Plane trigonometry, by S.L. Loney.

SOLUTION OF TRIANGLES.

197

we
so 1

2 n \)2 4& COS2 A
2(b+c)[- 4be+. a2=(b )21 (b + )2
4be A
Hence, if sin2 0 = ( cos 2
have a2 = (b + c)2 [1 - sin2 ] = (b + c2 cos2 0,
that a = (b + c) cos 0.
If then sin 0 be calculated from the relation. 2 VJbc A
sin 0 c= - -cos -
b+c 2'
have a=(b +c)cosO.

'we

(2) We have

a2 = b" - 2bc + c2 - 2bc (cos A - 1)
=(b -)2 + 4bc sin2 A
2
-(bc)[+ (b - C)24c i 2]
4be A
(- sin2" = tans 2,
(b - C)2 2

Let

2^/bc. A
tan 0b= — c sin-,
b - c 2'

so that
and hence 0 is known.

(b - C )2
Then a2 =(b-c)[1 + tan2 0]=( C)
so that a=(b —c) sec,
and is therefore easily found.
An angle, such as 0 or 0 above, introduced for the purpose of
facilitating calculation is called a subsidiary angle (Art. 129).
EXAMPLES. XXX.
y 1. If b = 90, c = 70, and A = 72~ 48' 30", find B and C, given
log 2= 30103, L cot 36~ 24' 15"=10-1323111,
L tan 9~ 37' = 92290071
and L tan 9~ 38' = 9'2297735.