By Euc. II, 13, 2dc=a2 +c3-b2, and by substitution, 2ac: a2+c2-b2::R: cos B; and, converting the proportion into an equation, 2ac cos B=R(a+ca-ba). Dividing by 2ac, cos B=RX (a2+c2-b2)÷2ac, and 2ac cos B÷R=a2 + c2-b2. 2ac cos B÷R-a2 —c2 ——b2 ; and, by changing signs, b2=a2+c2-2ac cos B÷R; and, extracting the square root, b=√a2+c2-2ac cos B÷R. -- Draw from A (Fig. 23.), AD perpendicular to CB; make CA radius, and let CD be represented by d then, b;d:: Rcos C, and 2ba: 2da::R⚫ cos C. As 2da b2+a2 — c2, by substitution, 2bab2+a2-c2::R; cos C, and by multiplying extremes and means, 2ba cos C=RX (ò2 +a2 -c2) and dividing by 2ab, cos C=R×(b2+a2 —c2)÷2ab. But 2ba cos C÷R=b2+a2-c2; therefore c2=b2+a2 — (2ba cos C)÷R, and c=√b2+a2—(2ab cos C÷R.) 2bca2-b2-c3::R: cosA(Fig. 24.); and, multiplying extremes and means, cos A 2bc=RX (a2 —b2 —c2); and, dividing by 2bc, cos A=R×(a2-b2-c2)÷2bc. cos A 2bc ÷R=a2 —b2 — c2, and by transposing and changing signs, a2=b2+c2+(cosA2bc)÷R, and by extracting the square root, a√ba+c2+(cosA2bc)÷R. By CASE IV. (Art. 221.)p=√2h×2(h—a)×2(h—b) ×2 (h-c) 2c: multiplying each side by 2c, 2cp=√2hX √2(h-a)×2(h-b)×2(h—c), and pc=2√2h × 2(h − a) × 2 √(h-b)×2(h−c). SinA×C=R×2√2h ×2(h—a)×2(h—b) ✔×2(h−c)÷c. Multiplying each side by c, and then divi2R ding by bc we have sin A= bc √2h×2(h—a)×2(h−b) ×2 ✔(h−c). Theorem D. By Art. 210, 2 sina R2-RXcos A: But in Theorem B, cos A=RX(ba+ca-a3)÷2bc; therefore, by substitution, 2sin2A=R2-R2 × (b2 +c2 —a2)÷2bc. And as R2 R2X2bc÷2bc, and-R2 × (b2 +c2 — a2)÷2bc=R2X (a2-ba-c2)÷2bc: By substitution, 2sin2A=R2×2bc÷ 2bc+R3× (a2-b2-c2)÷2bc, and 2sin3 A = R3× (2bc+ a2-b2-c2)÷2bc. But 2bc +a2-b2-c2=a2 (b2 + c2) 2bc+a2-b2— —(b2 2 +2b3c=a2(b+c); and by Alg. 235, a2-(b+c)2 = (a−b+c)×(a+b-c). Let h=(a+b+c); then 2h=a+b +c, and a-b+c=2h-2b, and a+b-c=2h-2c. By substitution 2sin2a=R2 × 2(h—b)×2(h-c)÷2bc. Dividing by 2, sina (h-b)× (h-c)÷bc, and extracting the square root, sin A=R√(h—b)×(h—c)÷bc. = Theorem E. By Art. 210, 2 cos21A=R2+R×cos A: But by Theorem B, cosA=RX (b2+c2-a2)÷2bc; hence, by substitution, 2cos2A=R3 +R2 × (b2+c2 —a2)÷2bc. And, as before shown, R2 R2 × 2bc2bc, by substitution, 2 cos 21A 2bc÷2bc + R2 × (b2+c2-a2) ÷ 2bc =R2X(2bc+b2+c2-a2)÷2bc. But by Alg. 235, 2bc+ b2 +c2 a2 = (b+c)2—a2 = (b+c−a) × (b+c+a). Let h=(a+b+c), then a+b+c=2h, and by substitution, 2cos21 AR2 x2(h-a)× 2h÷2bc, and cos2A=hX (h—a)÷bc, and by extract. the sq. root, cos21A=√h× (h− a)÷bc× R. Theorem F. By Art. 216, tangent= radius X sine sine. Then, by Theorems D and E, tan A R sin¦ A co cos A R2 (h-b) (h-c) bc RA (h—b)(h—c) __ h(h—a) — (Alg. 162.)R、 (h-b)(h-c) bc bc be bc AREAS OF FIGURES BOUNDED BY RIGHT LINES. Problem I. To find the area of a parallelogram, square, rhombus, or rhomboid. Example 1. 23.5=length Ex. 2. 66=length Example 3. 22 × 11 × 2=484=square feet in the larger two sides; and, 17×11×2=374=square feet in the smaller two sides. 484+374-858=the number of square feet in the four sides of the room. Example 1. Art. 5. By Trigonometry (Art. 134.) R : BC: sinB CH (Fig. 2.); or, R¦ sin B::AB × BC the area; and, by the tables, 10.0000000: 9.9498809:: 3.38667733.3365582=2170.492253873=the area. Example 2. By Trigonometry (Art. 134.) R; BC: :sin BCH; and, by the tables, 10.0000000; 9.9805963:: 3.6521496: 3.6327459-4292.851778656126482=the area. Example 2. Art. 6. 4 11' the hight of the window 3 5' the width of the window 20′7′′ Example 1. Art. 7. 63-36-137-13 width of the cloth. Example 2. The square root of 289, according to the rule in common arithmetic 17 a side of the piece of land. Example 3. 