The side 10 = 900:30:20"} nuse AC. { AC. side BC. BY CONSTRUCTION. PRACTICAL EXAMPLES. The side ab=50°.30':30" To find the hypotheAnswer. Ac=610.4.56", acute. (R. 212.) 2. In the right-angled spherical triangle ABC. The side AB= 54°22' To find the hypothenuse Given The side bc=1429.31') The angle A=370.25' Required the BY NAPIER'S RULE. E D Rad X cos a = cos BC x sine c. ::: (10+log cos A) -log sine c=log cos BC. I or, sine 2c =810.12 9.99486 G :: cos LA = 370.25' 9.89995 : cos BC = 369.31' 9.90509 Here Bc is acute. (S. 212.) B H Н BY RULE I. Sine id : sine FG::sine cd : sine cr. But ip is the measure of the angle c, Fg the complement of hg which is the measure of the angle A, cd radius, and cf the complement of Bc. Hence, sine c : cos A:: rad : cos BC. BY CONSTRUCTION. (Plate V. Fig. 12.) 1. With the chord of 60° describe the primitive circle, through the centre p draw cpe, and rnm. at right angles to it. 2. Set one foot of your compasses on 90° in the line of semitangents, extend the other towards the left hand, till the degrees contained between the points of the compasses be equal to (81°12') the Lc, and apply this extent from m to n. 3. Through the three points cne draw a great circle, and find its pole p. (N. 159.) C an arc; with the secant of the LA, and centre p crosses it in o; with o as a centre and radius op, describe the great circle ABP. 5. Then ABC is the triangle required, right-angled at B. To measure the required parts. 6. The hypothenuse ac, measured by a scale of chords, =789.20. 7. The sides Bc and ac (C. 169.) = 36o.31', and 75°.25'. PRACTICAL EXAMPLES. B { Required the 1. In the right-angled spherical triangle ABC. Required the side AB 2. In the right-angled spherical triangle ABC. Given {The angle = 31.517 side BC. Example. In the right-angled spherical triangle ABC. The angle a=370.25' To find the hypothe BY NAPIER'S RULE. Rad x cos ac=cot A X cot c. JE 10.00000 : cot LA =370.25' - 10:11633 ::cot LC =819.12 9:18979 : cos AC =780.20 - 9:30612 Here ac is acute. (S. 212.) niise AC. H н BY RULE I. Tang ID : tang Fr::sine ci ; sine cg. But id is the measure of the angle c, FG the complement of the angle A, ci radius, and cg the complement of Ac. Hence, tang 2c: cot 2 ::rad : cos AC. BY CONSTRUCTION. See the construction of CASE XV. (I. 225.) PRACTICAL EXAMPLES, 1. In the right-angled spherical triangle aBc. Given { To find the hypoS The angle A = 47°.54'.20" | The angle c=61°.50'.29"} thenuse AC. Answer. Ac=61°.4'.56", acute. (S. 212.) 2. In the right-angled spherical triangle ABC. Required the hypoThe angle c=1049.8') ibenusé AC. Answer. AC=1130.55', obtuse. (S. 212.) CHAP. VII. II. PRACTICAL RULES FOR SOLVING THE DIFFERENT CASES OF RECTILATERAL, OR QUADRANTAL, SPHERICAL TRIANGLES, WITH THEIR APPLICATION BY LOGARITHMS. (L).1. When one side of a spherical triangle is 90°, or a quadrant, it is called a quadrantal triangle. 2. If two sides of a quadrantal triangle be each 90°, the triangle will be isosceles, each of the angles at the base will be 90°, and the base itself will be the measure of the vertical angle. If the three sides be each 90°, the three angles will be each 90°. 3. A quadrantal spherical triangle may be changed into a right-angled spherical triangle, and the contrary. (X. 138.) B (M) RULE I. Subtract the angle opposite to the quadrantal side from 180°, and call the remainder the hypothenuse of a new triangle. The angle opposite to the quadrantal side must then be considered as a right angle, and the two remaining angles must be represented by the sides of the quadrantal triangle which are opposite to them. The sides and angles of the right-angled spherical triangle, will represent the angles and sides of the quadrantal triangle. Thus in the annexed figure, where ac is supposed to be a quadrant, the triangle abc may be considered as a right-angled triangle, whose hypothenuse ac=180°- B, the base AB= Lc; and the perpendicular sc=the LA; et contra. A (N) RULE 11. 