Second proof. In Fig. 82, let ABC be any triangle. triangle circumscribe a circle. B (c)° About the Let O be the center. Draw the b E and COROLLARY. The constant ratio of a side of the triangle to the sine of the opposite angle is equal to the diameter of the circumscribed circle. AM 77. Cosine theorem. In any triangle the square of a side equals the sum of the squares of the other sides minus twice the product of these sides by the cosine of their included angle. Proof. In each triangle of Fig. 81, (Notice that in (a) DC is positive, in (b) negative, and in (c) it is zero because D falls on C.) 1. Are the formulas [341], [342], and [343] adapted to solving by logarithms? 2. Derive [342] and [343] independently. 3. Solve each of the three formulas for the angles in terms of the sides. 4. Solve a2 = b2c2-2 bc cos a for b. Ans. b c cos a ± √a2 c2 sin2 a. 78. Case I. The solution of a triangle when one side and two angles are given. In this case, it is evident that the third angle can always be found from the equation α+β+γ The sides can then be found by using the relations stated in the sine theorem, namely, Since in each of these there are four parts of the triangle involved, therefore, if any three of these parts are known, the fourth can be found. Also since, in any example, only two sides are to be found, two of the above relations may be used for solving and the third for checking. The same suggestions as were given in Art. 32 for the solution of right triangles should be carried out here. Draw the triangle, state the formulas, make out a careful scheme for all the work, and, lastly, fill in the numerical part by the use of the tables. Remember that in computations time and accuracy are of very great importance. Time will be saved by carefully planning the arrangement of the work. Accuracy can be secured by checking the work at every step. Verify at every step the additions, subtractions, multiplications, and divisions. Check interpolations when using tables, by repeating the work at each step. From geometry, the area of a triangle equals one half the product of the base and altitude. Using b for base, h for altitude, 1. Given a, ẞ, and c; to find y, a, and b. Give formulas and scheme for solution. 2. Give the formula for area, when a is the given side. When c is the given side. 3. Given a = 60° 25′ 31′′, ß = 69° 26′, and c = 173; find α = 196, b = 211, and y = 50° 8' 29". 4. Given B find = απ 5. Given a = find 43° 44′ 18", y = 75° 2′ 42′′, and b = 81.5; 11° 11′ 18′′, y = 57° 37' 24", and c = 444.79; 6. Given B find b 7. Given a find В 8. Given B find = = = = = 20° 20.2', y = 12° 28.5', and a = 673.75; 28° 14' 44", y = 109° 32′ 30′′, and b = 730.8; απ 9. Given B find == = 14.397; 79. Case II. The solution of a triangle when two sides and an angle opposite one of them are given. It is known from geometry that when two sides and an angle opposite one of them are given the triangle may not be uniquely determined. With these parts given: (1) it may not be possible to construct any triangles; (2) it may be possible to construct just one triangle; (3) it may be possible to construct two triangles — the ambiguous case. Corresponding to Exercises 1, 2, and 3 above, we have the following which should be compared with the corresponding constructions in Fig. 84. (1) No solution when: (a) Angle is acute and opposite side less than adjacent side times the sine of the angle. (b) Angle is obtuse and opposite side not greater than adjacent side. |