Ciszterci rendi katolikus gimnázium, Eger, 1860

24

23521

1562

47042

141126

117605

23521

s = 3673.98 frt.

27. Adatván három oldal (1. első ábra) a, b, c; kerestessék ki a három szög, és a háromszög területe.

sin £ = l/ (s-b) • (b-c) ^ r be

logsin - = log (b—'b) + log (b-c) — log (b+c) 2 2

. a log 18,7 -4- log 95,5 — log(b-j-c)

logsin - = -s--------!----5—------------±—l---a = 114,2 b = 206.4 c = 129.6 « = 29« 56< 52" ß = 115° 32' 14" y = 34° 30' 54"

T = 6676,7

a =-■ 114, 2 b = 206, 4 c = 729, 6

log (s—b) = log 18.7 = 1.27184 | b = log 206,4 = 2,31471 7 log (s—c) = log 95.5 = 1.98000 ) c = log 129,6 = 2.11261 \ a + b -f c = 450, 2 23.25184 I log (b-f-c) = 4.42732 s = 225,1- log (b+c) = 4.42732

18.82452 — 20 : 2 = 9,41226—10

logsin — = 9.41226 == 14° 58' 26"

sin « = 29° 56' 52‘‘

s — b = 18, 7 s — c = 95, 5 s — a = 110, 9

v * 1 1

logsin j = 14« 58< 26" 2

sin+= 29° 56' 52''

• ß

sin

~ = 1/(s~a) (g—c)

^ ac

lofrsin l = lQg (s a) —|— log (b - c) — log (a-f- c)

8 2 2

loe-sin t = log H 0,9 +log 95,5 — log (a-fc)

8 2 -2

log 114,2 = 2,05767 1 log (s - a) = log 110,9 = 2. 04493 log 129,6 = 2,11261. > log (s - c) = log 95.5 = 1. 98000

24. 02493'

— log (a-f-c) = 4. 17028

19. 85465 — 20 : 2 = 9,92732

log (a-fc) = 4,17028

sin ^ = 57° 46' 7"

u

logsin £ = 9.92732 = 57« 46' 7"

u

sin ß = 115° 32' 14"

sin ß = 115° 32' 14'