describe the arc DE, and from the point D draw DF perpendicular to CA. Then DF will be the tangent, and CF the secant of the angle C. Because the triangles CAB, CDF are similar, we have CD: CA:: DF: AB, or Also, or R: CA :: tang. C : AB. R: CA: sec. C: CB. (43.) In every plane triangle there are six parts: three sides and three angles. Of these, any three being given, provided one of them is a side, the others may be determined. In a right-angled triangle, one of the six parts, viz., the right angle, is always given; and if one of the acute angles is given, the other is, of course, known. Hence the number of parts to be considered in a right-angled triangle is reduced to four, any two of which being given, the others may be found. It is desirable to have appropriate names by which to designate each of the parts of a triangle. One of the sides adjacent to the right angle being called the base, the other side adjacent to the right angle may be called the perpendicular. The three sides will then be called the hypothenuse, base, and perpendicular. The base and perpendicular are sometimes called the legs of the triangle. Of the two acute angles, that which is adjacent to the base may be called the angle at the base, and the other the angle at the perpendicular. We may, therefore, have four cases, according as there are given, 1. The hypothenuse and the angles; 2. The hypothenuse and a leg; 3. One leg and the angles; or, 4. The two legs. All of these cases may be solved by the two preceding theo rems. CASE I. (44.) Given the hypothenuse and the angles, to find the base and perpendicular. This case is solved by Theorem I. Radius: hypothenuse:: sine of the angle at the base perpendicular; :: cosine of the angle at the base: base C Ex. 1. Given the hypothenuse 275, and the angle at the base 57° 23', to find the base and perpendicular. The natural sine of 57° 23' is .842296; Hence 1: 275 :: .842296 : 231.631=AB. 1 : 275 :: .539016 : 148.229=AC. The computation is here made by natural numbers. If we work the proportion by logarithms, we shall have Ex. 2. Given the hypothenuse 67.43, and the angle at the perpendicular 38° 43′, to find the base and perpendicular. Ans. The base is 42.175, and perpendicular 52.612. The student should work this and the following examples. both by natural numbers and by logarithms, until he has made himself perfectly familiar with both methods. He may then employ either method, as may appear to him most expeditious. CASE II. (45.) Given the hypothenuse and one leg, to find the angles and the other leg. This case is solved by Theorem I. : Hypothenuse radius:: base: cosine of the angle at the base. Radius: hypothenuse: sine of the angle at the base: perpendicular. When the perpendicular is given, perpendicular must be substituted for base in this proportion. Ex. 1. Given the hypothenuse 54.32, and the base 32.11, to find the angles and the perpendicular. By natural numbers, we have 54.32 1: 32.11 .591127, which is the cosine of 53° 45' 47", the angl› at the base. Also, 1: 54.32 .806580: 43.813-the perpendicular. The computation may be performed more expeditiously by logarithms, as in the former case. Ex. 2. Given the hypothenuse 332.49, and the perpendicu lar 98.399, to find the angles and the base. Ans. The angles are 17° 12′ 51" and 72° 47' 9"; the base, 317.6. CASE III. (46.) Given one leg and the angles, to find the other leg and hypothenuse. This case is solved by Theorem II. Radius: base:: tangent of the angle at the base: the perpendicular. ::secant of the angle at the base: hypothenuse. When the perpendicular is given, perpendicular must be substituted for base in this proportion. Ex. 1. Given the base 222, and the angle at the base 25° 15', to find the perpendicular and hypothenuse. By natural numbers, we have 1:222 .471631: 104.70, perpendicular; :: 1.105638 : 245.45, hypothenuse. The computation should also be performed by logarithms, as in Case I. Ex. 2. Given the perpendicular 125, and the angle at the perpendicular 51° 19', to find the hypothenuse and base. Ans. Hypothenuse, 199.99; base, 156.12. CASE IV. (47.) Given the two legs, to find the angles and hypothenuse. This case is solved by Theorem II. Base: radius: perpendicular: tangent of the angle at the base. Radius: base:: secant of the angle at the base : hypothenuse. Ex. 1. Given the base 123, and perpendicular 765, to find the angles and hypothenuse. By natural numbers, we have 123: 1 :: 765: 6.219512, which is the tangent of 80° 51' 57", the angle at the base. 1: 123: 6.300479 774.96, hypothenuse. The computation may also be made by logarithms, as in Case I. Ex. 2. Given the base 53, and perpendicular 67, to find the angles and hypothenuse. Ans. The angles are 51° 39′ 16′′ and 38° 20′ 44′′; hypothenuse, 85.428. Examples for Practice. 1. Given the base 777, and perpendicular 345, to find the hypothenuse and angles. This example, it will be seen, falls under Case IV. 2. Given the hypothenuse 324, and the angle at the base 48° 17', to find the base and perpendicular. 3. Given the perpendicular 543, and the angle at the base 72° 45', to find the hypothenuse and base. 4. Given the hypothenuse 666, and base 432, to find the angles and perpendicular. 5. Given the base 634, and the angle at the base 53° 27', to find the hypothenuse and perpendicular. 6. Given the hypothenuse 1234, and perpendicular 555, to find the base and angles. (48.) When two sides of a right-angled triangle are given, the third may be found by means of the property that the square of the hypothenuse is equal to the sum of the squares of the other two sides. Hence, representing the hypothenuse, base, and perpendicular by the initial letters of these words, we have h=√b2+p2; b= √h*—p2; p= √ h2—b3. Ex. 1. If the base is 2720, and the perpendicular 3104, what is the hypothenuse? Ans., 4127.1. Ex. 2. If the hypothenuse is 514, and the perpendicular 432, what is the base? SOLUTIONS OF OBLIQUE-ANGLED TRIANGLES. THEOREM I. (49.) In any plane triangle, the sines of the angles are proportional to the opposite sides. Let ABC be any triangle, and from one of its angles, as C, let CD be drawn perpendicular to AB. Then, because the triangle ACD is right angled at D, we have or R : sin. A :: AC: CD; whence RXCD=sin. A×AC. For the same reason, R sin. B: : Therefore, BC: CD; whence RX CD sin. BX BC. sin. AXAC=sin. BX BC, sin. A sin. B:: BC: AC. THEOREM II. B (50.) In any plane triangle, the sum of any two sides is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference. Let ABC be any triangle; then will CB+CA : CB-CA :: tang. A+B : tang. 2 A-B Produce AC to D, making CD equal to CB, and join DB. Take CE equal to CA, draw AE, and produce it to F. Then AD is the sum of CB and CA, and BE is their difference. D The sum of the two angles CAE, CEA, is equal to the sum of CAB, CBA, each being the supplement of ACB (Geom., Prop. 27, B. I.). But, since CA is equal to CE, the angle CAE is equal to the angle CEA; therefore, CAE is the half sum of the angles CAB, CBA. Also, if from the greater of the two angles CAB, CBA, there be taken their half sum, the remainder, FAB, will be their half difference (Algebra, p. 68). A E F B Since CD is equal to CB, the angle ADF is equal to the angle EBF; also, the angle CAE is equal to AEC, which is equal to the vertical angle BEF. Therefore, the two triangles DAF, BEF, are mutually equiangular; hence the two angles at F are equal, and AF is perpendicular to DB. If, then, AF be made radius, DF will be the tangent of DAF, and BF will be the tangent of BAF. But, by similar triangles, we have |