Page 266 - Demo

260 P%u00ebrgjigjet N%u00eb trek%u00ebnd%u00ebshin BDC, sinB = ha , h = a sinBb a sinB = b sinA. Duke pjes%u00ebtuar me sinA dhe sinB marrim asinA = bsinBc Syprina = 12 %u00d7 AB %u00d7 h = 12 %u00d7 c %u00d7 b sinA ose 12 %u00d7 c %u00d7 a sinB Syprina = 12 bc sinA ose 12 ca sinB*13 a b2 = h2 + x2 ( Teorema e Pitagor%u00ebs)a2 = h2 + (c %u2212 x)2 = h2 + c2 %u2212 2cx + x2 (Teorema e Pitagor%u00ebs) Duke zbritur marrim a2 %u2013 b2 = c2 %u2013 2cx. K%u00ebshtu q%u00eb a2 = b2 + c2 %u2013 2cx N%u00eb trek%u00ebnd%u00ebshin ADC, cosA = xb , ndaj x = b cosA. Duke z%u00ebvend%u00ebsuar te formula e par%u00eb, marrim a2 = b2 + c2 %u2212 2bc cosAb 2bc cosA = b2 + c2 %u2212 a2 Pjes%u00ebtojm%u00eb me 2bc dhe marrim c a2 = h2 + (x + c)2 = h2 + x2 + 2cx + c2 (Teorema e Pitagor%u00ebs). b2 = h2 + x2 Zbritja na jep a2 %u2013 b2 = c2 + 2cx, k%u00ebshtu q%u00eb a2 = b2 + c2 + 2cx N%u00eb trek%u00ebnd%u00ebshin ADC, cos(180%u00b0 %u2013 %u0486A) = xb , k%u00ebshtu q%u00eb x = b cos(180%u00b0 %u2013%u0486A). Duke z%u00ebvend%u00ebsuar te formula e par%u00eb, marrim a2 = b2 + c2 + 2bc cos (180%u00b0 - %u0486A). Por, cos(180%u00b0 %u2013 %u0486A) = %u2013 cosA. K%u00ebshtu, a2 = b2 + c2 - 2bc cosA.D A BCa h bx cc + x7.4A1 a AC = 20 cm b AG = 25 cm c %u0486GAC = 37%u00b0 d %u0486FAB = 43%u00b0 2 a PS = 15 cm b i %u0486SPR = 28%u00b0 (i rrumbullakosur n%u00eb nj%u00eb grad%u00eb) ii cos %u0486QPR = %u0486QPR = 56,144%u2026%u00b0 %u0486SPR = 56,144%u2026%u00b0 : 2 = 28%u00b0 3 a 7,2 cm b %u0486PSU | 64%u00b0 c %u0486PMU = 71%u00b0 4 a i QV = 20 2 cm ii PV = 20 3 cm b i %u0486VQR = 45%u00b0 ii %u0486VPR = 35%u00b0 5 a h = 40 cm b AC = 78 cm c EC = 88 cm d %u0486ECA = 27%u00b0 e %u0486EBA = 53%u00b0 6 a i AQ = 61 cm ii 25%u00b0 iii 909 cm2b30 cmP QA XB70 cm60%u00ba i cos 60%u00b0 = XB30 , XB = 15sin 60%u00b0 = QX30 , QX = 15 3! !n%u00eb trek%u00ebnd%u00ebshin AQX, AX = 70 %u2212 15 = 55AQ2 = 552 + (15 3)2 = 3700, AQ = 61 cm ii N%u00eb trek%u00ebnd%u00ebshin AQX, tg %u0486QAB = QXAX = 15 355 , %u0486QAB = 25%u00b0 iii Syprina AQB = 12 %u00d7 70 %u00d7 15 3 = 909 cm2 c i AR = 209 cm ii %u0486RAC = 8,1%u00b0 7 a Brinja m%u00eb e shkurt%u00ebr = 41 mm Brinja m%u00eb e gjat%u00eb = 54 mmb 130%u00b0 (n%u00eb grad%u00ebn m%u00eb t%u00eb af%u00ebrt) 8 a RPS = 22%u00b0 (n%u00eb grad%u00ebn m%u00eb t%u00eb af%u00ebrt) b Syprina e trek%u00ebnd%u00ebshit RPS = 3,5 cm2c %u0486RQS = 25%u00b0 (n%u00eb grad%u00ebn m%u00eb t%u00eb af%u00ebrt) d 3,1 cm2*9 a 1,4 m b 56%u00b0 7.4Z1 Po, h = 50,154 > 50m2 %u0486ABC = 161%u00b0 (n%u00eb grad%u00ebn m%u00eb t%u00eb af%u00ebrt)3 a d = 22 km, kursi = 065%u00b0 (n%u00eb grad%u00ebn m%u00eb t%u00eb af%u00ebrt)b Po, distanca nga P = 9,440%u2026km < 10 km.4 a i sinD = 35 ii cosD = 45b i cosD = 513 ii tgD = 125c i sinD = 0,6 ii tanD = 0,755 OAH, (sinD)2 + (cosD)2 = (OH)2 + (HA)2= O2 +A2H2 = H2H2 = 1