Page:Spherical Trigonometry (1914).djvu/41

§44] (3) Suppose that one of the sides which contain the angle $$A$$ is a quadrant, for example $$AB$$; on $$AC$$, produced if necessary, take $$AD$$ equal to a quadrant and draw $$BD$$. If $$BD$$ is a quadrant, $$B$$ is a pole of



$$AC$$ (Art. 11); in this case $$a= \tfrac 1 2 \pi$$ and $$A=\tfrac 1 2 \pi$$ as well as $$c= \tfrac 1 2 \pi$$. Thus the formula to be verified reduces to the identity $$0=0$$. If $$BD$$ is not a quadrant, the triangle $$BDC$$ gives $$\cos a=\cos CD \cos BD + \sin CD \sin BD \cos C \hat D B,$$

and $$\cos C \hat D B=0$$, $$\cos CD=\cos (\tfrac 1 2 \pi -b)=\sin b$$, $$\cos BD=\cos A$$;

thus $$\cos a=\sin b \cos A$$;

and this is what the formula in Art. 42 becomes when $$c=\tfrac 1 2 \pi$$.

(4) Suppose that both the sides which contain the angle $$A$$ are quadrants. The formula then becomes $$\cos a=\cos A$$; and this is obviously true, for $$A$$ is now the pole of $$BC$$, and thus $$A=a$$.

Thus the formula in Art. 42 is proved to be universally true.

44. The formula in Art. 42 may be applied to express the cosine of any angle of a triangle in terms of sines and cosines of the sides; thus we have the three formulae,