T R I
C H5 ]
From the Sum of the Co-tangent of C, and the Tangent of A B, fubtraft the whole Sine ; the Remainder is the Sine of A G.
ii° Given the Legs AB 20° 12' 6", and AC J7 48' 26". to find the Angle C, oppofite to one of them.
From the Sum of the whole Sine, and Sine A C, fubtraft the Tangent ot BA5 the Remainder is the Co-tangent of C.
12" Given the Hypothemife EG 60°, and the oblique Angle C 23 30', to find the adjacent Leg A C.
Since C is the middle Part, and BC and AC conjoynt Farts 5 the whole Sine, with the Co-fine of C, will be equal to the Co-tangent of A C.
Therefore from whole Sine 100000000 Co-fine of C 99623978
Sum 199623978 Subtract Co-tangenr of B C 97614394 Remains Tangent of A C J02009584. The Number correfponding to which, in the Tables, is 57 48' 26".
13° Given the Leg AC $7° 48' 26", and the adjacent Angle C 23° 30', to find the Hypothemife BC.
From the Sum of the whole Sine, and the Co-fine of C, fubtradt the Tangent of A C, the Remainder is the Co- tangent of B G'
14° Given the Hypothemife BC 60°, and the Leg A G 57 48' 26", to find the adjacent Angle C.
From the Sum of the Co-tangem of B C, and Tangent of A C, fubtracl the whole Sine ; the Remainder is the Co-fine of C.
1 ;" Given the Hypothemife B C 60°, and one Angle C 23 30'j to find the other , B.
Since BC is the middle Part, and Band C disjunct Parts, the whole Sine, and the Go-fine of B C will be equal to the Co-tangents of B and C.
Therefore from whole Sine ioooooooG Co-fine of B C 96989700
Sum 196989700 Subtract Co-tangent of C 10366981
Remains Co-tangent of B 93372719 ; The corre- fponding Number to which, in the Canon, is 12 15' 56" 5 therefore B is 77" 44' 4".
16 Given the oblique Angles B 77 44' 4" andC 23° 30', to find the Hypothemife.
From the Sum of the Co-tangents of C andB, fub trail the whole Sine ; the Remainder is the Co-fine of B C.
Solution of oblique-angled fpherical Triangles.
i° In an oblique-angled fpherical Triangle ABC, (Tab-trigonometry Fig. 32.) two Sides AB and BC being given, together with an Angle A, oppofite to one of them, to find the other C: The Rule is,
As Sine of the Side B C, is to the Sine of the oppoGte Angle A ; lb is the Sine of the Side 8 A, to the Sine of the oppofite Angle C,
Suppofe, for Example, BC 39° 2?'; A 43° 20'; BA 66° 4$' ; then will
Sine of BC 98033572 Sine of A 983647
Sine of B A 99632168
197996939 ,
Sine of C 99963367. The correfponding Number to which, in the Tables, is 8 2° 34' 7".
2 . Given two Angles C 82 34' 7", and A 43° 20', toge- gether with the Side AB 60° 45", oppofite to one of them C ; to find the Side B C oppofite to the other of them A. Say,
As Sine of Angle C, is to Sine of the oppofite Side A B ; fo is Sine of Angle A, to Sine of oppofite Side B C. — The former Example may fuffice for the prefent Cafe.
3°. Given two Sides AB 66° 45', and BC 39 29', toge- ther with an Angle oppofite to one of them A 45 20' ; to find the Angle included by them B.
Suppofe the Angle C to be acute, fince the other, A, is alfo acute 5 "the Perpendicular B E falls within the Triangle, In the rectangled Triangle ABE, therefore, from the given An- gle A, and Side AB, find the Angle A BE. Since B E is affiirrfd as a lateral Part in the Triangle A E B, the Angle E B C is the middle Part, and the Side B C a conjoynt Part : The Cofine ot the Angle EEC will be found by fub- tracling the Cotangent of AB from the Sum of the Cofine of the' Angle ABE, and the Co-tangent of B C. If then the Angles' ABE and EBC be added together; or, in cafe the Perpendicular fall without the Triangle, be fubtrafted from each other, you will have the Angle requir'd B.
E.gr. whole Sine
Cofine of A B
T R I
IOOCOOOO
95963154
Sum Gotang. of A
19*963154 100252805
Cotang. of A B 95710349. The Number fpondingto which, in the Tables, is 20 25' 35". therefore is 69* 34' 25".
