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<br /> r H -Al ' 3 g y TRIGONOMETRIC FORMULIE
<br /> 36,z7- 339r 33d B I3
<br /> 4 Z 1 ? � � 39B B
<br /> 3 o 1 .7 y & 391241 38, .Y � 3637 c a a a a
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<br /> 9.yr i Y7 37 91 9 .d'¢'¢t 3 o�L9 :.�_ C ��b C A b '"
<br /> >ZigBt�Frlsngle Oblique Triangles
<br /> 3 r 3 •` Z 7 1, / 3 31--.0 Solution of Right Triangles
<br /> 2.3 / z• f 6-27 For Angle d,� sin =a,cos= b,tan= a,cot = b,sec=a, cosec= a
<br /> 2 32 b 2 ZJ' 3 95 N r7 ��_ a a b a b a
<br /> y o1' l 5 6 87 (uvea Iiephired
<br /> ._,u 7 . 3 3-- 7s 3 4, b 1!;B,c tan d=b= cot B,a = a 2 = a 1 } 2
<br /> 3,z't J J'7 �f.3q_ a. a A,;B, b sinA-Q=coeB,b=V/ c+a o--a)
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<br /> Z 35� 0
<br /> 7 3. _. S 3¢'' A,a B b, a B=90°-A,b = a cot A,o= a
<br /> `mss j `• y/ �'�3 J/.� .�/!6 sin A.
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<br /> 3 ti.L6 _ .
<br /> J-.i7 z¢� G� �9 / s `�3�7 ? A, b B;u, c B=90°-A,a = btanA,c= cosA.
<br /> 9b /
<br /> t'ZI,1f-- J�� A,a Bi'-a, b B=90°-A,a =c sin A,b=o cos A,
<br /> s'e 7 s°o Solution of Oblique Triangles
<br /> 3-G.,s 7 3191• g1 6•y J�'03 Given Required a sin B
<br /> 3 L1 -3, 67. h Lf - - d, B,a b, a, C b = , C = 180°-(A+B), o = asinC
<br /> '341',1 5" - - c5-¢,3,7 sin A sin A
<br /> 37, >^ pt �' 9 aa/ bs •
<br /> n C
<br /> 3 �j 3100 A,-,%, ba B,..o, C sinB= aA,C= 180°-(A+B),c = sin
<br /> 3 9 3 3 7 8 / 3.24
<br /> G a, b, C. d,B, o A+B=180°-C,tan},(A-B)= a-b)caani?bA+B)�
<br /> J o y .fir,'.' 4S-06 ZL;9x,,39' _ a sin C
<br /> y 3 r ) 7.1 1 L� - 3 3.7 2.b! 67,02 a sin A
<br /> _3 4 S --� �, yL a, b, a 'd, B, C. a=a+b+a,sinA=�ir
<br /> jig' 391/ y � Cf I� J7 � SS- 2 be
<br /> � 7a. 33i _
<br /> j---' 3 7 r '� 5 3 3 •i•-� z S3 go�z sini'B=�ls 4141 ,C=180°-(A+B)
<br /> 3l• �-- s a+b+c
<br /> 3 t B 3 o y 3 3 e-• 7 a. 5,f,G a, b, c Area s= 2 , area
<br /> 'P-0 A, b c Area be sin A
<br /> 3 9 3 ,� ) 2,a i , area =
<br /> z r z7 7 . !".6 S-,s•.b 9 a2 sin B sin C
<br /> 7_ `?x,4 7 1•L A,B,C,a Arep area = 2 sin A
<br /> _ 91
<br /> e3 7Z' REDUCTION TO HORIZONTAL
<br /> ti 0 y 3 91 3 3 ' (-4 3 O 3` 63.4 7 Horizontal distance=Slope distance multiplied by the
<br /> 9 r 9 c� 3 7 7 2.6z ae cosineoftheve tioalangle.Thusalopedistance=319.4ft.
<br /> 3 4.f 3 s-b p' � Vert. angle=5 NY. From Table,Page IX.cos b°lor=
<br /> 3 T, 3 �. o oPe 9969. Horizontal distance;319.4X.9959=318.09 ft.
<br /> 3 9 3 9 3 3 g1 �rg1e iorizontal distance,also=Slope distance minus slope
<br /> y 3 0 ����.� �� �-� e distance times (1-cosine of vertical angle). With the
<br /> -3 3 3 •L �-=y ��'' same figures as In the preceding example,the follow-
<br /> 1 p f 3 Horizontal distance ing result is obtained.Cosine 60 10?=.9958.1-.9959=.0041.
<br /> -_ `s - 8l&4X.0041=L3t.319.4-1.31=31&09 ft.
<br /> �F n 4 / G 3 7 f ,�7� 3S•3 When the rise is known,the horizontal distance is approximately:-tbe slope dist-
<br /> -73.4 ;:ante less the square of the rise divided by twiceahe slope distance. Thus:rise=l4 fL,
<br /> 3 3•' - 3 9
<br /> 3C �,l .72 slope distance-302.6 ft. Horizontal distanoe=302.8-14 X 14=3029-0.82=302.28 ft.
<br /> 2 X Sots
<br /> .MADE IN U.l.A.
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