affumed, as A, to D, and draw the right Circles DCI and ACH; at the Point D, by Problem 48, draw an oblique Circle, DB I, fo that it may make an Angle, ID A, of 59.50 with the primitive Circle; take 65°. 34 from the Line of Chords, and fet from A, both Ways, to b and c; by Prob. 42, Cafe 2, lay 65°.34 from A to a, on the right Circle ACH; thro' the three Points b, a, and c, by Prob. 17, draw the parallel Circle ba c, interfecting the oblique Circle DI in B; laftly, by Prob. 17, draw an oblique Circle thro' the Points H, B, and A, and the Triangle ABD is the Triangle required. But here it is to be obferved, that the parallel Circle bac cuts the oblique Circle DI in two Places, viz. in B and s, whence the Cafe is ambiguous; for it can't be determined whether the Triangle ABD or the Triangle As D be the true Solution, fince the Side As is 65°. 34 as well as the Side A B, as may be found by measuring them, as taught in Prob. 46; and this is always the Cafe when the Side oppofite the given Angle is lefs than the other given Side. In the oblique-angled Triangle A B D, AB oppofite the given Angle at D, and the the required Angle at B; therefore, As the Sine of the Side A B, 65°.34, we have the Side Side A D oppofite Co. Ar. 0.040747 59.50, 9.936799 9.983558 To the Sine of its oppofite Angle ABD, 66°.7', 9.961104 The Angle at B, being made by the Interfection of the two oblique Circles HBA and DBI, may be measured by Prob. 56. Prob. 82. In the oblique-angled Spherical Triangle ABD, given the Side A D 45°. 36', the Side A B 48°. 50′, and the included Angle at A 54°. 19', to find the Side D B. Conftruction. With the Chord of 60°, draw the primitive Circle ADEF; from any Point affumed, as A, fet off I 45°.36′ 45°.36′ to D, and draw DF and AE; by Prob. 48, draw an oblique Circle, E B A, to make an Angle of 54°. 19 with the primitive Circle, in the Point A; and, by Prob. 45, lay 48°.50′ on this oblique Circle, from A to B; laftly, thro' the Points D, B, and F, by Prob. 17, draw the oblique Circle D BF, and the Triangle ABD will be the Triangle required. In order to folve this Problem, and most of the following Cafes, it is neceffary to let fall a Perpendicular, which must always fall from the End of a given Side oppofite to a given Angle. The Side on which the Perpendicular falls is ufually called the Bafe: And, if the Angles at the Base are, either both more, or both lefs, than 90°, the Perpendicular will fall within the Triangle, and divide it into two right-angled Triangles But, if one of the Angles of the Bafe is less than 90°, and the other greater, the Perpendicular will fall without the Triangle, and, by that Means, add a right-angled Triangle to the given oblique Triangle: In both Calcs, the Parts may be found by the Rules laid down in the feveral Cafes of right-angled fpherical Triangles, from Prob. 64 to Prob. 79, inclufive. In the present Cafe, a Perpendicular, being let fall from D on the Bafe A B, will divide the Triangle ABD into two right-angled Triangles, ADG and BDG, both right-angled at G. To do this, find p, the Pole of the oblique Circle EBA; then, by Prob. 17, draw the oblique Circle Dp F, which falls from D, the End of the given Side AD, oppofite the given Angle at A, and is perpendicular to EBA, the Circle on which it is to fall, because it paffes thro' its Pole p. Now, in the right-angled Triangle AD G, we have the Side A D 45° 36′, and the Angle at A 54°. 19', to find the Perpendicular DG; and, the Angie A D G lying between the given Side AD and the required Side DG, it is an Extreme disjunct, and D G is the middle Part; whence, by the third Rule, and its Note, As To find the Side A G in the fame Triangle ADG; the fame Things being given, it is an Extreme conjunt, and the Side AG is one of the Extremes; whence, by the fecond Rule, and its Note, Then fubtract A G 30°. 13' from A B 48°. 50, and the Remainder is B G 18°37′: Then, to find the Hypothenufe D B, in the right-angled Triangle BDG, we have the Perpendicular D G 34°. 38', found by the first Proportion, and the Side B G, by the laft: Now, the Angle at B lying between B G and B D, it is an Extreme disjunct, and the Hypothenufe BD, the middle Part, is required; therefore, by the third Rule, and its Note, The Side D B, being Part of the oblique Circle DB F, may be measured by Prob. 46. A Perpendicular might have been let fall from the angular Point to the Side A D, by drawing a right Circle thro' the Pole of the primitive Circle, as the Side AD, on which it would fall, is a Part of the primitive Circle: By this Means, the Side D B might have been found in the fame Manner. Prob. Prob. 83. In the oblique-angled Spherical Triangle A EF, given the Side A E 72°34, the Side A F 53°. 12, and the included Angle at A 47°. 32', to find the Angle at E. There being the fame Things given as in the laft Problem, the Conftruction is the fame. If a Perpendicular is let fall from the angular Point at F, on the Side AE, which is done by drawing a right Circle thro' C, the Pole of the primitive Circle, and F, then will the oblique Triangle EFA be divided into two right-angled Triangles, AFH and E FH, both right-angled at H: In the Triangle AFH there is given the Hypothenufe AF 53°.12', and the Angle at A 47°. 32, to find the Perpendicular F H. The Angle AFH lying between the given Part A F and the required Part FH, it is an Extreme disjunct, and the Perpendicular F H is the middle Part; therefore, by the third Rule, and its Note, As the Radius Is to the Sine of AF, 53°. 12. 10. 9.903487 9.867862 To the Sine of the middle Part FH, 36°. 12', 9.771349 To find the Side A H, the fame Parts being ftill given as in the laft Calculation, it is now an Extreme conjunct, and A H, one of the Extremes, is required; by the fecond Rule, and its Note, As the Co-tangent of the Hypothenufe AF, Ca. Ar. 0.126043 53°. 12', Is to the Radius, 10. So is the Co-fine of the middle Part FAH, 47°. 32, 9.829407 To the Tangent of A H, 42°. 4', 9.955450 Which being fubtracted from A E 72°. 34', there remains HE 30°. 30'; therefore, in the right-angled Triangle E F H, we have given the Side F H 36°. 12', and the Side HE 30°. 30, to find the Angle at E; which is an Extreme conjunt, and the Angle at E, being one of the Extremes, is found by the second Rule, and its Note: As the Tangent of FH, 36°. 12', Co. Ar. o.135555 So is the Sine of the middle Part H E, 30°. 30 ́, To the Co-tangent of the Angle at E, 55°. 15', 10. 9.70469 9.841024 The Angle at E, being made at the Periphery of the primitive Circle by the oblique Circle D F E, may be measured by Prob. 54. Prob. 84. In the oblique Spherical Triangle RST, given the Angle at R 82°. 19, the Angle at T 59°. 10, and the Side TS, oppofite one of the given Angles, 69°. 18′, to find the third Angle at S. The fame Parts being given here as in Prob. 80, the Conftruction will be the fame. And, if a Perpendicular is let fall from either End of the given Side TS, as S, on the oppofite Side TR, which Perpendicular will be a right Circle, drawn thro' the Pole of the primitive Circle and the Angle at S, the oblique Triangle RST will be divided into two right-angled Triangles, TS a and RS a, both right-angled at a: In the Triangle TSa, there is given the Side TS, and the Angle at Ꮩ T S a R T, to find the Side Sa; therefore it is an Extreme disjunct, and the Side Sa, which is the middle Part, is found by the third Rule, and its Note: To the Sine of the middle Part Sa, 53°.26′, 9.904840 |