2nd method.—Having stated the terms of the proposition, resolve it like any other proportion, in which a fourth term is to be found from three given terms, by multiplying the second and third terms together, and dividing the product by the first.
Note.—Every triangle has six parts—viz., three sides, and three angles; and, in every case in trigonometry, there must be given three of these parts to find the other three. Also of the three parts that are given, one of them at least must be a side; because, with the same angles, the sides may be greater, or less, in any proportion.
Computation.
Case 1.—When a side and its opposite angle are two of the given parts.
The sides of any triangle having the same proportion to each other, as the sines of their opposite angles; then—
As any one side, is to the sine of its opposite angle; so is any other side, to the sine of its opposite angle.
To find an angle, begin the proportion with a side, opposite to a given angle; and, to find a side, begin with an angle opposite to a given side.
Case 2.—When the three sides of a triangle are given, to find the angles.
Let fall a perpendicular from the greatest angle, on the opposite side, or base, dividing it into two segments; and the whole triangle into two right-angled triangles: then the proportion will be—