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4 LAND SURVEYOR-IN-TRAINING SAMPLE EXAMINATION TRIGONOMETRIC FORMULAS B 6 a ¢ A B G b c right triangle Solution of Right Triangles os a & a b ¢ e Por angle A:sinA=—- cosA=— tanA=— cotAm— secA=~ cosecA= — ¢ e & a b a Given Required a Le a,b A, Boe tan = Zo col ea Ve = all tos a ae ABb sind ="-csB b= /eFale~al= ¢ Aja Boe B=or—A baacwtA c= —~ sin A Ava Brae B=90—A a=bianA c= b cos A Ave B, a, B= 90° -—A a=csinA bs coosA Solution of Oblique Triangles Given Required A,Ba ac p= 28k C=180°—(A4B) oe SRE sin A sin A bsinA inc Ava,b Bye C sin B= * C= 180° (A4B) cx 2809 a sin A A-B - a,b,c ABoe A+FB=180°—C tan (A>) = (- *) [isa ()] 2 a+b 2) asin c= sinA abe ABC sa Stite sin (4) - [ode 9 2 2 he (5) ° sin | = 2 ete cree = fle TIC TIC LS 2 besin A 2 ae sin Bain C C= 180° ~ (A +B} a, bc area A, bc area area = A,B, G,a@ area area = 2sinA Material on this page copyright © 1998 by the National Council of Examiners for Engineering and Surveying, PROFESSIONAL PUBLICATIONS, INC. = Beimont, CA LAND SURVEYOR-IN-TRAINING SAMPLE EXAMINATION HORIZONTAL CURVE FORMULAS D = degree of curve, are definition 1° = 1-degree of curve 2° = 2-degree of curve PC = point of curve PT = point of tangent PI = point of intersection I = intersection of angle; angle between two tangents L = length of curve from PC to PT T = tangent distance # = external distance R= radius LC = length of long chord WM = length of middle ordinate e=: length of subchord d = angle of subchord Le L LC R= sae T = Rtan 3) = TTT 2sin G) 2cos G) 2 2 ; i 5729.58 8729.58 + = Rein (2) DP =R=579058 DP = pore 2 2 2 R r I I M=Ri1~ os (* = R—~ Reos { - [ 2 2 B+R (5) R-M G) == sec | = = cos | ~ R 2 R 2 d e= 2Rsin (5) 2 ‘Material on this page copyright (©) 1993 by the National Council of Examiners for Engineering and Surveying. PROFESSIONAL PUBLICATIONS, INC. * Belmont, CA