94. Résumé of Cases. To summarize, the cases are: Case I. Given two sides and the included angle: Use the cosine law to find the third side; then use the sine law to find the other angles, determining the correct quadrant by (19), p. 111, or by means of the cosine law. See § 92. Case Ip. (Polar of Case I.) Given two angles and the included side. Solve the polar triangle. Case II. Given three sides. Find one angle by the cosine law; then proceed as in Case I. Case IIp. (Polar of Case II.) Given three angles.* Solve the polar triangle. Case III. (Ambiguous Case.) Given two sides and an angle opposite one of them: Determine the third side by the cosine law, as in Example 1, p. 117, deciding the ambiguity as in Examples 1-4, § 93. Then proceed as in Case I. Case IIIp. (Polar of Case III.) Given two angles and a side opposite one of them.† Solve the polar triangle. The student should note that we cannot construct a spherical triangle nor compute the lengths of its sides in ordinary linear units unless the radius of the sphere is given. 2. Find the distance, in degrees, from St. Petersburg (lat. 60° N., lon. 30° E.) to Hopedale (lat. 55° 30′ N., lon. 60° W.) and the angle which the arc joining the two places makes with the meridian through each. [HINT. Use the north pole for the third vertex of a triangle.] * This case does not arise in plane triangles, since a plane triangle is not determined by its three angles. †The corresponding case in the plane is not ambiguous because the sum of the angles of a plane triangle is precisely 180°. 3. Reduce the distance computed in Ex. 2 to miles, assuming that the radius R of the earth is 3956. mi. 4. Find the distance, in degrees, from San Francisco (lat. 37° 48′ N., lon. 122° 28′ W.) to Manila (lat. 14° 35'.5 N., 120° 58'.1 E.). Express the same distance in miles. 5. By means of the cosine law and the polar triangle, show that, for any triangle, cos A = cos B cos C+ sin B sin C cos a. [Polar Cosine Law.] PART III. SPECIAL METHODS FOR RIGHT TRIANGLES 95. Right Spherical Triangles. The methods of § 94 will solve all cases, including right triangles; but on account of their frequent occurrence and the simplicity of the formulas in case one angle is 90°, a separate treatment seems desirable. A right spherical triangle can be solved when any two of its parts (in addition to the right angle) are given; in fact, if two parts are given, the other three parts can be computed in succession from the given data. All possible cases are provided for in ten formulas which, for a reason to be explained presently, are collected into two groups. с b FIG. 89 B a C=90° Let ABC be a spherical triangle, rightangled at C; we shall call a and b its sides, c its hypotenuse, To prove 1, 2, use law of sines; to prove 3, use law of cosines; to prove 4, 5, apply the law of cosines to the polar triangle, remembering that c' = 90°. To prove 6, substitute in 1 the values of sin b and sin A from 2 and 5; similarly prove 7, 8, 9, 10, by substituting in 2, 3, 4, 5. 96. Napier's Rules. The preceding formulas are simple, are adapted to the use of logarithms, and are easy to apply; but there are so many of them that they become a burden to the memory. To avoid this difficulty an ingenious device was given by Napier, a Scotch mathematician (1550-1617) who invented logarithms and made important contributions to trigonometry. He states two rules which enable one to write down the particular formula needed in any given computation without attempting to remember the whole set of formulas. These two rules of Napier employ what are called "the five circular parts" of a right spherical triangle. These are as indicated in the diagram, Fig. 90, the two sides, the complement of the hypotenuse, and the complements of the two angles. The right angle is not counted and is not indicated on the diagram. If five objects are placed on a closed contour, e.g. five persons sitting at a round table, each has two neighbors which are adjacent, one on the right, the other on the left; and the two remaining are nonadjacent or opposite. So on the diagram of the five circular parts of a right spherical triangle, if we select any one, there are two others which are adjacent and the remaining two are opposite; moreover, if any three are selected, one of these may always be chosen and called the middle part so that the other two are either both adjacent or both opposite. Napier's rules refer to these circular parts and are as follows: RULE 1. The sine of the middle part is equal to the product of the cosines of the opposite parts. RULE 2. The sine of the middle part is equal to the product of the tangents of the adjacent parts. These rules may be remembered by the alliteration of the first vowel in the words cosine and opposite, tangent and adjacent. When applied in succession to the five circular parts, rule 1 gives the formulas of group I; and rule 2, those of group II. To apply these rules to the solution of a right spherical triangle, use the diagram of the five circular parts; mark the given parts and the part to be found; of these three choose the middle part (which may or may not be the unknown) and note whether the other two are opposite or adjacent; then apply the appropriate rule to write down the formula, remembering that any function of the complement of an COB a angle is the corresponding cofunction of the angle itself; e.g. sin co c = cos c, tan co B = ctn B, cos co A = sin A, etc. = Co B *Coc Example 1. The hypotenuse c of a right spherical triangle is 124° 50' and the angle A is 37° 25'. Solve the triangle. To find a, mark co A, co c, and a; a is the middle part and the others are opposite; then as you say to yourself, "The sine of a is equal to the product of the cosines of co A and co c;" write "Co A sin α = sin c sin A. Before beginning a computation by logarithms, b FIG. 91 attention must first be paid to the signs of the factors. They are here both positive, so we proceed to compute : Since side a is in the same quadrant as its opposite angle, a = 29° 50'. To find b, mark co A, co c, and b; co A is the middle part, the others are adjacent. Say, "Sin co A is equal to the product of tan co c and tan b;" write cos A = ctn c tan b, Co B a + whence tan b = cos A tan c. Cos A is positive, tan c is negative; therefore, tan b is negative and b is in the second quadrant. To make all the factors positive we change the signs of the negative factors and write the formula tan (180° - b) = cos A tan (180° — c), and compute as follows: log cos 37° 25' = 9.89995 10 log tan 55° 10' = 0.15746 log tan (180° - b) = 0.05741 180° b = 48° 46'.5, b = 131° 13'.5 To find B, mark co A, co c, and co B; co c is the middle part, the others are adjacent : + cos c = ctn A ctn B, whence ctn B = cos c tan A. Signs of the factors indicate that B is in the 2d quadrant; changing signs of negative factors, the formula becomes ctn (180° - B) = cos (180° - c) tan A = *COB log cos 55° 10' 9.75678 - 10 *Coc *Co A log ctn (180° - B) = 9.63045 – 10 b FIG. 93 180° - B = 23° 7'.4, B = 156° 52'.6 97. Ambiguity. When an unknown part of a right spherical triangle is to be determined by its sine, the following statements are useful to determine whether the angle is in the first or second quadrant. 1. An angle and the opposite side are always in the same quadrant, by (4), § 95, since sin B is surely positive. 2. When the two sides are in the same quadrant, the hypotenuse is in the first quadrant; when the two sides are in different quadrants, the hypotenuse is in the second quadrant; and conversely. This follows from (3), § 95. 3. The Ambiguous Case. In case there are given one of the angles of a right spherical triangle and the side opposite, there are, always, two solutions which are colunar, except that if the given side is also 90°, the triangle is birectangular and the two solutions coincide. These facts are obvious geometrically, since the lune whose angle is the given angle is divided by the given side into two colunar triangles, either of which is a solution. In no other case, when two parts besides the right angle are given, is there more than one solution. 98. Quadrantal and Isosceles Spherical Triangles. A quadrantal spherical triangle is one that has one side equal to a quadrant or 90°. Such triangles are readily solved by means of their polar triangles, which are right angled. An isosceles spherical triangle may be divided into two symmetric right triangles by an arc perpendicular to the base through the vertex of the angle included between the equal sides. |