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In any plane triangle, the sum of any two sides is to their difference :
as the tangent of half the sum of the angles at the base, is to the tangent of half their difference. Fig 54. plate 4.
Dem. LET ABC be a plane triangle, AB+BC:AB_BC :: tan. ang. A+Ang. C:tan. ang. C-ang. A, upon A as cena
tre with AB, the longest side for a radius, describe a circle meeting AC produced in Eand F; produce BC to D, join DA, FB, EB, and draw FG parallel to BC, meeting EB in G.
The angle EAB is equal to the sum of the angles at the base and the angle EAB at the centre is double the angle EFB at the circumference, therefore, EFB is half the sum of the angles at the base; but the angle ACB, is equal to the angles CAD, and ADC, or ABC together, therefore, FAD is the difference of the angles at the base, and FBD is half that difference, but FBD is equal to the alternate angle BFG; since the angle FBE in a semi-circle, is a right angle, FB being radius, BE, BG will be tangents of the angles EFB, BFG; but it is plain, that EC is the sum of the sides, BA and AC, also that CF is theirdifference; and since EG and BC are parallel, EC :CF as EB : BG, that is, the sum of the sides is to their difference: as the tangent of half the sum of the angles at the base, is to the tan gent of half their difference.
In a plane triangle, the base is to the sum of the sides, as the
difference of the sides, is to the difference of the segments of the base, made by the perpendicular upon it from the vertex. Figo 55. plate 4.
Let ABC be a plane triangle, if from B the vertex a perpendicular BD be dropped on the base, AC: AB+BC :: BC-AB: DC-AD. Upon B as centre with BC, the greater fide for a radius, describe a circle meeting BA and CA, produced in F and E. It is manifeft, that AF is the difference of the fides, and that EA is the difference of the segments of the base, for ED and DC are equal, and AG is the sum of AB and BC; but, because FG and EC cut each other within a circle in the point A, the rectangle contained by the segments of the one, is equal to the rectangle contained by the segments of the othera that is, FAXAG=EAXAC, and by Euclid vi. 16. AC: AG::FA:EA.
Wherefore, in any plane triangle, the base is to, &c. Note, 'The fum and difference of two magnitudes being given,
to find each of them.
Rule, To half the fum, add half the difference, the sum will be the greater, and from half the fum, subtract half the difference, the remainder will be the lefs.
In plane triangles may be given,
The three angles and one side.
The angles and one side given, to find the two remaining fides.
Plate 3. fig. 47.
Ang. C 52°157 Req. AC and BC Ex, 1. Given Ang. A 59°
AB 276.50 5180—52°15+59°=68°43'=an.B.
To find AC.
To find BC.
As fine ang. C52° 15' 9.89801 | As fine ang. C 52° 15' 9.89801
is to A B 276.5 - 2.44170 is to AB 276.5 - 2.44170 So is fine an. B68° 45' 9.96942 So is fine ang. A 59° 9.93307
TO AC 325.9
EXAMPLE 2, plate 4. fig. 56.
Req. AC BC.
Ang. C 41° 15'J180--47°30+41°15'=91°15'ang.B.
The fine, tangent, secant, &c. of any arch, is the fine, tangent, fecant, &c. of its fupplement. Hence the line of 91° 15' may be obtained thus, 180°-21° 55'=88° 45' =the fupplement of g1° 15'.
I wo sides and the angle opposite to one of them being given, to find
the other angles and the third fide. Fig. 48. plate 3.
39 42 ang. B 91°
Req. ang. A; ang. C and BC.
180°-ang. C+ang. B.=ang. A 47° 30'. To find angle C.
To find BC. 1.59572 | As fine ang. C41° 15' 9.81911 is to sine ang. B 91°15' 9.99990
is to AB 26
1.41497 So is AB 26 1.41497 Sois sine ang. A 47°30' 9.86763
As AC 39.42
11.41487 To fine C 41° 15" 9.81915 | To BC 29.07
Tavo fides and the angle contained being given, to find the remaining
angles, and the third side. Fig. 49. plate 3.
АС Ex. Given BC
Req. ang. A, ang. B. and AB.
Ta To find the angles.