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TRIGONOMETRIC FORMULAE
<br /> B B B
<br /> y v
<br /> [' a a
<br /> f
<br /> { a A a
<br /> Right Triangle a `-----Oblique Triangles C
<br /> Solution of Right Triangles
<br /> J a b
<br /> 5 > - For Angle A. sin —— cos— tan— a o
<br /> e . e � b ,.cot= —,sec= 'cosec
<br /> = —
<br /> '< (liven Required a
<br /> J ay A, B,e tanA=a- cotB,c= a + •=a 1 + _
<br /> a9
<br /> A, B, b sin A——c
<br /> = a a
<br /> , c— =c —
<br /> t_
<br /> b, c' B=90°—A,b =a cotA,c= sin A.
<br /> 9. J b��� /u T B',a, o B=90°A '
<br /> ,a = btan.4c=
<br /> /- l cos.A.
<br /> -B,a, b B=90-A,a=csin A,b—ccosA,
<br /> !>tiss : Required
<br /> Solution of Oblique Triangles
<br /> asinB
<br /> �
<br /> B,a b, c, 0 b snA ' C= 180°—(A+B), c=asinC
<br /> sin A
<br /> bin A
<br /> ? dr cj,.b B,c, C sin B= ,C= 180°—(A+B),c _ a sin C
<br /> �- _ a sin A
<br /> �L o; b, 0 A,B, c A+B=180°—C,tan (A—B)—— a—b tan (A+B)
<br /> a+ b
<br /> V V a sin C
<br /> yc =
<br /> sin A
<br /> !9 ,...y b, a A, B, C a—a °�sinjA—�
<br /> � _ "«► sin}B=� —oa ,C=180°=(A+B)
<br /> a, b, c Area 8=E ,area = s a—a s— a—c
<br /> A, b, a Area b e sin A
<br /> f -1 area = 2
<br /> d,B,C,a Area area a as sin B sin Ci
<br /> '
<br /> `� ? 2snnA
<br /> REDUCTION TO HORIZONTAL
<br /> > Horizontal dtstanoe=Slope distance multiplied by the
<br /> --2— cosine otthf ver icalantde.Thus:siopedietsnee=E10.41t.
<br /> � ° Yert angle—b°16". From Tabid,Pass XX cos 561W=
<br /> 9660. Horizontal distance-19.4XAYo—Rd9I
<br /> J.o0.Li$ p 11O�°ntal distance also=Slope=istance minus slo
<br /> dlstaace times(1-cosine o! cal an91e). With t�e
<br /> saasa as in the preceding example,JIM
<br /> distance lna'r iIssobtained.Costae 6°IN=.9960.1—.9969=.6641.
<br /> N&4XAU�l.SL alarl u=>ns oo to
<br /> 'A'hmae rue is thwhorlsontiti dhRasce —flue slope dist
<br /> dace less the sdnsrs of b!twtee diNanewe' ' ss:rLw=14 i4,
<br /> elope distance I Le it. horizontal atstanoe.ft6-1-�4
<br /> 2 X 8626
<br /> MADE IN Y.S.A.
<br /> OF
<br /> r
<br /> 1, _. - • ,
<br /> f •
<br />
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