pothenuse and one leg be given, extracting the square root of the difference of the squares, will give the other leg. Let h the hypothenuse } p the perpendicular of a right angled triangle. Then h2=b2+p2, or (Alg. 296.) h= √b2 +p2 b = √h2 — p2 Ex. 1. If the base is 32, and the perpendicular 24, what is the hypothenuse? The square of 32 is 1024 of 24 is 576 The sum of the squares is 1600 The root of this sum is 40, the hypothenuse. 2. If the hypothenuse is 100, and the base 80, what is the perpendicular? The square of 100 is 10000 6400 The difference of the squares is 3600 The root of which is 60, the perpendicular. 3. If the hypothenuse is 300, and the perpendicular 220, what is the base? 2 Ans. 300-22041600, the root of which is 204 nearly. 141. It is generally most convenient to find the difference of the squares by logarithms. But this is not to be done by subtraction. For subtraction, in logarithms, performs the office of division. (Art. 41.) If we subtract the logarithm of b2 from the logarithm of h2, we shall have the logarithm, not of the difference of the squares, but of their quotient. There is, however, an indirect, though very simple method, by which the difference of the squares may be obtained by logarithms. It depends on the principle, that the difference of the squares of two quantities is equal to J the product of the sum and difference of the quantities. (Alg. 235.) Thus h2 —b2 =(h+b) × (h—b) as will be seen at once, by performing the multiplication. The two factors may be multiplied by adding their logarithms. Hence, 142. To obtain the difference of the squares of two quantities, add the logarithm of the sum of the quantities, to the logarithm of their difference. After the logarithm of the difference of the squares is found; the square root of this difference is obtained, by dividing the logarithm by 2. (Art. 47.) Ex. 1. If the hypothenuse be 75 inches, and the base 45, what is the length of the perpendicular? 2. If the hypothenuse is 135, and the perpendicular 108, what is the length of the base? Ans. 81. SECTION IV. SOLUTIONS OF OBLIQUE ANGLED TRIANGLES. ART. 143. THE sides and angles of oblique angled triangles may be calculated by the following the orems. THEOREM I. In any plane triangle, the sines of the angles are as their opposite sides. Let the angles be denoted by the letters A, B, C, and their opposite sides by a, b, c, as in Fig. 23 and 24. From one of the angles, let the line p be drawn perpendicular to the opposite side. This will fall either within or without the triangle. 1. Let it fall within as in Fig. 23. Then, in the right angled triangles ACD and BCD, according to art. 126. R:6:: Sin A:p R:a:: Sin B:p Here, the two extremes are the same in both proportions. The other four terms are, therefore, reciprocally proportional: (Alg..387.*) that is, a:b::Sin A: Sin B. 2. Let the perpendicular p fall without the triangle, as in Fig. 24. Then, in the right angled triangles ACD and BCD; R:b:: Sin Ap R:a:: Sin B:p Therefore as before, a:b::Sin A: Sin B. Sin A is here put both for the sine of DAC, and for that *Euclid 23. 5. of BAC. For, as one of these angles is the supplement of the other, they have the same sine. (Art. 90.) The sines which are mentioned here, and which are used in calculation, are tabular sines. But the proportion will be the same, if the sines be adapted to any other radius. (Art.119.) THEOREM II. 144. In a plane triangle, As the sum of any two of the sides, So is the tangent of half the sum of the oppo- To the tangent of half their difference. Thus the sum of AB and AC (Fig. 25.) is to their difference; as the tangent of half the sum of the angles ACB and ABC, to the tangent of half their difference. Demonstration. Extend CA to G, making AG equal to AB; then CG is the sum of the two sides AB and AC. On AB, set off AD equal to AC; then BD is the difference of the sides AB_and AC. The sum of the two angles ACB and ABC, is equal to the sum of ACD and ADC; because each of these sums is the supplement of CAD. (Art. 79.) But, as AC=AD by construction, the angle ADC=ACD. (Euc. 5. 1.) Therefore ACD is half the sum of ACB and ABC. As AB=AG, the angle AGB=ABG or DBE. Also GCE or ACD=ADC =BDĚ. (Euc. 15. 1.) Therefore, in the triangles GCE and DBE, the two remaining angles DEB and CEG are equal; (Art. 79.) So that CE is perpendicular to BG. (Euc. Def. 10. 1.) If then CE is made radius, GE is the tangent of GCE, (Art. 84.) that is, the tangent of half the sum of the angles opposite to AB and AC. If from the greater of the two angles ACB and ABC, there be taken ACD their half sum; the remaining angle ECB will be their half difference. (Alg. 341.) The tangent of this angle, CE being radius, is EB, that is, the tangent of half the difference of the angles opposite to AB and AC. have then, We CG the sum of the sides AB and AC; GE=the tangent of half the sum of the opposite angles; EB the tangent of half their difference. 145. If upon the longest side of a triangle, a perpendicular be drawn from the opposite angle; As the longest side, To the sum of the two others; To the difference of the segments made by the In the triangle ABC (Fig. 26.) if a perpendicular be drawn from C upon AB; AB: CB+CA:: CB-CA: BP-PA* Demonstration. Describe a circle, on the centre C, and with the radius BC. Through A and C, draw the diameter LD, and extend BA to H. Then by Euc. 35, 3, And AH=HP-PA=BP-PA (Euc. 3. 3.) If then, for the three last terms in the proportion, we substitute their equals, we have, AB:CB+CA:: CB-CA: PB-PA 146. It is to be observed, that the greater segment is next the greater side. If BC is greater than AC, (Fig. 26.) PB is geater than AP. With the radius AC, describe the arc * See note G. |