or the sine of square root of adjacent to the either angle of a triangle = rad. into the the sum of the three sides one of the sides angle, into the sum the other adjacent side, divided by the rectangle of the adjacent sides. This applied to each angle separately will determine all three. CASE IV. Two sides and the included angle being given to find the other parts. Let A be the given angle and b and c the given sides, then b+c: b c:: tan (B+ c) : tan 1⁄2 (B — c) . (Art. 75.) or the sum of the two given sides: their difference :: the tan of the sum of the two unknown angles: the tan. of their difference. B+C 180° - A Three terms of the above proportion being known, the fourth, tan (в —c), may be found, then and ↓ (B + C) + 1 (B — C) — B (B+C) (B — C) — C All the unknown parts will thus be determined, except the side opposite the given angle, which may be found by the proportion. The sines of the angles are as the opposite sides. We add one other formula for the cosine of an angle in terms of the three sides for its miscellaneous use, rather than to be employed in the solution of triangles. RX sin of the middle part the rectangle of the tangents of the adjacent parts the rectangle of the cosines of the opposite parts. (Art. 88.) Observe that the complements of the hypothenuse and ob lique angles are used. The middle part is either between two others which are adjacent to it, or else separated from two which are adjoining each other. N. B. The two given parts must always be employed with one of the required parts in applying these rules; by this means an equation will be formed from which the value of the required part contained in it must be derived. 2. Oblique angled spherical triangles. CASE I The three sides being given to find the angles. Or the sine of either angle of a spherical triangle =ra dius into the square root of the sine of the sum of the three sides minus one of the sides adjacent the required angle, into the sine of the sum minus the other adjacent side, divided by the rectangle of the sines of the adjacent sides. This rule applied to each of the angles will serve to determine them all successively. or the cos. of either side of a spherical triangle = radius into the square root of the sum of the three angles minus one of the angles adjacent the required side into the sum minus the other adjacent angle, divided by the rectangle of the sines of the adjacent angles. By this rule the three sides may be separately found. CASE III. Two sides and the included angle being given, employ Napier's Analogies. sin(a+b): sin (a—b) :: cot c: tan cos (a+b): cos(a--b) :: cot c: tan (A+B) (A-- B) 1 (A + B) + 1 (1 — B) = A (A+B) — (^— B) = B The first proportion is read thus: Cos of the sum of the two given sides: cos of their difference: cot. of the given included angle : tan. of the sum of the unknown angles. The other is read in a similar manner. To find the remaining side sin A sin a :: sin c: sin c or to avoid ambiguity COS (A B) : COS 1 (A + B) : : tan 1 (a + b) : tan c observe that log. tan (a + b)=10+ar. comp. log. cos. (a+b)-ar. comp. log. sin (a + b) to save one reference to When two angles and the included side are given. Napier's Analogies. COS (A+B): Cos } (A—B) : : tan c: tan (a+b) sin a or better sin a sin c sin c cos (ab): cos (a+b) : ; tan † (A+B) : cot § c. (Art. 87.) CASE V. Two of the three given parts being a side and its opposite angle. Find the part opposite the other given part by the proposition, the sines of the angles are as the sines of the opposite sides. Then there will remain unknown in the triangle, an angle and the opposite side. Let fall from the vertex of the unknown angle, an are perpendicular upon the unknown side, the given triangle will thus be divided into two right-angled spherical triangles, in each of which the hypothenuse and one angle are known, apply Napier's rules to find each of the partial angles which compose the required angle of the given triangle, and the partial sides which compose the required side. This case is ambiguous, since there will be two values corresponding to the fourth term of the first proportion," the sines of the angles are as the opposite sides." Both of these solutions answer to the conditions of the problem. COS A = a We add the fundamental formula of spherical trigonometry, that of the cosine of an angle, in terms of the three sides, which applied to each angle furnishes three equations, and the six parts of the triangle, from which equations any two parts may be eliminated, and the result will contain four parts, any three of which being given the fourth may be found. R? dos a R cos 6 cos c sin b sin c or, the cosine of a spherical triangle is equal to radius square into the cosine of the side opposite minus radius into the rectangle of the cosines of the adjacent sides, divided by the rectangle of the sines of the adjacent sides. Note.--In assuming hypothetical cases, care must be had not to suppose such as are impossible. The following are the governing principles to be observed. In plane triangles, 1. One side must be less than the sum of the other two, (Geom., B. 1, Prop. 7). 2. The greater side of a triangle is opposite to the greater angle, (Geom., B. 1, Prop. 13). 3. The sum of the angles must be exactly two right angles. In spherical triangles, the first two principles also apply, (Geom., B. 9, Props. 1 and 14). 4. The sum of the three . angles must not be less than two, nor greater than six right angles (Geom., B. 9, Prop. 16). 5. The sum of the three sides must be less than a circumference (B. 9, Prop. 3). 6. Each side must be less than a semicircumference (B. 9, Def. 1). 7. Each angle must be less than two right angles, (B.9, Prop. 16, Schol.). a Part IV. is a short treatise containing the application of plane and spherical trigonometry, to Navigation and Nautical Astronomy. It will occupy the student, properly acquainted with trigonometry, but a few days. Should it be omitted, Part V. will be found to contain a few additional general formulæ, necessary for the study of Analytical Geometry. |