66 TRIGONOMETRICAL PROPOSITIONS.
produced; its arc C D subtract from, or add to, the given side B D, and you have B C. Then multiply the sine of the complement of A D by the sine of the complement of B C; divide the product by the sine of the complement of C D, and the sine of the complement of A B will be produced; hence you have A B itself.
Given A D, & the angle D with the side B D, to find the angle A.
This follows from the above, but the problem would require the “Rule of Three” to be applied thrice. Therefore substitute A for B and B for A, and the problem will be as follows:—
Given B D & D, with the side A D, to find the angle B.
This is exactly the same as the sixth problem, and is solved by the “Rule of Three” being applied twice only.Multiply the sine of A D by the sine of D; divide the product by the sine of A B, and the sine of the angle B will be produced.
Multiply radius by the sine of the complement of D, divide the product by the tangent of the complement of A D, and the tangent of the arc C D will be produced. Then multiply the sine of the complement of C D by the sine of the complement of A B, divide the product by the sine of the complement of A D, and you have the sine of the complement of BC, Whence the sum or the difference of the arcs B C and C D will be the required side B D.
