The first two cases may be solved by the law of sines, and the last two by the law of cosines. The last two may also be solved by using first the law of cosines and then the law of sines. CASE I. Given a side and two angles, the third angle is found from A + B + C = 180°, and the other two sides by the law of sines. EXAMPLE 1. Two forts by the sea, A and B, are 12 miles apart. At A the angle between B and a target C anchored off the coast is 37°.24, and at B the angle between A and C is 42°.87. Find the distance from each fort to the target. C 37.24° 42.87 A B c=12 = From the law of sines, 180° – (A + B) = 99°.69. Check. Find c from a, b, C. By the law of cosines, c2 = a2 + b2 - 2ab cos C and therefore c = · 7.372 + 8.282 — 2 × 7.37 × 8.28 × (− 0.168) 143.5, 11.98. This value agrees reasonably well with the given value of c, especially when we take into account the fact that the table of squares does not enable us to use all four figures in a and b. CASE II. Given two sides and an angle opposite one of them, the angle opposite the other given side is found by the law of sines, then the third angle is found from the relation A+B+C and finally the third side is found by the law of sines. 180°, EXAMPLE 2. Two straight roads diverge at an angle of 35°. An automobile starts from the fork in the road and runs along one road until a cross road is reached, the odometer showing the distance to be 2.1 miles. It turns into the cross road and after running 1.4 miles comes to the other road. How far is the automobile from the starting point, and au what angles does the cross road meet the other two? If A denotes the fork in the road, B the point 2.1 miles from A, and C the third point reached, a triangle is determined by the parts whence C = 59°.43 or 120°.57 (compare Example 1, page 197). Since BCC' Both of these values agree very well with the given value of c. In order to illustrate the method of solution, the computations have been carried out with all the accuracy permitted by the tables. Do the given data warrant us in saying that the automobile is either 2.433 miles or 1.009 miles from the fork in the road? If two sides of a triangle and the angle opposite one of them are given, there can be TWO SOLUTIONS only if the angle is acute and the side opposite it is less than the other given side. Under these conditions, there will be two solutions unless in solving for the second angle (C, in Example 2), it is found that its sine is equal to or greater than unity (sin C=1). In the first case there is but one solution, a right triangle; and in the second case there is no solution, since the sine of an angle cannot exceed unity. A complete statement of the possible solutions is given in Exercise 3 below. CASE III. Given two sides and the included angle, the third side is found by the law of cosines, and then the remaining angles by either the law of cosines or the law of sines. EXAMPLE 3. To find the distance across a bay, AB, a point C is taken whose distances from A and B can be measured. It is found that AC is 70 yards, BC is 120 yards, and ZACB is 50°. Find AB and also the angles at A and B. A triangle is determined whose given parts are a = 120, b = 70, C = 50°. By the law of cosines c2 = a2 + b2 2ab cos C To find A and B, we use the law of cosines: Substituting the values of B whence and hence a-120 с FIG. 121. A 180° - 85°.55 Substituting the values of the sides in 4900 = 8500 + 14,400 – 2 × 92.2 × 120 × cos B, Check. The angles having been found without using the fact that the sum of the three angles is two right angles, we have A + B + C = 94°.45 + 35°.55 + 50° 180°. Having found A as above, B might have been obtained from A + B + C = 180°. But then it would have been necessary to use the law of sines as a check, and little would have been gained. CASE IV. Given the three sides, the three angles are found by the law of cosines, or one may be found by the law of cosines and then the others by the law of sines. EXAMPLE 4. The lengths of a triangular lot are found by pacing to be 80 feet, 50 feet, and 100 feet. Find the angles at the corners and the area. A triangle is determined by a = 80, b = 50, c = 100. Substituting these values in the three forms of the law of cosines we get To find the area, draw the altitude from C and denote it by h. Then in the right triangle ACD, h = b sin A = 50 × 0.7924 = 39.62. Then the area is ch = 50 × 39.62 = 1981 square computations, and the angles to four significant figures for the purpose of obtaining drill in interpolation. In the next chapter, a labor saving device to assist in the computations will be considered, and further exercises given. EXERCISES 1. In Example 4, find B and C by the law of sines. What point in the solution might be overlooked, which one is more likely to notice in using the law of cosines? 2. Is a triangle always determined if values are given for a side and two angles? For two sides and the included angle? For the three sides? 3. If two sides, say a and b, and the angle opposite one of them, say A, are given, show that (a) If A is obtuse, there is one solution if a>b, and there is no solution if ab. (b) If A is right, there is one solution if a > b, and there is no solution if ab. (c) If A is acute, there is one solution if a ≥ b, there are two solutions if b sin A <a<b, there is one solution if a = b sin A, and there is no solution if a <b sin A. 4. Determine the number of solutions, by Exercise 3, if, 5. Solve the following triangles, and check the solution: (a) B = 62°.74, C = 87°.20, a = 10. 6. A tunnel is to be built through a hill from a point A to another point B. A point C, at the same level as A and B, is 1000 feet from A and 800 feet from B and ZACB = 42°. Find the length of the tunnel. 7. A man starts from camp and walks N.E. for 5 miles, and then 22°.5 east of south until he reaches a point from which the camp is visible in a direction due west. How far has he walked, and how far is he from camp? 8. Two batteries of artillery, A and B, are four miles apart. An enemies' battery is located at a point C such that ≤BAC = 64°.22 and LABC 43°.17. Find the range for each battery. 9. A field is bounded by two roads intersecting at right angles, and by two other straight lines. Find its area if the lengths of the sides, beginning at the corner at the crossing of the roads and measured in order around the field, are 30, 60, 70 and 40 rods. |