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Whence it follows that in every right-angled spherical triangle, radius is to the cosine of the hypothenuse, as the tangent of one of the oblique angles is to the colongent of the other angle.
APPLICATION OF THE PRECEDING PRINCI
PLES AND PROPORTIONS.
91. By means of the propositions which we have just demonstrated, we can very easily solve any spherical triangle, provided three of its parts are given. We shall commence with right-angled triangles.
And first, as the right-angle is always given, it is sufficient to know two of the other five parts, that make up these triangles.
Again the number of combinations of m quantities, taken two by two, is expressed generally by
m. (m—1) (Page 111). It is evi
2 dent, therefore, 1*, that in this point of view, right-angled spherical triangles present ten of these combinations. But, as for each combination we have three quantities to determine, it is clear 2dly, that all the possible varieties in the solution of right-angled spherical triangles are 30 in number.
The following Table contains them all, the angle A is supposed
E to be the right-angle; and the other two angles are denoted indifferently by B or C.
B 92. The construction of this
G Table is wholly founded upon two propositions already demonstrated. We shall place them once more F under the eye of the student.
Prop. 1. In every right-angled spherical triangle, radius is to the sine of the hypothenuse, as the sine of one of the oblique angles is to the sine of the side opposite to it (81).
Prop. 2. In every right-angled spherical triangle, radius is to the sine of one of the sides of the right-angle, as the tangent of the oblique angle opposite to the other side is to the tangent of that side (89)
Sometimes these proportions are applied immediately to the triangle ABC, and sometimes we must have recourse to one of the coma plemental triangles CDE, BFG, in order afterwards to transport the results to the triangle ABC, as will be shewn in several examples.
TABLE For the solution of all the possible cases of a spherical triangle ABC,
right-angled at A.
Tre part required is less than peo.
с R: cos BC::tang B : cot C If BC and B are of the same species AB R: sin BC::sin C : sin AB If C
B R: cos BC::tang C : cot B If BC and C are of the same species
B R; cos AC::sin C : cos B If AC < 90°
C cos AC : cos B::R: 'sin C Ambiguous
B sin BC: sin AC::R: sin B lif AC < 90°
B cos AB : cos C::R: sin B Ambiguous
R: cos B::cot AB : cot BC If AB and B are of the same species
с R: sin AC::cot AB : cot C If AB < 90°
BR : tang AB::cot BC : cos B. If BC and AB are of the same species
с sin BC : sin AB::R: sin C If AB < 90°
To facilitate the application of this Table, let the hypothenuse BC=81° 18', and the angle B=37° 19'; required the side AC, opposite to the angle B. To solve this case, we must employ the first proportion in the Table, and say
R: sin BC::sin B : sin AC
Therefore AC=36° 48', or 143° 12', which is its supplement. To decide which of these values is the right, we must call to mind that the side AC must be of the same species as the angle B oppo to it (74.) By way of reminding the calculator of the conclitions upon which depend the results which he seeks for, they are inserted in the last column of the Table.
93. Supposing still the same hypothenuse BC and the same angle B to be given, and that the adjacent side AB were required; it would be easy in the first place to find the side AC, as before, and afterwards to employ the proportion
tang B : tang AC ::R: sin AB. But this process introduces two proportions into a computation which may be managed by only one. For in the complemental triangle CDE we have
R: sin DE: : tang D : tang CE, and therefore transporting this proportion to the triangle ABC, it will become
R: cos B :: cot AB : cot BC, or perhaps better,
R: cos B :: tang BC: tang AB, But log radius taken negatively log cos B (37° 19)
9.900529 log tan BC (81° 13)
= 10.811042 log tang AB
= 10.711571 This last logarithm corresponds to 799 or 101°; but as BC and Bare of the same kind, we must take the first value: Consequently the side AB is 79°, or calculating as far as seconds, 79° 0 20".
N. B. When the logarithm of radius is to be subtracted, we shall simply write it 10; and in the aggregation of the three terms which form the propor. tion, the unit surmounted with the negative sign must be considered as negative and subtracted.
