Now, to find the distance BC make the proportion, As sine B 84° 47′ 51′′ ar.-comp. Is to sine A 68° 40′ 55′′ So is AC 1201.744 To BC 1124.145 log. 0.001793 9.969218 3.079811 3.050822 4. Wanting to know the distance between two inaccessible objects which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and found to be, of the nearest, 57°; of the most remote, 25° 30' required the distance between them. : Ans. 173.656 feet. 5. In order to find the distance between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distance of a third point C from each, was measured, viz. CA=588 feet and CB 672 feet, and also the contained angle ACB-55° 40': required the distance AB. Ans. 592.967 feet. 6. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°: then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45': required the height of the tower. Ans. 83.9983 feet. 7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen, at a distance from each other equal to 200 yards, from the former of which A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE equal to 200 yards, and the following angles were taken, viz. AFC=83° ACF= 54° 31', ACD=53° 30, BDC=156° 25', BDE-54° 30', and BED=88° 30': required the distance AB. Ans. 345.46 yards. 8. From a station P there can be seen three objects, A, B and C, whose distances from each other are known, viz. AB= 800, AC 600, and BC=400 yards. There are also measured the horizontal angles, APC=33° 45′, BPC=22° 30′. It is required, from these data, to determine the three distances PA, PC and PB. Ans. PA=710.193, PC=1042.522, PB=934.291 yards. o here SPHERICAL TRIGONOMETRY. I. It has already been shown that a spherical triangle is formed by the arcs of three great circles intersecting each other on the surface of a sphere, (Book IX. Def. 1). Hence, every spherical triangle has six parts: the sides and three angles. Spherical Trigonometry explains the methods of determining, by calculation, the unknown sides and angles of a spherical triangle when any three of the six parts are given. II. Any two parts of a spherical triangle are said to be of the same species when they are both less or both greater than 90°; and they are of different species when one is less and the other greater than 90°. III. Let ABC be a spherical triangle, and O the centre of the sphere. Let the sides of the triangle be designated by letters corresponding to their opposite angles: that is, the side opposite B the angle A by a, the side opposite B by b, and the side opposite C by c. Then the angle COB will be represented by a, the angle COA by b and the angle BOA by c. The angles of the EC 6 H a F spherical triangle will be equal to the angles included between the planes which determine its sides (Book IX. Prop. VI.). From any point A, of the edge OA, draw AD perpendicular to the plane COB. From D draw DH perpendicular to OB, and DK perpendicular to OC; and draw ÂH and AK: the last lines will be respectively perpendicular to OB and OC, (Book VI. Prop. VI.) The angle DHA will be equal to the angle B of the spherical triangle, and the angle DKA to the angle C. The two right angled triangles OKA, ADK, will give the proportions R: sin AOK :: OA: AK, or, RxAKOA sin b. R: sin AKD :: AK: AD, or, R× AD=AK sin C. Hence, R2 × AD=AO sin b sin C, by substituting for AK its value taken from the first equation. In like manner the triangles AHO, ADH, right angled at H and D, give R: sin c: AO: AH, or Rx AH=AO sin c R: sin B:: AH: AD, or Rx AD=AH sin B. Hence, R2x AD=AO sin c sin B. Equating this with the value of R2 × AD, before found, and dividing by AO, we have The sines of the angles of a spherical triangle are to each other as the sines of their opposite sides. IV. From K draw KE perpendicular to OB, and from D draw DF parallel to OB. Then will the angle DKF=COB=a, since each is the complement of the angle EKO. In the right angled triangle OAH, we have R: cos c: OA: OH; hence AO cos c=R × OH=R× OE+R.DF. In the right-angled triangle OKE R: cos a: OK : OE, or R× OE=OK cos a. But in the right angled triangle OKA R: cos b:: OA: OK, or, R× OK=OA cos b. R: sin a But in the right : KD: DF, or Rx DF-KD sin a. R sin b OA: AK, or RX AK OA sin b R2 cos c=R cos a cos b+sin a sin b cos C. or Similar equations may be deduced for each of the other sides. Hence, generally, R2 cos a=R cos b cos c+sin b sin c cos A. } (2.) That is, radius square into the cosine of either side of a spherical triangle is equal to radius into the rectangle of the cosines of the two other sides plus the rectangle of the sines of those sides into the cosine of their included angle. V. Each of the formulas designated (2) involves the three sides of the triangle together with one of the angles. These formulas are used to determine the angles when the three sides are known. It is necessary, however, to put them under another form to adapt them to logarithmic computation. Taking the first equation, we have cos A: R2 cos a- -R cos b cos c Adding R to each member, we have R2 cos a+R sin b sin c-R cos b cos c R+cos A= 2 cos 2A But, R+cos A Ꭱ (Art. XXIII.), and R sin b sin c-R cos b cos c=-R2 cos (b+c) (Art. XIX.); 2 cos2AR2 (cos a—cos (b+c)). hence, R sin b sin c sin (a+b+c) sin} (b+c—a) 2 R (Art. XXIII). sin b sin c Putting s=a+b+c, we shall have s=(a+b+c) and s—a= (b+c-a): hence cos A=Rsin ( s )sin († s—a) sin b sin c (s) (s—b) cos + B=R1\/sin } ( s ) sin (3s cos CR sin a sin c sin(s) sin ( s- (3.) Had we subtracted each member of the first equation from R, instead of adding, we should, by making similar reductions, have found Putting s=a+b+c, we shall have }s—a—} (b+c—a), }s—b=} (a+c—b), and s—c=}(a+b—c) VI. We may deduce the value of the side of a triangle in terms of the three angles by applying equations (4.), to the polar triangle. Thus, if a', b', c', A', B', C', represent the sides and angles of the polar triangle, we shall have A=180°-a', B=180°-b', C=180°- -c' ; a=180°-A', b=180°-B', and c-180°-C' (Book IX. Prop. VII.): hence, omitting the ', since the equations are applicable to any triangle, we shall have cos a=Rcos (A+B—C) cos } (A+C—B) sin B sin C cos +b=R1cos (A+B—C) cos § (B+C—A) > (6.) cos c=R cos &c=R√ sin A sin C |
