Theodosius <Bithynius>; Clavius, Christoph, Theodosii Tripolitae Sphaericorum libri tres

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513501 niatur ergo, per 16. problema, triang. ſphær. arcus AB, recto angulo D, op-
poſitus, ex duobus angulis B, BAD;
eritq́; proinde &
359[Figure 359] AC, illi æqualis, cognitus.
Deinde, per problema 14.
triang. ſphær. ex inuento arcu AB, rectum angulum ſub-
tendente, &
dato angulo B, reperiatur arcus BD; eritq́;
propterea &
CD, illi æqualis, cognitus; ideoq́; & totus
BC, notus.
Inuentiq́; iam erunt omnes tres arcus AB,
AC, BC.
_PER_ ſolos autem ſinus ita rem exequemur. _P_er 1. praxim
11Per ſolos G
nus, quãdo
duo dati an
guli ſunt
æquales.
problematis _4._
triang. ſphær. inquiratur arcus _BD,_ ex duobus
angulis _B, BAD;_
eritq́; idcirco & _CD,_ illi æqualis, cognitus, proptereaq́; & totus
_BC,_ notus.
_D_einde, per problema _3._ triang. ſphær. ex arcuinuento _BD,_ & an-
gulo ei oppoſito _Bad,_ reperiatur arcus _AB,_ recto angulo oppoſitus:
quia præter da-
ta conſtat etiam ſpecies alterius anguli _B,_ cum datus ſit:
eritq́; propterea & arcus
_AC,_ ipſi _Ab,_ æqualis, cognitus.
18. DATIS omnibus arcubus trianguli non
22Quætũtur
omnes an-
guli.
rectanguli, inueſtigare omnes eius angulos.
SINT omnes arcus in triangulo ABC, dati, ſitq́; primo loco inquiren
33Quãdo duo
dati arcus
ſunt inæ-
quales, &
quadrante
minores.
dus angulus A, &
duo arcus AB, AC, eum continentes ſint inæquales, quadran
teq́;
minores, quicquid ſit de arcu BC. Productis arcubus AB, AC, vt fiant
quadrantes AD, AE, deſcribatur per D, E, arcus circuli maximi DE, occur-
rens arcui BC, producto verſus maiorem arcum, qui ſit AC, in puncto F.
Sta-
44Prop. 63.
triãg. ſphęr.
tuantur ſinus complementorum arcuum datorum AB, AC, pro terminis pro-
portionis ſinus arcus BF, ad ſinum arcus CF.
Atque ex hac proportione, &
arcu dato BC, qui differentia eſt arcuum BF, CF, inueſtigetur, per proble-
ma 8.
triang. rectil. vterque arcus BF, CF.
360[Figure 360] Deinde, per problema 8. triang. ſphær. inue-
ſtigetur tam arcus DF, ex arcu inuento BF,
rectum angulum D, ſubtẽdente, &
arcu BD,
qui complementum eſt dati arcus AB;
quam
arcus EF, ex arcu inuento CF, rectum angu
lum E, ſubtendente, &
arcu CE, qui comple
mentum eſt dati arcus AC.
Subducto enim
arcu EF, inuento, ex inuento arcu DF, no-
tus remanebit arcus DE, anguli, A;
ac proin
de angulus A, notus erit.
Poſt hæc, per pro-
blema 11.
triang. ſphæ. ex arcubus notis BD,
DF, circa rectum angulum D, inueniatur an-
gulus DBF, ac proinde &
reliquus duorum
rectorum ABC.
Eadem denique ratione, ex arcubus CE, EF, notis circa an-
gulum rectum E, eruatur angulus ECF, atque adeo &
angulus ACB, ei ad
verticem æqualis.
Atque ita iam omnes tres anguli A, B, C, inuenti erunt.
_PER_ ſolos ſinus ita progrediemur. Vterque arcus _BF, CF,_ reperiatur per _3._
55Per ſolos fi-
nus, quãdo
dati duo ar
cus sũt in-
æquales, &
praxim problematis _8._
triang. rectil. _D_einde, per _1._ praxim problematis _8._ triang.
ſphar. tam arcus _DF,_ ex arcu inuento _BF,_ rectum angulum _D,_ ſubtendente, & ar-
cu _BD,_ complemento dati arcus _Ab,_ inueniatur, quam arcus _EF,_ ex inuento

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