To find the angle C, we must have recourse to the triangle CDE, in which we have
sin CE : R:; tang DE : tang C, whence we obtain by substitution,
cos BC :R:: cot B : tang C
R: cos BC :: tang B:cot C. By logarithms, we obtain the angle C as follows: neg. log of radius
log cos BC (814 13') 9.183834
= 9.065935 which gives C=83° 22', and not 96° 38', since 'BC and B are of the same kind.
94. If instead of the hypothenuse being given, we had known the adjacent side AB, with the same angle B, and had desired the
opposite side AC, we must have employed the following proportion :
R: sin AB :: tang B : tang AC
9.874048 Therefore we conclude that the value of A36° 48', as we had before found it,
95. Let us now seek for the hypothenuse BC, supposing that we know the same quantities AB and B.
Neither the first nor the second of the proportions announced above (92) can be immediately applied to the triangle ABC; but in the complemental triangle CDE, we know D and DE, we have therefore R: sin DE :: tang D: tang CE, or by substitution,
R: cos B :: cot AB: cot BC.
9.288652 log cot EC
9.189181 Therefore BC is 81° 13', since B and AB are of the same species.
To obtain the value of the angle C upon the same suppositions, we must have recourse to the complemental triangle BFG, in which
R: sin BF:; sin B : sin FG, which by substitution gives
R: cos AB:: sin B: cos C. neg log radius
9.063139 Hence the angle C is 83° 22', or calculating as far as seconds, 83° 21'
33". This small difference arises from our supposing the side AB only 79°, whereas we have already found its value to be 790 0 20".
96. After having in two different cases determined the three unknown parts of the right-angled triangle ABC, it remains for us to examine the case in which this determination is impossible.
And first, if we suppose the side AB and the opposite angle C to be known, it is evident that we cannot know of what species the hypothenuse is; for then the first proportion gives
sin C :sin AB :: R: sin BC But this last sine belongs indifferently to an hypothenuse less than 90°, or to its supplement, and to determine us in our choice we ought to know of what species the two oblique angles are.
For want of this knowledge the case is doubtful; and therefore it is marked as such in the fourth column of the table.
We fall into the same dilemma, if we desire to find the side AC, the same things being given. For, by the second proportion (92) we have
tang C :tang AB ::R: sin AC, but nothing in this case determines of what species the side AC is.
The same difficulty respects the angle B, of which the value is deduced from the first 'proportion (92) applied first to the triangle BFG, and afterwards transported to the triangle ABC. For in the triangle BFG, we have
sin BF:R:: sin FG : sin B, which in the triangle ABC gives
cos AB: R:: cos C :sin B, from which we cannot ascertain of what species is the angle B.
This is the first case of which the varieties offer three ambiguous solutions. Another case perfectly similar, is that in which we suppose the side AC, and the opposite angle B to be known. This case also presents three other ambiguous solutions. Generally, when in a right-angled spherical triangle we only know one of the oblique angles, and the side opposite to it, the value of the three other parts cannot be determined.
97. Hence there are six imperfect solutions among the thirty problems which any right-angled spherical triangle offers for solution: We might reduce these thirty to sixteen questions, by suppressing those which are absolutely similar. But it is desirable to find in a table, the proportion for which we have occasion, without being under the necessity of making any change.
EXAMPLES OF RIGHT-ANGLED SPHERICAL
1. In the right-angled spherical triangle
C ABC, given the hypothenuse BC=640 40, and the base AB=42° 10', required the remaining side and the angles ? Answer, C=47° 57' 47", B=64° 36'40', AC=54° 44' 23".
B 2. In the right-angled spherical triangle ABC, given the hypothenuse BC=50°, and the base AB=44° 20; required the rest ?
Answer, C=65° 49' 10", B=34° 56' 8", AC-26° 1' 10". 3. In the right-angled spherical triangle ABC, given the hypothenuse BC=63° 56', and the angle C=45° 41; required the other sides and angles.
Answer, AC=55° 0' 1", AB=39° 59' 39", B=640 46' 15". 4. In the right-angled spherical triangle ABC, given the side AC=55°, and the opposite angle B=65° 46'; required the remaining parts?
Answer, C=45° 41' 33", AB=40° 0 12", BC-680 56 12