Since 30° and 60° are complementary, any ratio of the one is equal to the co-ratio of the other. 109. The ratios of and of a right-angle. Describe the isosceles triangle ABC, right-angled at A. Pro duce AB to D, making BD = BC. Join DC. Then, as in Art. 107, ABC is a right-angle; and the angles BDC, BCD being equal by Euc. I. 5, each of them is of a right-angle by Euc. I. 32. Take AB = AC=1. Then BD = BC = √2, as in Art. 107. Again DC2 = DA2 + AC2 = (2 + 2 √√2 + 1) + 1 = 4+2 √√2. Similarly the other ratios may be written down, and their denominators may be rationalised by an algebraic process. may The same figure gives the ratios of 671°. By the following simple geometrical construction; the other ratios be at once written down with rationalised denominators. With centre B and radius BD or BC, describe the semi-circle DCE, cutting BA produced in E. Join CE. 1 Then DCE in a semi-circle is a right-angle. Also L .. LACE=complement of DCA= 2 ADC=22°. .. CE2=AE2+AC2=(2−2√2+1)+1=4−2/2, .. from ▲ EDC, sin 22°√(2-2) and cos 22°=√(2+√2), and from ▲ ECA, tan 223°=√2-1 and sec 2210=√(4−2√2). Let ABC be the half of an equilateral triangle, so that, as in Art. 108, BCA = 60°, ▲ ABC = 30°, ▲ BAC = 90°. AB to D, making BD = BC. Join DC. Then, the angles BDC, BCD being equal, each of them is half angle ABC, i.e. 15°. Produce [Euc. I. 5. [Euc. I. 32. Take AC=1. Then BD = BC= 2, and AB= √3, as in Art. 108. DC2 = DA2 + AC2 = (4 + 4 √√3 + 3) + 1 = 8+ 4 √√3 = 6 + 2 √/12 + 2, Similarly the other ratios of 15° and 75° may be written down, and their denominators may be rationalised by an algebraic process. By the same construction as in the preceding article, the other ratios may be at once written down with rationalised denominators. With centre B and radius BD or BC, describe the semi-circle DCE, cutting BA produced in E. Join CE. Then DCE in a semi-circle is a right-angle, .. LACE complement of DCA= 2 ADC=15°. = .. CE2=AE2+AC2=(4−4√3+3)+1=8−4√/3, .. from A EDC, sin 15° (√6-2) and cos 15°=(√6+√2), = and from ▲ ECA, tan 15° =2-√3 and sec 15° = √6−√2. 111. We may proceed to bisect the angles 221° and 15° in The student will easily show that the same way. In precisely the same way the student may prove geometrically the general formula 112. The ratios of 1, 2, 3, 4 of a right angle. [18°, 36°, 54°, 72°.] Bisect any straight line AB in C. Draw BD at right angles and equal to AB. Join CD. Produce AB to E, making CE equal to CD. With centres A and B and radii each equal to AE describe two circles cutting in F. Join FA, FB, so that FA and FB are each equal to AE. The angle AFB shall be 36°. AB is bisected in C and produced to E, For .. AE. BE + BC2 = CE2 = CD2 = BD2 + BC2, .. AE. BE = BD2 or AB2. From FB, which is equal to AE, cut off FO equal to AB; then the remainder BO = remainder BE. .. the above equation becomes BF. BO=AB2, :. BF: AB=AB : BO, .. the triangles FBA, ABO, having the sides about their common angle B proportional, are similar. .. LAOB= L FAB = L FBA and AO = AB = FO. But AO FO, .. LAFO = ▲ FAO, :. ¿AOB = twice LAFO. J. T. 5 |