30x221=675=the area of the floor in feet, which divided by 9=75=the area of the floor in yards, 14 yards 5 quarters, and 75 yards=300 quarters, 300÷5-60 the answer. Example 4. 936×104-97344=the area of the parallel. ogram, whose square root=312=the answer. Example 5. 2 feet 8 inches 32 inches, and 5 feet=60 inches, 32 × 60=1920. 8×10=80=the number of square inches in one pane, and 1920÷80=24=the number of panes in the window. PROBLEM II. To find the area of a triangle. Example 1. Art. 8. 65×31.2=2028, and 2028÷2 = 1014. Example 2. 3ft. 2′×2ft. 9′=8ft. 8′ 6′′ 8.102, which divided by 2=4ft. 51=the answer. Art. 9. By Trigonometry (Art. 184, Fig. 2.) R; sin B:: ABXBC the area; and, by the tables, 10.0000000: 9.9030894::3.4039780: 3.3070674=2028-twice the area of the triangle, and divided by 2,=1014=the area of the triangle. Art. 9. b. See Figure 5. R: sinB::BC: CP, and, sin ACB:sinA::AB:BC; therefore (Alg. 390, 382.)R×sin ACB sinAX sin B::AB× BC: CPX BC; and, multiply. ing analogous terms by AB, RXsin ACB sinAxsinB:: AB2 × BC: CP × AB × BC; and dividing analogous terms by BC,RXsinACB; sinA× sinB::AB2: AB×BC Taking sines &c. from the tables, we have by this last proportion 19.9729858 19.8217840: 3.5117417 : 3.3605479 =2294, which divided by 2=1147=the area. By Trig. 221, S=‡√(b+c+a)×(b+c−a) × (a+b−c) VX(a-b+c). Let h=1(a+b+c), then 2h=a+b+c, 2h-2a =b+c-a. 2h-2c=a+b-c, and 2h-2b-a-b+c; then by substitution, S=‡√2h×2(h—a)×2(h—b)×2(h—c), and by removing a factor from under the radical sign, S= h(h-a)(h-b)(h-c). This formula, as reduced is used in the solution of the next four examples. Example 2. Putting h for half the area of the triangle, S=√161X53X81X27=√18661671-4319 nearly. By logarithms, 2.2068259+1.4313638+1.7242759+1.90848 50=7.2709506. By Trig. 47, the square root of a number is its logarithm divided by 2; hence, 7.2709506÷2=3.635 4753-4319, which is the area a little too small. Example 3. By the above formula, the area=√430×59 ✓ 166×205/863341100=29380, which is a little smaller than the true answer. By logarithms, 2.6334685+1.77 08520+1.2201081+2.3117539-8.9361825, which divided by 2 4.4680912=29382.67 nearly. Example 1. Art. 11. a2 √3-1342 × 1.732=1156 × 1.732=289×1.732=500.548 a little more than 5001= the area. = Example 2. a2 √3-11002×1.732=10000×1.732= 2500×1.732=4330.000÷160=27 acres. PROBLEM III. To find the area of a trapezoid. Figure 4. R: BC::sinB: BC; and, by the tables, 10. 0000000 1.4913617: 9.9729858: 1.4643475= 29.13. (46+38)=184-42, and 42x29.13=1223.46=the area of the trapezoid. Example 2. (65+38)=103-51.5, and 51.5×27=139 0,5 the contents of the field. Example 1. Art. 13. 18×14.6-262.8, and 16x7.3= 116.8. The sum of 262.8 and 116.8 is 379.6=the area. Example 1. Art. 14. See Figure 8. By Trig. 90, R: sin N::2 prod. diagonals: 2 area of the trapezium. Substituting numbers, 1 sin54°::2294 twice the area; and, by the tables, 10.0000000; 9.9079576::3.3605934: 3.26855 10=1856 twice the area. 1856=928=the area. Example 2. Using the proportion as in the preceeding example (Trig. 90.); and, by the tables 10.0000000; 9.98284 16: 3.8979019: 3.8807435=7599 twice the area and 7599 divided by 2=3799.5=the area. 2 Fig. 33. Trig. Let the side AB be represented by a, BC by b, CD c, AD by d; and let s represent the sum of either the angles BAD, BCD, P'CD, also let h represent of (a+b+c+d). Making AD radius, R¦s::AD ; DP, and DP=s× AD=ds; also, making CD radius, R; s:: CD ; DP' and DP'=CD×s=cs. By Euc. I, 47. AP2 =AD3 —DP2 = d2 — d2s2, and CP2= CD2 —DP'2 = c2 — c2s2. By extracting the square root in the preceding, AP = √ d2 — d2 s2=d√1-s2, and CP'= √ c2 —c2 s2 = c √ 1 − s2. Again, BP=a—d√1—s2, and BP'=b+c√1—s2; therefore BP2+DP2=DB2 = BP2+DP2. By substitution, 2 |