96. Case V. Given the three sides j or a, h **«••• (-

and e. (Fig. 9.) Wo have three methods
f'or computing the half angles :

1st. Bj the sinesy from (31), remembering
that A

««J(a + 6+c)



203




sin



^^ ;. sin(a-6)sin(g-gK
' N \ sin 6 sin (? /



-* N \ sin c sm a /



sin



1 (7= / (?!?_(i. - ^) s^^ (* - ^)



C0S1



sin a sin 5

2d. By the cosines, from (33),

1 J — / / sin « sin (« — a) \
^i^-slK sin b sine /

biL 4 sin (« — J) \
sin (7 sin a /

sin a sin (a — <?) >^

cos i (7 = J { -^^ ; ^^ — 7 — - )

^ N \ Sin a sm 6 /

8d. By the tangents, from (34),
tan



)



cos






1 J — / / sin (« — J) sin (a — e)
^ N \ sin a sin (a — a)



tan^f



-k



tan J (7 =



sin (a — (?) sin (a — a)
sin a sin (a — 6)



/ /sin (a — a) sin (a — 6) \
^ V ain « ain ^« — c\ /



(164)



(165



(166)



sin a sin (a — c)

When only one of the angles is required, the simplest method will
be by (165), but if the required angle is less than 90°, it will be
found more accurately by (164), for then ^A <46°, and the sine
varies more rapidly than the cosine. And, for a similar reason, if
the angle is greater than 90°, we should prefer (165). By (166) we
always have an accurate result, although the formula is not quite so
simple.
