T R I
BC = 4000 E G = 1089 CG- 1822 CG = i8«
BG = 2i 7 8 CE = !,n
[ 2 44 1
T R I
BE= 1089
Log. of A B = 3-5563025 Log, of whole Sine = 10.0000000 Log. of EB = 3.03 702 79
Log. of Sine of EAC = 9. 8108297; to which the cor- refpondent Number in the Tables is 40 18'; therefore A CE 49 42' ; and CAB 57° 54'.
Solution of right angled fpherical Triangles, by the common Rules,
I. In a right tingled fpherical Triangle, any two Tarts leftde the right Angle, being given, to find any of the reft.
l u Conlider whether the Parts, which come to the Quellion, be conjunct or disjunct (See Fart.) If the di-junct be oppolite to each other; as, if the Hypothenuie B C, and the Angle C, Fig. 3 1. be given for the oppofite Leg A B ; then the Rule is ;
As the whole Sine is to the Sine of the Hypothenufe B C ; fo is the Sine of the Angle C, to the Sine of the oppofite Leg AB.
2° If the disjunct Parts be not oppolite to each other ; as, if A B, and the adjacent Angle B be given for the oppofiie Angle C ; the Sides of the Triangle are to be con- tinued one Way, 'till they become Quadrants, that you may thus have a new Triangle, wherein the Parts that come into the Queftion, are mutually oppofite to each other ; as, in our Cafe, the Triangle E B F, wherein we have given BF, the Complement of the Leg A B, and the Angle B for E F, the Complement of the Angle C. The Rule then is ;
As the whole Sine is to the Sine ofBF;fois the Sine of the Angle B to the Sine of E F, or Co-fine of C.
3 It the Hypothenuie be not among the conjunct Parts, as if the Legs A B and A C be given for an Angle oppofite to one of them ; the Rule is,
As the Sine of AC is to the whole Sine ; fo is the Tangent of A B, to the Tangent of O.
4 But if the Hypothenuie be found among the conjunct Farts ; as if the Hypothenufe B C, and the Angle C be given to rind the adjacent Side A C .- The Sides of the Triangle are ro be continued one way, till they become Qua- drants, that we may have a new Triangle, wherein the Hy- pothenufe is not among the Parts that come into the Queftion; e.gr. in our Cafe, the Triangle E B F, wherein are given the Complement E B of the Hypothenuie B C, and the Complement of the Angle C, and the Angle F the Complement of the Leg A C. Since then, in the Triangle EFB, the Hypothenufe does not come in the Queftion ; the Rule is as before.
As the Sine of E F, or Co-fine of C, is to the whole Sine ; fo is the Tangent of E B, or Co-tangent of B C, to the Tangent of F, or Co-tangent of A C.
5° When the Sides of a Triangle are to be continued, 'tis the fame thing, which way ibever rhey be produced, pro- vided no acute Angle come into the Queftion; otherwise, the Sides are to be continued through the other oblique one. If both be in the Connection, the Sides are to be continued through that adjacent to the Side in Queftion.
By this means, a Triangle is always obtained, wherein the Thing required is found, either by the Rule of Sines or Tangents.
Solution of right angled fpherical Triangles, by one catholick Rule.
Confider, as before, whether the Parts that come in Queftion be conjunct or disjunct. See Part.
If either one, or both the Sides, including the right Angle, come into the Queftion ; for ir, among the Data, write its Complement to a Quadranr. - — Since, then, by the catholic Rule, deliver'd under the Article Trigonometry ; the whole Sine, with the Sine Complement of the middle Part, is equal to the Sines of the disjunct Parts, and the Co-tangents of the conjunct Parts ; from the Sum of thofe Data, f'ubrract the third Datum ; the Remainder will be fome Sine or Tangent, the Side or Angle correfponding to which, in the artificial Canon of Triangles, is the Side or Angle fought.
This univerfa! Rule being of great Service in Trigonometry, we {hall apply it to the various Cafes thereof, and illuftrate it with Examples ; which Examples, in the Cafe of fe- parate Parts, will at the fame time illuftrate the common Method ; but in the Cafe of contiguous Farts, admit of other Solutions.