Cofine of A B E 95428300
Cotang. otBC 100141529
cori'f- A B
Sum
Cotang. of B
196269829
96330085
Cofine of EBC 9938544. The Number corre- fponding to which, in the Tables, is 8o° 26'. Therefore ABC is 79 9' 59".
4 . Given two Angles A 43 20', and B 79 9' 59", to- gether with the adjacent Side A B 66° 45' 5 to find the Side B C oppofite to one of them.
From one of the given Angles B, let fall a Perpendicular EB, to the unknown Side Ac 5 and, in the reftangled Tri- angle ABE, from the given Angle A, and Hypcthenufc A B, find the Angle ABE; which fubtracTed from the Angle ABC, Je,aves the Angle EBC- But if the Perpendicular fhould fall without the Triangle, the Angle ABC ftioukl have been fubtrac/ted from A B E. Since as the Perpendicular BE is taken for one of the lateral Parts, the middle Part in the triangle A B E, is the Angle E, and the coujoym Part BC ; the Co-tangent of the Side BC is found by iltbtracV ing the Gofinc ot £B A, from the Sum of the Co-tarigent of AB, and the Cofine of E B C — The Example of the preced- ing Cafe is eafily apply M to this.
5°. Given two Sides AB 66° 45', and BC 39 29', with the Angle A oppofite to one of them, 43 20' ; iofind the third Side A C.
Letting fall, as before, the Perpendicular B E ; in the re5fangled Triangle ABE, from the given Angle and Hypo- themife AB, find the Side AE. Since, affuming BE for a lateral Part in the Triangle AEB, AB is the middle Part, and. AE the feparate Part ; and in the Triangle BE C, BC, is the mean Part, and E C a disjunct Part ; the Cofine of E C is found by fubtracting the Cofine of A B from the Sum of the Cofines of A E and C tf . If then the Segments A E and E C be added together ; or in cafe the Perpendicular fall without the Tri- angle, be fubtracled from each otherj the Side A C will be had.
6°. Given two Sides A C 65° 30' 46", tind A B 66° 45', together with the included Angle A 5 to find the third Side B C oppofite thereto.
Letting fall the Perpendicular B E, find, in the rectangled Triangle, the Segment A E ; which fubtracted from A C, leaves EC- If the Perpendicular fall without the Triangle? AG is to be fubtracted from AE; fince by atTuming the Perpendicular BE for a lateral Part in the Triangle & E B, A B becomes a middle Part, and A E a feparate Part : In the Triangle EBC, C B is the middle Part, E C a feparate Part : The Cofine of B C is found, by fubtracting the Cofine of A E from the Sum of the Cofines of A B and E C.
7 . Given two Angles A 43 20', and B 79 9' 59", toge- ther with the Side C B 39 29', oppofite to one of them ; to find the Side A B adjacent to both.
Letting fall the Perpendicular CD from the unknown An- gle C, to the oppofite Side A B 5 and that falling within the Triangle, from the given Angle B, and the Hypothenufe BC, feek in the rectangled Triangle BCD for the Segment B D. Since afTuming the Perpendicular C D for a lateral Part in the Triangle CD B, D B is the mean Part, and the Angle B a conjunct Part • and in the Triangle CD A, AD is the mid- dle Part, and the Angle A a conjunct Part : The Sine of the Segment AD is found, by fubtracting the Co-tangent of the Angle B from the Sum of the Sine of D B, and the Co-tan- gent of the Angle A. If rhenthc Segments A D and D B be added,or in cafe the Perpendicular falfwithout the Triangle, be fubtracted from each other, the Refult will be Side A B re- quired.
8°. Given two Sides A B 66° 45', and BC 39 i9> «&& the included Angle 79° 9' 59" ; to find the Angle A oppofite to one of them.
Letting foil the Perpendicular C D, find the Segment D B, as in the 'preceding Problem. This fubtracted from A B leaves AD. If the Perpendicular fill without the Triangle, A B is to be added to DB ; fince by afluming the Perpendicular CD for a lateral Part in the Triangle CDB, BD is the middle Part, and the Angle B a conjoint Part ; and in the Trian- gle CD A, AD is 'the middle" Parr, and the Angle A a con- joynt Part : The Co-tangent of the Angle A is found by fub- tracring the Sine of D B from the Sum of the Co-tangent of the Angle B, and of" the Sine of AD.
9 . Given two Angles A. 43° 20', and B 79 9' 59", to- gether with the adjacent Side A B 66° 45' 5 to find the Angle C oppofite to the fame
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