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TRIGONOMETRIC 'FORMUL E
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<br /> b O A C
<br /> s - 7 / Right Triangle C?611QW'}i'ri les 3 3
<br /> For solUU46► Triangllee& `.y
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<br /> Angle A. sin = __ � a b
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<br /> o b.cot a _ bb,
<br /> cosec =
<br /> Qivea•• Ytpaired a
<br /> `}, S b
<br /> / 3r a' t B,otan A
<br /> -ftl.cot B,a a } 9 =a 1+ a!
<br /> —606B, o a o—,a =c
<br /> y i. _ .QJ 5 A,a B, b, a B 9°—A,b=acotA a=— gv
<br /> sin A.
<br /> b _-- —
<br /> - •�f y d, b B,a, o =90°d,a = b tan A,c= oos A. 37. 4 $
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<br /> A,o B, a, b Bl 9—W-A,a=0sinA,b=aCosA, 8S6 31 -o
<br /> e, a
<br /> Solution of Oblique Triangles
<br /> l �` — ' ✓ `- 7 6 aC b= sin , C= 180°—(A+B), o=asin Ck-1 a
<br /> Ur
<br /> 63A A
<br /> (J$ (h p l 2 `�•< `. .t7' sin B= b elan A C= 180°—(A-F-B) o �a sin C
<br /> 2; y sin A
<br /> A+B=1800—C,tan}(A—B)=
<br /> asinC a+ b �,9 •d �
<br /> ti�3 �GG o sin A
<br /> d, B, C s=a+2+a,.in}A=
<br /> 213 3
<br /> gg sin)B=�8 a c ,C=180°—(AtB)
<br /> 8—a 2b±a,area
<br /> A, b, c A area = b.a sin A
<br /> ? .a9 * 2
<br /> J l 6 A,B,C,a A area:., sin B sia C 7,3,6 '
<br /> ' 2 siu A
<br /> 9 "`r RF.DLbdiiON TO HOl ZONTAL
<br /> llbrizontal diatance+OSlope distance multiplied by the
<br /> �. p�ine of the verticalinirle.Thus:slope distance=719.4 rt.
<br /> �,tt NerL angle==6°W. From Table,Page IX"cos b°1(y=
<br /> 12-14 7'2 1 o4e a 9868 Horizontal distanoe=819"4X.9669=81&o9 ft.
<br /> 5 MS Horizontal distance also=Slope distance minus slope
<br /> 9/ �'r; ` " / /g I g j j k 1 " v distance a In th 3n of vertical angle). With the
<br /> r q same figures as In the recedi pIs,the
<br /> Z _ I1esizontal distance j,1ing result is obtained.Cosine b°Id=Z886p 1—9968 0111
<br /> g •..........
<br /> - . _ When the rise is lm 819,4X"0041=1.31.$I& , .91=51&09 tL
<br /> S' owp,the borizo- distance is approximately:—tbe slope dist..,,
<br /> r as less the square of the rLe divided by twice the slope distance. Thus:rise=14 fL,
<br /> pe distance-3026 ft Horizontal dlstanae-3OB 6—14X
<br /> ` ... s :> 3 g�3029-0 32=802 29 fL
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