The Projection. Draw the Primitive BIFH, and the Oblique Cir cle BDF, to make the Angle at B = 36° 08'; then draw the Paral H lel ADE, to cut off the Arch BD = 42° og'; then through the Points C, D, G, draw the Oblique Circle - CDĠ ; so will the Triangle be projected, as was required. K BCD 1. To find the Side D C. This is found as in Cafe 2, and is equal to Cf= 2. To find the Angle D. This is also done as in Cafe 2 ; for it is equal to cd = 46° 18'. 3. To find the Angle B C D. Measure K L on a Line of Half-tangents from go downwards, and you will find it to be 769 00', whose Complement 1049 is the Angle required. Cafe Cafe 5. B 420 E K 3000 12404 a BD = 42 09, BC 30 00, CD = 24 04: The Proje&tion. Describe the Primi tive BIFH, then make the. Arch ! BD="429 og'; draw the Parallel Q ACE, to cut off the Arch H R T ; also draw the Paral lel OCQ, to cut off the Arch D C = 24° 54'; then thro' the Point of Intersection C, draw the two Oblique Circles BCF, D GG, lo shall the Triangle BDC, be made or projected, as required. 1. To measure the Angle at D. Measure K L on the Hals-tangents, and 'twill be found 469 18'. 2. To measure the Angle B. Measure RI in the same Manner, and it will be found 369 08'. 3. To measure the Angle C. This is done as taught in Cafe 1, and is found to be 1049 Cafe 36 os, B 36 oec Case 6. B С 104 oo. This Case is to be converted into the last foregoing Case, by changing those Angles into Sides, by Theorem 13, taking for the greater Side 760, the Complement of the Angle C = 1049 to 2 Right-angles. As in this adjacent Scheme, the Triangle BCD has its Sides, BE, CD, equal to the given Angles B, and D; and the Side BD, the Complement of the Obtuse Angle C to 180°. Wherefore if the Angles of this Triangle be measured, they will be found as follows ; viz. B=bc= 300 oo'; D = dc = 249 04'; and the Complement of BCD = b d = 429 09'. So that the sides fought in this case are those which constitute the Triangle b dc, and are to be measured as before taught. And thus much for the Projection of Oblique Spherical Triangles. The Learner may vary the Method of Projection, and Form of the Triangle, sometimes if he pleases ; but this is here omitted, and left for his Exercise. 7600 4618 С НАР, CH A P. XXI. . of the Menfuration of the Area of a Sphe tical Triangle, and the Completion of LEMMA 1. T HE Lunary Superficies of Hemispherics, is Demonstration. wa 2 Those |