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Solution of the Cases of Right Angled Triangles.
· GENERAL PROPOSITION. In a right angled triangle, of the three sides and three angles, any two being given besides the right angle, the other three may be found; except whien the two acute angles are given; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them.
It is manifest from 47. 1. that of the two sides and hypothenuse, if any two be given, the third may also be found. It is also manifest from 32. 1. that if one of the acute angles of a right angled triangle be given, the other is also given for it is the complement of the former to a right angle.
If two angles of any triangle be given, the third is also given, being the supplement of the two given angles to two
right angles. Fig. 15. The other cases may be resolved by help of the preceding
propositions, as in the following table.
Given. w. Sought. i Two sides, AB, The angles AB:AČ:: R:T, B, off AC.
B, C. which C is the complement.
2 AB, BC, a sidel The angles BC: BA::R;S, C, of and the hypo-B, C., which B is the complement. thenuse.
8 AB, B, a side The other
R:T,B:: BÁ: AC.
s, C:R::BA: BC.
4 AB and B, a The hypo-
6 BC and B, the The sidel R:S; B :: BC:CA.
These five cases are resolved by Prop: 1...
Solution of the Cases of Oblique Angled Triangles.
GENERAL PROPOSITION. In an oblique angled triangle, of the three sides and three angles, any three being given, the other three may be found, except when the three angles are given; in which case the ratios of the sides are only giveit, being the same with the ratios of the sines of the angles opposite to them.
Given. ' Sought.
1 A,B, and there- BC, AC. | S, C:S, A :: AB : BC. Fig. 16. 17. forę C, and the
and also S, C: S, B :: AB side AB.
1: AC. (2.)
2 AB, AC, and The angles AC:AB :: S, B: S, C.
B, two sides and A and C. . 12.) This case admits of two lan angle oppo
solutions; for C may be site to one of
Igreater or less than a quathem.; .
drant. (Cor. to def. 4.) 1
BAB, AC, and The angles AB+AC: AB-AC::T,
A, two sides, B and C. C+B C and the in- en
-:T, 2 (3). it cluded angle.
the sum and difference off
I :T, I . (4.) | therefore B and C are given as before. (7.)
2 AC x CB : ACq+CBq/ -ABq :: R: Cos, c. 1 ABq+CBq be greater than ABq. Fig. 16.
2 AC RCB: ABq-ACT
-CBq :: R: CoS, C. 4 AB, BC, CA, A,B, C, the ABq be greater than ACq+ the three sides. three angles. CBq. Fic. 17. (4.)
L . Otherwise,
Let AB+BC+AC=2P. P R P - AB :P – AC x| P-BC :: Rq: Tg, c and hence C is known. (5.)
Otherwise, Let AD be perpendicular to BC. 1. If ABQ be less than ACq+CBq. Fig. 16. BC : BA+AC : : BAAC : BD-DC, and BQ the sum of BD, DC is given; therefore each of them is given. (7.)
2. If ABq be greater than JACq+CBq. Fig. 17. BC: JBA +AC:: BA-AC:BD
+DC; and BC the differjence of BD, DC, is given, therefore each of them is given. (7.)
And CA : CD ::R:Co S, C. (1.) and C being found, A and B are found by case 2 or 3.