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BY RULE II. (N. 228.) Make CD=AC, then CAD, CDA, and ADB are each of them a right angle, and ad measures the angle c(L. 227, and M. 194.); hence, if we take 90° from Bc we get 239.18'=BD; and ADB is 'a right-angled triangle. 1. To find the LB, by Case IV. right-angled spherics.
rad : cot AB::tang BD ; cos , B=78°.20ʻ.
But LB is obtuse (0. 211), and therefore=1010.40%. 2. To find av, by Case V I. right-angled spherics.
Cos BD : rad::cos AB : cos ad=62o.26'.
But Ad is obtuse (0.211), and therefore=1170.34'. 3. To find the , DAB, by Case V. right-angled spherics.
Sine AB : rad::sine BD : sine 2 DAB=250.55'.
BY RULE 1. (M. 227.)
1. To find the hypoth. ac, by Case XVI. right-angled spheries, Rad:cot Za=cot BC::cot Lc=cot AB:cosae=sup Z B=789.20'. Here ac or suppt. Z B is acute (S. 212.); therefore B is obtuse =1019.40'.
2. To find ab, by Case XV. right-angled spherics. Sine L A=sin bc:rad::cos L.C=COS AB:cosab=cos Lc=62°26'. Here ab= Lc is obtuse (S. 212.), and therefore=1170.34'.
3. To find be, by Case XV. right-angled spherics. Sine Lc=sin AB:rad::cos Larcos BC:cos bc=cos LA=6495'. Here besla is obtuse (S. 212.), and therefore=1150.55'.
If the two preceding cases be thoroughly understood, there can be no difficulty in solving those which follow; for which reason, they are given as practical exercises.
(Q) CASE III. Given a quadrantal side, and its two adjacent angles, to find the rest.
The quadrantal side ac=90° Given
43540.43 Required AB, BC,
and the angle B. Answer. AB=48°, Bc=640.35', and the angle B=1150.20'.
(R) CASE IV. Given a quadrantal side, one of the other sides, and the angle comprehended between them, to find the rest.
The quadrantal side aç= 90 Required BC, Given The side
and the anThe angle CAB
gles B and c. Answer. bc=1130.18', the angle B-101°.40', c=1170.34'.
(S) CASE V. Given a quadrantal side, its adjacent angle, and a side opposite to that angle, to find the rest.
The quadrantal side ac=90° Required BC, Given Its adjacent angle
and the anAnd the opposite side AB=48°.00'
gles A and B. Answer. Bc=649.35', the angle a=540.43'. and B=1150.20'.
But the required parts are ambiguous, and therefore are either acute or obtuse.
(T) CASE VI. Given the quadrantal side, one of the other sides and an angle opposite to the quadrantal side, to find the rest.
The quadrantal side ac= 90° Required BC, Given The side AB
and the anThe angle B
gles A and c. Answer. BC = 113.18', the angle A = 1150.55', and c= 1170.34'.
III. PRACTICAL RULES FOR SOLVING ALL THE CASES OF
OBLIQUE-ANGLED SPHERICAL TRIANGLES, WITH A PERPENDICULAR; AND THEIR APPLICATION BY LOGARITHMS.
(U) The rule Prop. xxv. (X. 173.) will solve ten cases; see also the rules P. 178.
(W) When the three sides are given, to find the angles. Any one of the three sides may be called the base. Then,
The tangent of half the base,
As the tangent of half the difference between the sides,
base, the perpendicular falls within or without the triangle. When the sum of the two sides is less than 180°, the
perpendicular falls nearest to the less side, when greater than 180° it falls nearest to the greater side; consequently the greater segment is joined to the greater side in the former case, and to the less side in the latter. The sum of half the base, and the fourth term found by the above proportion, gives the greater segment, their difference gives the less.
The triangle being thus divided into two right-angled triangles, the remaining parts must be found by the proper rules.
(X) When the three angles are given, to find the sides.
The co-tangent of half the sum of the angles at the base,
As the tangent of half the verticle angle, Is to the tangent of the excess of the greater of the two vertical angles (formed by a perpendicular), above half the aforesaid vertical angle. (W. 180.)