1° Given the Hypothenufe BC 60° and the Angle C 23 30', to find the oppofite Leg AB (Fig. 32,)
Since A B is the middle Part, C and B C are disjunct (See Part) ; the whole Sine, with the Co-fine of the Complement A B, i. e. with the Sine itfelf of A B, is equal to the Sinesof C and BC.
Therefore, from Sine of C 96006997 Sine of B C 99375306
Sum 195382303 Subtract whole Sine 100000000 Remain Sine of A B 95382303. The cor- refponding Number to which, in the Canon, is 20 12' 6".
2° Given the Hypothenufe BC 60° and the Leg KB io* 12' 6" to find the oppofite Angle C.
'Tis evident from the preceding Problem, that from the Sum of the whole Sine, and the Sine of the Leg A B, the Sine of the Hypothenufe B C is to be fubtracted' ; the Re- mainder is the Sine of the Angle C. The Example, there- fore, of the former Cafe is eafily converted into an Example of rhis.
3 Given the Leg A B 20° 12' 6" and the oppofite Angle C23 30', to find the Hypothenufe BC.
'Tis evident from the firll Cafe, that from the Sum of the whole Sine, and the Sine of A B is to be fubtracted the Sine of the Angle C, and the Remainder is the Sine of the Hypothenufe B C.
4° Given the Hypothenufe B G 6a", and one Leg A B 20 12' 16" tofind the other Leg.
Since B C is the mean Parr, and A B and A C are dis- junct Parts, the whole Sine, with the Co-fine of the Hypothe- nufe B C, are equal to the Sines of the Complements ; i. e. to the Co-fines of the Legs AB and A C.
Therefore from whole Sine 100000000 Co- fine of B C 96989700
Sum 196989700
SubrractCo-fine of AB 99724279
Remains Co fine of AC 97265421. The Corre- fponding Number to which, in the Canon, is 32" n' 34'; therefore AC 57° 48' 26".
5° Given the Legs A C 57° 48' 26" and AB 20° 12' 6", tofind the Hypothenufe B C.
'Tis evident from the preceding Cafe, that the whole Sine is to be fubtracted from rhe Sum of the Co-fines of the Legi A B and AC; the Remainder is the Co-fine of the Hypo- thenufe BC. The Example, therefore, of the preceding Cafe is eafily applied to this.
6° Given the Leg A C 50° 48' 26" and the adjacent Angle C 23 30' to find the oppofite Angle B.
Since B is the middle Part, and A and C disjunct Parts ; the whole Sine, with the Co-fine of B, is equal to the Sine of C, and the Sine of the Complement, i. e. to the Co-fine of A C ;
Therefore from Sine of C 96006997 Co-fine AC 97265421
Sum 193272418
Subtract whole Sine iaooooooo
Remains Co.fine of B 93272418. The Number correfponding to which, in the Canon, is 12 ij' 56" 5 there- fore B 77 44' 4".
7 Given the Leg A C 57° 48' 26" and the oppofite Angle B 77 44' 4" tofind the adjacent Angle C.
'Tis evident from the preceding Cafe, that the Co-fine of A C is to be fubtracted from the Sum of the whole Sine, and the Co-fine of B ; the Remainder is the Sine of C. The former Example, therefore, is eafily accommodated to the prelenr Cafe.
8° Given the oblique Angles B 77° 44'4'tfBiC 23 30', to find the Leg adjacent to the other, A C.
From Problem the Sixth, 'tis evident that the Sine of C is to be fubtracted from the Sum of the whole Sine, and the Co-fine of B ; and that the Remainder is the Co-fine of A C. The Example of the fixth Problem is eafily applied to this.
9° Given the Leg A C 57° 48' 26" and the adjacent Angle C 2 3 30', tofind the oppofite Leg A B.
Since AC is the mean Part, and C and A B conjunct Parrs ; the whole Sine, with the Sine of A C, is equal to the Co-tangent of C, and the Tangentof A B.
Therefore from whole Sine Sine of A C
100000000 99275039
Sum 199275039 Subtract Co-tangent of C 103616981 Remains Tangentof A B 95658058- To which the correfponding Number in the Canon, is 20° 12' 6".
io° Given the Leg hB 20° 12' 6" and the oppofite Angle C 23° 30', tofind the adjacent Leg A C-
From