1. In the triangle ABC, if one of the sides ac be a quadrant, and another side Bc less than a quadrant. Produce the side bc till it becomes a quadrant, and ads will be a right-angled triangle; wherein ad is the measure of the angle c, dB is the complement of BC, and as is the hypothenuse. 2. In the triangle arc, where one of the sides ac is a quadrant, and another side ce greater than a quadrant. Subtract 90° from ce, and the remainder will be the side DE of the right-angled triangle ADE; AE is the hypothenuse, and Ad is the measure of the angle c. The same reasoning will apply whether AB and AE be less or greater than quadrants, for in either case ADB and ads will be right-angled triangles. (O) Case I. Given a quadrantal side, its adjacent and opposite angle, to find the rest. The quadrantal side ac=90°.00! Required the Given Its adjacent Lo= 42°.12' sides AB, BC Its opposite B= 1150.20 ) and the LA. BY RULE II. (N. 228.) Because ac and cd are quadrants, the angles CAD and cpa are each of them a right angle (L. 227.), and ad is the measure of the angle c. The angles ABC and ABD, being supplements to each other, may be considered as the same angle, since they have the same sine, tangent, &c. 1. To find ab, by Case XII. right-angled spherics. sine ABD : sine ad=sine Zc:: rad : sine ab. Here AB is acute=48o. 2. To find BD and hence BC, by Case X. right-angled spherics. rad : cot AB)) :: tang ap=tang 4c: sine BD. Here ed is acute=250.25', and its complement=bc=640.35'. 3. To find the L BAD, by Case XI. right-angled spherics. cos Ad=cos Lc: rad :: COS ABD=cos 64o.40' : sine BAD. Here Bad is acute=350.17', and its complement=bac=54°.43'. BY RULE 1. (M. 227.) Here the supplement of the angle B= 64°.40', must represent the hypothenuse ac, the quadrantal side ac=the right angle b. The angles c and a must represent the sides ab and bc, and the sides BC and AB B must represent the angles a and c. B 1. To find the Lc, in the triangle abc, by Case V. rightangled spherics. sine ac = sine suppt. LB: rad::sin ab=sin Lc:sin Lc=sin AB. Here the Lc or AB is acute=48°. 2. To find the La, in the triangle abc, by Case IV. rightangled spherics. rad:cotac=cot suppt. / B::tang ab=tang 2c:cos La=cos BC. Here cosine of La=640.35' acute. 3. To find bc, in the triangle abc, by Case VI. right-angled spherics. cos ab=cos Lc: rad::cosac=cos suppt. LB : cosbc=cos Z A. Here bc is acute=540.43'. BY CONSTRUCTION. (Plate V. Fig. 13.) 1. With the chord of 60 degrees describe the primitive circle; and through the centre p draw ard, and cpe at right angles to it. 2. Set one foot of your compasses on 90 degrees on the line of semi-tangents, extend the other towards the beginning of the scale, till the degrees between them be equal to the angle c=42°.12', and apply this extent from p to A. 3. Through the three points cae draw a great circle, then AC will represent the quadrantal side. 4. With the tangent of the complement of the angle B, and centre c, describe an arc; with the secant of the complement of the same angle, and centre A, cross it in o: with o as a centre, and radius on, describe the great circle bab. Then ABC is the triangle required. To measure the required parts. And the angle a=54°.43'. (G. 164.) Note. The right-angled spherical triangle abc (Plate V. fig. 14.) into which the quadrantal triangle ABC is transformed, may be constructed and measured exactly in the same manner as Case IV. of right-angled spherical triangles was constructed and measured. See Plate V. fig. 9. (P) Case II. Given a quadrantal side, and the other two sides, to find the rest. S The quadrantal side ac= 90°+ Required the Given =1150. 9' angles A, B, The side BC 1130:18'S and C. + This example was formed from Example 2, Case VI. of right-angled spherics. The ZB was made equal to the supplement of the hypothenuse, the two sides AB and bc were made equal to the angles of the right-angled triangle; and hence the legs of the right-angled triangle become angles in the quadrantal triangle. In a similar manner the other examples were made. |