If the sum of the base angles be less than 180°, the perpendicular and the less segment are nearest the greater base angle, if
greater than 180° they are nearest the less base angle. The sum and difference of this fourth term, and half the vertical angle, gives the greater and less vertical angle formed by the perpendicular. The triangle being thus divided into two rightangled triangles, the remaining parts must be found by the preceding rules.
(Y) CASE I. Given two sides and an angle opposite to one of them, to find an angle opposite to the other.
The side ac=80°.19'
DETERMINATION OF THE SPECIES. A perpendicular in this case
.:D is unnecessary.
If ac + BC, A+(B acute), and A+(B obtuse), be each of the same species with respect to 180°, Á B is ambiguous: - But if only two of these sums be of the same species, that value of B must universally be taken which agrees with the sum of the sides, in all such cases B is not ambiguous.
BY FORMULAT 11. page 200.
(Z) Case II. Given two sides and an angle opposite to one of them, to find the angle contained between these sides.
The side AC-80°.19'
DETERMINATION OF THE SPECIES. 1. If ac and the L A be of
d..... C the same species, the _ ACD is
D acute. The perpendicular CD is of the same species as the ZA. If BC and dc be of the
B same species, the LBCD is acute.
2. If the / Bcd be less than the LACD, and their sum less than 180°, then the ACB is ambiguous; but if their sumn be not less than 180°, their difference is the true value of the L ACB, not ambiguous.
If the BCD be not less than the L Act, and at the same time their sum be less than 180°, this sum is the true value of the L ACB, not ambiguous.
1. In the triangle Adc, find the 2 ACD.
Thus, rad x COS AC=cot LAX cot Z AÇD.
Hence, cot LA ; rad::cos AC : cot ACD=78°.4' acute. 2. In the triangle cdB, rad x cos BCD=tang DC X Cot BC.
In the triangle Adc, rad x cos acd=tang DC x cot ac. Hence, cot Ac : cot bc::COS ACD : COS BCD= - 53o.28' acute.
Because the _ BCD is less than the L ACD, and / BCD + LACD less than 180°, the L ACB is ambiguous ; viz. L ACB= LACD+ BCD=1310.32', or _ ACB= LACD – LBCD=249.36'.
+ In using the Formulæ, the three angles of the triangle are represented by A, D, C, and their opposite sides by a, b, c. The perpendicular is not regarded.
BY FORMULA III. page 2013 Log tan p=(log cos b+log tang A)-10=11°.56'. acute. Log sine o+c=(log tang b+log sine p) - log tang a=360.32'. Then (P+c) -Q=360.32 – 11°.56'=249.36'= LC.
This example is ambiguous, though not shewn by the formula, vide Table I. page 207.
OR, Log cot p=(log cos 6 + log tang A)-10=78°.4' acute. Log cos ($+c)=(log cot a +log cos %) - log cot b=530.28'. Then (P+c)+p=539.28'+780.4'=131°.32'= Lc.
and L ACB=61°.40', not ambiguous. (A) CASE III. Given two sides and an angle opposite to one of them to find the other side.
The side AC-809.19
DETERMINATION OF THE SPECIES. 1. The segment ad is acute or obtuse, according as the LA is of the same, or of different species with ac. DC is of the same species with the LA.
DB is acute or obtuse, according as bc is of the same, or of different species, with dc.
2. If db be less than ad, and their sum less than 180°, then AB is ambiguous; but if their sum be not less than 180°, their difference is the true value of ab, not ambiguous.
If be not less than Ad, and at the same time their sum be less than 180°, this sum is the true value of AB, not ambiguous.
SOLUTION. 1. In the triangle Adc, find the side AD.
Thus, rad * cos LA = cot ac X tang AD.
Cot ac : rad::cos LA : tang Ad=749.40' acute. 2. In the triangle cDB, rad x cos BC=COS DC X COS DB.
In the triangle ADC, rad cos ac=cos DC X COS AD.