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First, (90) R: cos AC :: tang A : cot ACD
And then, (86) cos BCD : cos ACD :: tang AC: tang BC.

VI.

104. Suppose any two sides given, AC and BC for example, with one of the angles opposite to those sides, such as the angle A ; to find the angle B, opposite to the other side.

This problem may be resolved by the B well known proportion

sin BC: sin AC ::sin A : sin B If we assign to the given sides and angle A, the same values as in the preceding problems, we shall find the angle B by the follow

B
ing calculation :
arith; co. log sin BC (59° 40) = 10.063938
;

log sin AC (779 5') 9.988869
log sin A (61° 25')

9.943555
log sin B

9.996362 The angle B is therefore 82° 36', as we have before supposed it, or 97° 24', as we might also have supposed it.

This problem falls within the first case; they differ only by an inversion of the terms in the proportion which resolves both.

=

=

VII.

105. The same things being given, required the third side AB.

(87) R:cos A :: tang AC: tang AD

(83) cos AC : cos BC :: cos AD: cos BD. By the first of

C these two proportions we find AD, and by the second BD: therefore we A

A shall have AB=

B

B AD + BD, according to the position of the perpendicular arc.

If we had referred the known quantities in the triangle ABC to the supplemental triangle, we should merely have to resolve this problem (99). To find the third angle when we know the other two angles, and one of the sides opposite to them. For this problem once resolved, we should find the value of the required

E side AB, by taking the supplement of this third angle.

VIII.

we

106. On the same suppositions, to find the angle C, contained between the given sides AC and BC..

(90) Riersic :: tang A, OU ACD, and, (88' ting BC : tang AČ: ACD: BCD.

Whenc deduce / ('B= ACD+BCD, according as the sides AC and BC are of the same A

AS or of different

B
D

B species.

D The supplemental triangle is equally proper to resolve this problem by reducing it to the third case.

IX.

107. Given the two sides AC and AB, and the angle A, contained between them, to find the other side BC. (Preceding figure).

(87) R:cos A :: tang AC : tang AD
(83) cos AD: cos BD :: cos AC : cos BC.

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X.

108. The same things being given, to find one of the other two angles, the angle B, for example. (Preceding figure.)

(87) R: cos A :: tang AC : tang AD
(89) sin BD : sin AĎ:: tang A : tang B.

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XI.

109. If the three sides were given, how should we find one of the three angles, A for example? (Preceding figure). We know (83) that

A+B AC+BC AC-BC AD_DB lang : tang

::tang

: tang
2
2
2

2 This first proportion will give us the segment AD, and the following one will give the angle A.

(87) tang AC : tang AD::R: cos A. Resuming the values we have before supposed, we shall have

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log tang

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arith. co. log tang AB (38° 254)

10.100692
AC+BC
log tang

(68° 22')) = 10.401838
2
AC-BC
(8° 42')

9.185174
2

AD-io log tang

OC 37701

2 The half difference of the two segments is 25° 59', w. th being added to the half base 38° 25', gives 64° 24' for the greater seg. ment AD. This premised, we have then arith. co. log tang AC (77° 5) = 189.360474

log tang AD (64° 24') 10.319556
log R

10.
log cos A

9.680030 This logarithm corresponds to 61° 24'. Consequently the angle A=610 24. Yet we have elsewhere supposed it to be (99) 610 25', whence arises this small difference? It proceeds from the quantities neglected in the valuation of the logarithms; for if we had continued the calculation as far as seconds, we should have found the half difference of the segments 25° 58' 30", which added to 58° 25', would have given 640 23' 30", for the value of AD. But the logarithm of tang 64° 23' 30", is 10.319394. This logarithm written instead of 10.319556; gives for the result 9.679868, which answers to 610 25'.

110. This last case may be resolved by one proportion, which may be investigated in the following manner, and which will serve to exemplify the use of the trigonometrical formulas as given in the foregoing pages. In the oblique-angled spherical

А.
triangle ABC, call the side BC a;
AC, b; and the side AB, c.
Imagine O to be the center of

B the sphere, and from the point A,

P draw AP, AQ, tangents to the arcs AB, AC, Then by Euclid, def. 6, Book XI, the angle QAP is equal to the spherical angle A, and the angle at O is measured by the arc BC. Draw OQ and OP, then will OA be the secant of the are AC, and OP that of the arc AB. By (43) we have in general co: A

AQ? + APP-PQ?

or,

2 APX AQ
cos a_tang* 6+ tang* c—PQ” whence
A=b*

2 tang 6 tang c
PQ+=tang 6+tang - 2 tang 6 tang c cos A.

? +? For similar reasons,

PQ+=sec 6+sec ~2 sec 6 sec c cos a. and subtracting the preceding equation from this latter one, we have 0=1+1+2 tang 5 tang c cos A-2 sec b sec c cos a, because sec-tang=R2=1). Whence by transposition,

?

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2

C

D D

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Again, since cos 6 tang b=sin b, the expression becomes

cos acos 6 cos c cOS A sin b sin c

cos a−(cos b cos c—sin b sin c) Consequently I +cos A:

sin b sin c cos acos (b+c)

by (11) sin 6 cos c

P+Q P-Q But in general (17) we have cos Q- cos P=2 sin

2

2 2 sin (a+b+) x sin b+cma) 2

2 and therefore 1 +cos 'A=

sin b sin c 2 sin S. sin (S-a)

sin b sin c

x sind

by taking S=2+b+c

cos Therefore

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2

2A But we have shewn that 1 +cos A=2 cosa A sin S. sin (S-a)

and
2 sin b sin c
sin S. sin (S-a)

if radius=1
sin b sin c
А R?. sin S. sin (S-a)

if radius=R
sin b sin c

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or, cos

Example. In the oblique-angled spherical triangle ABC, (preceding figure) given AB=76° 50'; AC=77° 5'; and BC=59° 40'. Required the angle A. By the formula we have A R’. sin S. sin (S—BC)

But 2

bin AB sin AC

COS

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log R

760 50

= 20. 77° 5°

log sin S (106° 47') = 9.981095

'
59° 40'

log sin (S-BC) 9.864950
2|213o
35

co. log sin AB (76° 50') = 10.011570
106° 47'-S co. log sin AC (77° 5') = 10.011131
59° 40

2|19.868746 47° 7' =S--BC

A log cos

9.934373

2 A Whence -= 30° 42', and A=61° 24, the same as before.

2

XII.

:

111. Given the three angles, to find a side, AC for example. This case resolved by the supplemental triangle would not differ from the preceding problem. But it may also be resolved by the two following proportions, if we

AS suppose that the arc CE bisects the an

E

B gle ACB. For we have first A+B

B-A cot

: tang :: tang $C : tang DCB 2

2 and then,

tang A : cot ACD:;R: cos AC. The first proportion gives the value of the angle DCE, which being added to the angle ACE, will shew the angle ACD. And this once known, we directly obtain the required side AC, by the second proportion. Let then A=610 25', B=82° 36', C=82° 4', we have

A+
.

=
B-A
log tang (10° 35' 30") = 9.271829

"
2
log tang IC (41° 2')

9.939673
log tang DCE

9.699941 Therefore the angle formed by the perpendicular arc, and by that which bisects the angle C, must in this case be 26° 36' 59". Adding this value to the half of the angle C, we shall have 67° 38' 59", for the value of the angle ACD; and the second proportion gives

arith. co. log cot 3-4B (72° 0 80") = 70.488439

$

2

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