Limits.-1. The angle B may be any quantity from 0 to 180°. 2. The difference between AC and BC may be any quantity less than AB. PROBLEM III.—The difference between the diagonal of a square and one of its sides being given, to determine the square. Analysis. Suppose ABCD to be the required square, having the difference between the diagonal AC and the side AB equal to a given quantity; then produce the side AB until AE= AC, and BE will be equal to that given quantity. Join CE. Then in the isosceles triangle AEC, the angle CAE, being half a right angle, is 45°; hence, each of the equal angles AEC and ACE (180-45)=67°; and the triangles EBC and EAC are determined. = = Whence this Construction. In any line, make EB the given difference, the angle BEC 6740, and let EC meet a perpendicular erected at B, in C. On BC describe the square ABCD, which will be the one required. Demonstration.-Draw the diagonal AC. We have to prove that AC-AB=BE. In AEC, since <A=45°, being half a right angle, and <E=6710, by construction, the angle ACE must be the supplement of their sum, which is 671°. Hence <ACE= <AEC, and the side AC=AE. From each take AB, and we have AC-AB-AE-AB-BE the given quantity. Q. E. D. Calculation.-In EBC, Case 1, find BC=AB; then AC= AE=AB+ BE. = Limits. The difference BE may be any quantity whatever. PROBLEM IV.—In a plane triangle are given one angle, the side opposite thereto, and the difference between the other two sides, to determine the triangle. Analysis.—Let ABC represent the required triangle. Produce BA, making BD= BC; then AD is the given difference. Join DC. Then, since BD=BC, we have the angles BDC and BCD equal, and each half the supplement of the given angle B. Hence the triangle DAC is determined, and we have this B F A D Construction.-Draw an indefinite line, and on it lay DA= the given difference between AB and BC. Make the angle ADF equal to half the supplement of the given angle B; and to DF apply AC= the given side. Make the angle DCB=<ADC, and BAC will be the triangle required. Demonstration.-We have to prove that the angle B is equal to the given angle, and that AD—BC — BA. Since the angles BCD and BDC are equal by construction, and each half the supplement of the given angle, the angle B must be equal to the given angle, and the side BC= BD. Take BA from each; then BC — BA= BD-BA AD, the given difference. Q. E. D. = Calculation.-1. In ADC; Case 2, find <ACD; then <ACB =<BCD-< ACD=<BDC <ACD. 2. In ABC, Case 1, find AB and BC. Limits.-AC may be any quantity greater than AD. PROBLEM V.-In a plane triangle are given one angle, the side opposite thereto, and the sum of the other two sides, to determine the triangle. FIG. 2. Analysis. Let ABC (Fig. 1) represent the required triangle. The line AB being fixed in position, produce it until the produced part BDBC; then AD will be equal to the given sum of AB and BC. Now, since BD BC, the angles BCD and BDC are equal, and each half the given exterior angle ABC (25 I. cor. 6*). Hence, in the ADC we know AC, AD, and the angle D= the given angle; whence this = * I. 32. Construction.-Draw AD=the given sum of the sides AB and BC. Make the angle ADF=half the given angle, and to DF apply AC the given side. Then make the angle DCB=the <ADC, and ABC will be the required triangle. Demonstration.. We have to prove that the angle ABC= 2 ADF, and that AB + BC=AD. By 25 I. cor. 6*, < ABC= angles BDC+BCD=2 BDC=2 ADF. Also (12 I.†) BC=BD. Add each to AB, and we have AB + BC=AB+ BD=AD. Q. E. D. Calculation. In ADC, the first formed, find, Case 2, the angle ACD, whence we have angle CAD; then, in AABC, Case 1, find AB and BC. Limits, and the discussion of the general problem. 