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_ TRIGONOMETRIC. FORMUL M. <br /> 37 35 B B B <br /> r <br /> o c a e <br /> / <br /> le Oblique Trian gl e <br /> Solution of Right TiIiangles <br /> For $ =a b I = b co <br /> a o <br /> a tan— <br /> a <br /> ' T,cot �.sec=b, eeti <br /> a,b ,o tan A=a= cot B o = as s a <br /> b ��= a 1 � s <br /> a, og. b sin A—a=cos B,b=V a+a <br /> U, A,a b o —_ ° a os <br /> ..� B90 A,b p.aootA,a= <br /> A ---- <br /> a, o B=90°—A,a = btan A,a= b <br /> coo A. <br /> A b B-90'—A,a =o sin A,b=a coo A, <br /> Solution of Oblique Triangles <br /> i Given )tequired a sin B- <br /> B,a b, e, C b = sin A ' C= 180°—(A+B),o =a sin C <br /> g5- 4/ 40 sin A <br /> sob27. 2 ayb B,a, C sinB= bBin ArC= 180°—(A-( B).o = asinC <br /> .I7.» a sin A <br /> c a, 6, 'C' d;B,a tan A+B=180°—C, �(A—B)= a—b tan (A-f-B) <br /> �3S aB.Z S asin C a+ b <br /> - sin A <br /> Z q .z.3 , b, o , B, C s=a�b <br /> Z+c <br /> ,sinA= <br /> I <br /> d,,3 sing= J e—a)(e—o ,C.180°—(A+B) - <br /> a, c Area 8= 2 , area <br /> Z 3 S A, b, c Areaarea = b o sin A <br /> 2 <br /> 1 A,BX,a Area area =as sin B sin C <br /> 2 sin A <br /> 1 g,9 REDUCTION TO HORIZONTAL <br /> Horizontal distance=$ w81lo distance multiplied by the <br /> cosineottheverticalaa¢le. a:slopedistance9.4ft. <br /> Vett anile=6°101. From Tabid,Page IX.odd.b°W= <br /> �oPe 9860. Hosiaontal:distance=819.4X.9969.81&09 ft. <br /> S M10 Horizontal dtagnoe 930=Slope distance minus slo <br /> distance times(1—ceeine of_verticpi anse®)t Wfoll <br /> - same fiQaes�s in he a� _ + <br /> Horizontal distance i�y result iss obtained.Cosine a s�99N 1=8969=.o f <br /> When the rise is 819 4X.t1011=181.il&4-1:81=81&09 it <br /> knows.the horizontal distance is appcozimatelg.-4be slope dist- <br /> aaee less the square of tlie3fse divided by twice the slope distance. Thna:rue=14 ft. <br /> slope distance-Me ft Horizontal dispmee—Me—14 X 14-4ae--0.82=X&]Dft. <br /> - — 2Xaoae <br /> .ry.., MADS IN U.S.A.IV <br /> ?. <br /> -. • <br />