CASE II. When two sides and an opposite angle are given. 28. In a plane triangle, ABC, . there are given AC=216, CB=117, the angle A=22° 37', to find the other parts. GEOMETRICALLY. 29. Draw an indefinite right line ABB': from any point, as A, draw AC, making BAC=22° 37', and make AC=216. With C as a centre, and a radius equal to 117, the other given side, describe the arc B'B; draw B'C and BC: then will either of the triangles ABC or AB'C, answer all the conditions of the question. TRIGONOMETRICALLY. To find the angle B. 7.931814 : AC 216 . . . . . . . . . 2.334454 :: sin A 22° 37' . ........ 9.584968 i sin B' 45° 13' 55", or ABC 134° 46' 05" 9.851236. The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for either of the angles ABC, or AB'C, which are supplements of each other, and therefore, have the same sine (Art. 13). As long as the two triangles exist, the ambiguity will continue. But if the side CB, opposite the given angle, is greater than AC, the arc BB' will cut the line ABB', on the same side of the point A, in but one point, and then there will be only one triangle answering the conditions. If the side CB is equal to the perpendicular Cu, the arc BB' will be tangent to ABB', and in this case also there will be but one triangle. When CB is less than the perpendicular Cd, the arc BB' will not intersect the base ABB', and in that case, no triangle can be formed, or it will be impossible to fulfil the conditions of the problem. 2. Given two sides of a triangle 50 and 40 respectively, and the angle opposite the latter equal to 32° : required the remaining parts of the triangle. Ans. If the angle opposite the side 50 is acute, it is equal to 41° 28'.59"; the third angle is then equal to 106° 31' 01", and the third side to 72.368. If the angle opposite the side 50 is obtuse, it is equal to 138° 31' 01", the third angle to 9° 28' 59", and the remaining side to 12.436. CASE III. When the two sides and their included angle are given. 30. Let ABC be a triangle ; AB, BC, the given sides, and B the given angle. Since B is known, we can find the sum of the two other angles B for A+C=180° – B, and, (A+C)=3(180° – B). We next find half the difference of the angles A and C by Theorem II., viz., BC+BA : BC - BA :: tan 3(A+C).: tan }(A - C), in which we consider BC greater than BA, and therefore A is greater than C; since the greater angle must be opposite the greater side. Having found half the difference of A and C, by adding it to the half sum, }(A + C), we obtain the greater angle, and by subtracting it from half the sum, we obtain the less. That is, (A + C)+(A– C)= A, and 3(A+ C) – }(4 – C)= C. Having found the angles A and C, the third side AC may be found by the proportion, sin A : sin B :: BC: A C. EXAMPLES. 1. Ia the triangle ABC, let BC=540, AB= 150, and the included angle B= 80°: required the remaining parts GEOMETRICALLY. 31. Draw an indefinite right line BC, and from any point, ás B, lay off a distance BC=540. At B make the angle CBA = 80° : draw BA, and make the distance BA= 450; draw AC; then will ABC be the required tri. angle. TRIGONOMETRICALLY. BC+ BA=540+450 = 990; and BC-BA=540 – 450 = 90. A+C=180° – B=180° – 80° = 100°, and therefore, (A+C)=3(100°)= 50°. . To find }(A – C). 1.954243 1.95429 :: tan }(A + C) 50° . ...... 10.076187 : tan 3(4 - 0) 6° 11'. ..... 9.034795. Hence, 50° +6° 11'=56° 11'=A; and 50° — 6° 11'=' 43° 49' = C. sin C : sin B :: AB : AC To find the third side AC. 43° 49'. : ar comp. 0.159672 80" . . . . . . . . . 9.993351 450 ......... 2.653213 640.082 . . . . . . . . . 2.806236. 2. Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 04': required the other parts. Ans. Angles, 33° 34' 39"; 18° 21' 21"; side 2400. CASE IV. 32. Having given the three sides of a plane triangle, to find the angles. Let fall a perpendicular from the angle opposite the greater side, dividing the given triangle into two rightangled triangles: then find the difference of the segments of the base by Theorem IIJ. Half this difference being added to half the base, gives the greater segment; and, being subtracted from half the base, gives the less segment. Then, since the greater segment belongs to the right-angled triangle having the greater hypothenuse, we have two sides and the right angle of each of two right-angled tri. angles, to find the acute angles. EXAMPLES. 1. The sides of a plane triangle being given; viz., BC=40, AC= 34, and AB= 25: required the angles. BD GEOMETRICALLY. 33. With the three given lines as sides construct a tri angle as in Prob. IX. Then measure the angles of the triangle either with the protractor or scale of chords. 40 TRIGONOMETRICALLY. 59 X 9 40+13.27500 Then, - 2 '= 26.6375 -- CD, 40 – 13.275' And, "=13.3625= BD. 40 In the triangle DAC, to find the angle DAC. 8.468521 : DC 26.6375 . . . . . . . . 1.425483 :: sin D 90° . . . . . . . . 10.000000 : sin DAC 51° 34' 40" .. . 9.894014. In the triangle BAD, to find the angle BAD, AB 25 ar. comp. 8.602060 : BD. 13.3625 . . . . . . . 1.125887 :: sin D 90° ......... 10.000000 i sin BAD 32° 18' 35" ...... 9.727947. Hence, 90° – DAC-= 90° — 51° 34' 40" = 38° 25' 20" = C, and, 90° – BAD= 90° — 32° 18' 35" = 57° 41' 25" -= B, and, BAD+DAC= 51° 34' 40" +32° 18' 35" = 83° 53' 15" = A. "2. In a triangle, of which the sides are 4, 5, and 6, what are the angles ? Ans. 41° 24' 35"; 55° 46' 16"; and 82° 49' 09". SOLUTION OF RIGHT-ANGLED TRIANGLES. 34. The unknown parts of a right-angled triangle may be found by either of the four last cases; or, if two of the sides are given, by means of the property that the square of the hypothenuse is equivalent to the sum of the squares of the two other sides. Or the parts may be found by Theorems IV. and V. EXAMPLES. 1. In a right-angled triangle BAC, there are given the hypothenuse BC=250, and the base AC= 0 240: required the other parts. Ans. B = 73° 44' 23" ; C=16° 15' 37" ; A B = 70.0003. 2. In a right-angled triangle BAC, there are giveu AC=384, and B= 53° 08': required the remaining parts. Ans. AB= 287.96; BC= 479.979; C = 36° 52'. |