2. Solve the following spherical triangles : (a) One side is 90°, and the angles adjacent are 72° 15′ and 104° 26'. (b) The three sides are 125° 40', 90°, 115° 10'. (c) Two sides are 54° 20′ and 90°, and the included angle is 57° 55'. (d) Two sides are each 80° 28', and the angles opposite are each 33° 20'. 3. Quito is on the equator in longitude 78° 50′ W.; New York is in lat. 40° 43', lon. 74° W. Find the distance between them and the angle that the arc connecting the two places makes with the meridian at each. 4. The arc of the great circle connecting St. Petersburg and Hopedale (lat. 55° 30′ N., lon. 60° W.) is 44° 28′ long and makes an angle of 45° 33' with the meridian through Hopedale. Find the distance from St. Petersburg to St. Nicholas (lat. 34° 30′ S., lon. 60° W.) and the angle between the two arcs at St. Petersburg. 5. Prove that in any quadrantal spherical triangle, if c denotes the quadrantal side, i. e. if c = 90°, then : (a) cos C+cos A cos B = cos c. (b) cos C+ ctn a ctn b = cos c. (c) sin a cos B = cos b, sin b cos A = cos a. 6. Prove that in any right spherical triangle, if C denotes the right angle; then : sin a cos c tan b tan A. PART IV. SPECIAL LOGARITHMIC METHODS 99. Tangents of the Half-angles. When adapted to logarithmic computation, the law of cosines of spherical trigonometry yields two sets of formulas which are useful in the solution of triangles and are analogous to those of plane trigonometry given in §§ 26-27, pp. 34–36. N M Let ABC be a spherical triangle; a small circle can be inscribed in it as follows: draw arcs of great circles bisecting its angles; these bisectors meet in some point P. This is the pole of the inscribed circle; for, from P draw arcs of great circles PL, PM, PN, perpendicular respectively to the sides BC=a, CA=b, AB=c. The right triangles PMA, PNA are symmetric, having the hypotenuse and an angle of the one equal respectively to the hypotenuse and an angle of the other; therefore PM= PN, and similarly each of these is equal to PL. Let r=PL = PM=PN, the polar distance of the inscribed circle. Also from the same symmetric triangle, AM AN, BN = BL, and CL = CM; hence if we set 2s = a+b+c, AN+BL+CL=s, whence ANS- (BL+CL) = 8 a, and similarly, BL - b and CM =s - C. = 3 L FIG. 94 = 8 In the right triangle ANP, we have, by (6) § 95, or by Napier's Rule 2, and therefore (1) sin AN tan PN ctn PAN, = sin (s— a) = tan r ctn (A/2), Similarly from triangles BLP and CMP, (2) (3) tan (4/2) = tan r/sin (s — a). tan (B/2)= tan r/sin (sb). tan (C/2) = tan r/sin (s — c). Compare these with (1) § 27, p. 36. Eliminate tan r from (1) and (2) by division: Produce the sides AB and AC to form a lune, and apply (4) to the colunar triangle A'B'C': tan (B'/2) sin (s' — a')' or, since s'= 180° (s — a), s' — b' = s - C2 Compare this result with that of § 29, p. 39. Formulas (1), (2), (3), and (7) hold for all spherical triangles which have no side or angle greater than 180°.* Let 100. The Law of Tangents, or Napier's Analogies. ABC be a spherical triangle; suppose A> B. Then at B, the smaller of the two angles, lay off angle ABA' = A. Apply formula (4) of § 99 to angles B' and C' of the triangle A'B'C'. = B'A- B, C 180°C, s'= (a + b'+c')/2; but since triangle A'AB is isosceles, c'=b+b'; hence s'= (a + b)/2+b', *All these formulas can be deduced from the law of cosines by an alge braic process without reference to any particular figure. See, for example, Chauvenet's Treatise on Spherical Trigonometry. s'b' = (a+b)/2, s'-c' = (a - b)/2. Therefore Therefore, the If A=B, then a=b and (1) is satisfied. formula holds for all cases, since (4) of § 99 does. Produce the sides AB and BC to form a lune, and apply formula (1) to the triangle A'B'C', colunar to ABC. A Substitute 180° - A for A', B for B', and so on: If we divide (1) by (2) we obtain a formula which is analogous to the law of tangents for plane triangles. Formulas (1) and (2) are called Napier's Analogies. They furnish another proof of (19) (ƒ) and (g), p. 111. 101. Solutions of Triangles by Logarithms. When two sides and the included angle are given (Case I of § 94), the other two angles can be found by Napier's Analogies, (1) and (2) of § 100, and then the third side can be found by the law of sines. Example 1. Assuming the radius of the earth to be 3956 miles, find the shortest distance from Seattle (lat. 47° N. lon. 122° 20′ W.) to Manila (lat. 14° 35'.5 N., lon. 120° 58'.1 E.), and the direction of the course at each place. n N m Take the north pole for the third vertex; then m = 42° 24', s = 75° 24'.5; N = 116° 41'.9 To find the angles at M and S, we write, since s> m, sin [(s M FIG. 98 m)/2] N tan ctn (sm)/2 = 16° 30'.2 (s + m)/2 = 58° 54'.2 N/2 58° 20'.9 Arrange the computation so as to look up both sine and cosine of (s—m)/2 The course leaves Seattle going N. 60° 24'.5 W. and arrives at Manila going S. 37° 17'.9 E. To find n, use sin n/sin s = sin N/sin S: log sin n = 9.99749 10 Hence either n = 83° 51' or n = 96° 9'. By the law of cosines, cos n = (.74) (.25) — (.67)(.97)(.45) approximately; hence cos n is negative; therefore n = 96° 9'. Hence the length of the course in miles nπ R/180 = 96.15 × 3.1416 × 3956/180 = 6639. (to the nearest mile) is (Compute by logarithms.) |