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<br /> TRIGONOMETRIC FORMUL,,E
<br /> F�✓ ,cis'/,� -.:.
<br /> B B B
<br /> e
<br /> c
<br /> i a a c a
<br /> G
<br /> Right Triangle Oblique Triangles
<br /> Solution of Right Triangles
<br /> For Angle A. siu =a ,cos= L tan= a b e e
<br /> C c ' b ,Cot = a,sec=b, cosec= —
<br /> f/ (liven Required a
<br /> a, b A, B,c tan A=a= cot B c = a2 a _ s
<br /> T 1�-- a 1 +_
<br /> 4
<br /> •A.B, b sinA=a=cosB,b=%/ c+a) c--a =c�1—al
<br /> B. b, c B=90°—A,b=a cot A,e= a
<br /> sin A.
<br /> b B,a, B-90'—A,a = b tan A,c= b
<br /> cos A.
<br /> e B, a, b B=90°—A,a=e sin A,b a cos A,
<br /> Solution of Oblique Triantles
<br /> res . Required a sin B
<br /> B; b, c, C b = ¢inA ' C = 180°—(A+B), e = asinC
<br /> sin A
<br /> b sin A
<br /> a, b B,e, C sin B= ;C= 180°—(A-{-B),c = a sin C
<br /> ✓�-r Z G, .. _ a sin A
<br /> ay b,T A,B,c A+B=180°—C,tan?(A—B)—(a—b)tan (A+B)
<br /> • r'
<br /> e =
<br /> a sin C a b
<br /> sin A
<br /> e d B C a=a+b+e x .j;s- s—c
<br /> , , 2 ,sin sA—V b
<br /> be '
<br /> sinaB= ( Xs—c ,C-180°—(A+B)
<br /> ac
<br /> +
<br /> �-
<br /> o Area s.Q 2 c,area
<br /> v A, b, c Areaarea b e sin A
<br /> =
<br /> -2
<br /> A,B,C,a Area area =
<br /> a2 sin B sin C
<br /> _ 2.sin A
<br /> REDUCTION TO HORIZONTAL
<br /> " Horizontal distance=Slope distance multiplied by the
<br /> cosine of the vertical angle.Thus•slope distance=818.4 tt,
<br /> Vert. at►yde=
<br /> 0':5° From TOZ page It cos ti°10'=
<br /> UK' H04zoeta�Alstaooe-44AXAp6Y+iMOD ft
<br /> p dHiaorfzoataE'distanee also=Slope distae of vertical nce minus sl
<br /> lope
<br /> '
<br /> -Fame, res�I1i6 tohnnprecediing eexample,lthe With
<br /> l tat d3staaep fag reanit is obtained.Cosine 60 10'=A8 I—jm=.ONL
<br /> 81A4X fIDt1=1 .818.4-1.81=mac ft.
<br /> When the 1'fse is known,the horizoahl distance is approzoWeIy:-the slope dist-
<br /> ance less,the spare of the Ilse ..vfdgd 4Y twice the slo distance, Thus:rise=14!t;
<br /> s ope
<br /> aimd' lam
<br /> 3
<br /> — — -I ass -
<br /> 's v
<br /> we
<br /> V
<br />
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