B A.
This is the same throughout as problem 11, and is solved by applying the “Rule of Three” twice only.
The use and importance of half-versed
Sines.
1.GIven two sides & the contained angle, to find the third side.
From the half-versed sine of the sum of the sides subtract the half-versed sine of their difference; multiply the remainder by the half-versed sine of the contained angle; divide the product by radius; to this add the half-versed sine of the difference of the sides, and you have the half-versed sine of the required base.
Given the base and the adjacent angles, the vertical angle will be found by similar reasoning.
From the half-versed sine of the base subtract the half-versed sine of the difference of the sides-multiplied by radius; divide the remainder by the half-versed sine of the sum of the sides diminished by the half-versed sine of their difference, and the half-versed sine of the vertical angle will be produced.
Given the three angles, the sides will be found by similar reasoning.