1. In applying the given side AC to the line DF (see Fig. 2) in the above construction, if the arc were continued to the right, it would cut the line DF in another point C', and, by drawing C'B' parallel to CB, or, which is the same thing, by making the angle DC'B'=<ADC', we would have another triangle, AB'C', fulfilling all the required conditions of the problem; for AC' AC, < AB'C' =< ABC, and AB' + B'C'— AB' + B'D=AD. Hence the other parts in the triangles ABC and AB'C' are equal,—that is, AB=B'C', BC= AB', angle ACB=< C'AB', <CAB=< AC'B'; also < AC'D=180° —<AC'C=180°- ACD, and the angles AC'D and ACD are supplements of each other. In the preceding calculation, by Case 2, the angle directly obtained is the acute angle ACD, the supplement of which is the angle AC'D. When we use the angle ACD in the subsequent part of the calculation, we obtain the sides AB and BC, of which AB is less than BC. When we use its supplement AC'D, we obtain AB' and B'C', which AB' is greater than B'C'. of 2. On DF, let fall the perpendicular AC", and draw C"B" parallel to BC; then AC" is the inferior limit of the length of the given side; for, if the length of AC were given less than the perpendicular AC", it could not be applied from A to the line DF, and the problem would be impossible. = When the given side is just equal to the perpendicular AC", the AAB" C" becomes isosceles, having AB": B" C", and each equal to half the given line AD. For the = 90° - ADC"DAC"B"AC". B'D=AD. AC"B" = 90° B'C'D Hence AB" — B" C" = 3. AD is the superior limit of the length of the given side. Apply to DF, AC"" = AD; then angle AC""D=<ADC=< BCD. Hence AC" is parallel to BC; but the side AB then vanishes, or becomes equal to 0, and the triangle closes in the line AC"". 4. If to DF we apply a line A Civ, greater than AD, and draw Civ Biv parallel to CB, it will make the angle A'Biv Civ, with DA produced, equal the given angle ABC; then Biv D= Biv Civ, and AD becomes the difference of the sides of the triangle ABiv Civ, in which an angle Biv, equal to the supplement of ABC, is known, and the side A Civ, which leads exactly to Problem IV., page 23, where the angle Biv, the side A Civ, and the difference between ABiv and Biv Civ are given. PROBLEM VI.--In a parallelogram are given the angles, one diagonal, and the sum of all the sides, to determine the parallelogram. A = Ꭰ = Analysis. Let ABCD represent the required parallelogram. Produce the side AB until the produced part BE BC; then the side AE AB + BC= half the given perimeter, and is known. Hence, in the triangle ABC we have the angle B, the opposite side AC, and the sum of the sides AB and BC, to construct the triangle ABC just as in the last problem. Then complete the parallelogram ABCD, and it will be the one required. The construction, demonstration, calculation, and limits are precisely as in the last problem. NOTE. When the angle B is a right angle, the parallelogram becomes a rectangle. PROBLEM VII.-In a plane triangle are given the base, the difference of the angles at the base, and the sum of the other two sides, to determine the triangle. Analysis. Suppose ABC to be the required triangle; and having neither side given in position, produce the longer of the unknown sides AC until the produced part CE =CB. Then AE the given sum of AC and CB. Join EB, and on it, produced, let fall the perpendicular AD. Then ABC—< CAB=(ABE—CBE) - - CAB ABE = 25 cor. 6*) ABE AEB-EAB—ABE =2(90°- ABD)=2 BAD. (AEB+EAB)=(I. - ABD= = 180°- 2 ABD Hence BAD is half the given differ ence of the angles. Whence this = Construction.-Draw AB=the given base, and make the angle BAD the given difference of the angles at the base. Through B draw a line DF perpendicular to AD, to which apply AE= the given sum of the sides. Make the angle EBC=< BEA, and ABC will be the required triangle. Demonstration.-Since the angles EBC and BEC are equal by construction, the sides CB and CE are equal. To each add AC; then ACCB AC+ CEAE, the given sum. Also, by the analysis ABC-BAC—2 BAD= the given difference by construction. Q. E. D. = Calculation.-1. In triangle ABD, the first formed, Case 1, find < ABD; then ABE 180° — ABD. = 2. In triangle ABE, Case 2, find angles AEB and BAE; then ACB 2 AEB. 3. In triangle ACB, Case 1, find the sides AC and CB. Limits. The given difference may be taken any quantity less than a right angle; and the sum of the sides, any quantity greater than the base. * I. 32. |