Page:Spherical Trigonometry (1914).djvu/40

22 43. We have supposed, in the construction of the preceding Article, that the sides which contain the angle $$A$$ are less than quadrants, for we have assumed that the tangents at $$A$$ meet $$OB$$ and $$OC$$ respectively produced. We must now shew that the formula obtained is true when these sides are not less than quadrants. This we shall do by special examination of the cases in which one side or each side is greater than a quadrant or equal to a quadrant.

(1) Suppose only one of the sides which contain the angle $$A$$ to be greater than a quadrant, for example $$AB$$. Produce $$BA$$ and $$BC$$ to meet at $$B'$$; and put $$AB'=c'$$, $$CB'=a'$$.



Then we have from the triangle $$AB'C$$, by what has already been proved, $$\cos a' = \cos b \cos c' + \sin b \sin c' \cos B'AC ;$$

but $$a'=\pi-a$$, $$c'=\pi-c$$, $$B'AC=\pi-A$$; thus $$\cos a= \cos b \cos c+\sin b \sin c \cos A.$$

(2) Suppose both the sides which contain the angle $$A$$ to be greater than quadrants. Produce $$AB$$ and $$AC$$ to meet at $$A'$$; put $$A'B=c'$$, $$A'C=b'$$; then from the triangle $$A'BC$$, as before, $$\cos a=\cos b' \cos c +\sin b' \sin c' \cos A';$$



but $$b'=\pi-b$$, $$c'=\pi-c$$, $$A'=A$$; thus $$\cos a=\cos b \cos c+\sin b \sin c \cos A, $$