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TRIGONOMETRIC FORMUL:E �-Z
<br />�+�� .4-9 1) !� •�7 1�9, 7� -�_ 33 _ � �._¢_ Wil• ,��<�.
<br />�_ 3 `'' lZ e - [•5 a c `b 5 a c 1 .9 ..,a
<br />,e�73 3`2 I 47,12 %?Z �Z �_�, la, ybt�33- A
<br />1Z.1� t J� J .b .'r GY b C b �'i.
<br />3 . 1 U '- Right Triangle . Oblique Triangles
<br />Solution of Right Triangles b
<br />+-7 �• J 9 Q �� .g' For Angle A. sin = c ,cos = c ,tan = b cot = a ,sec = b , cosec =
<br />I137a
<br />�� _ (c• �y p ca �q - Given Required `
<br />[•5 .p - �} ! j .: :, a, b .1, Z1 L,c tan _�. = L =cot Ii c= 1/a=' T b2 = a 1 a
<br />a2
<br />a, c A, B, b sin =cosB,b=�/(c Fa)(c-a) =c 1-az
<br />,l
<br />Z A, a B, b, c B=90°-A, b = a cot q, c-
<br />I l 'AZ
<br />�f 6 _31
<br />sin .9.
<br />r ---7773 b
<br />I 5 �tc' ��3 �J A,b B,a, c 8=90°—A,a=btan.l,c=
<br />cos A.
<br />% A,c B, a; l+ B-90°—A,a=csin 11,b-ccos A, 1-2 7.
<br />3
<br />Solution of Oblique Triangles-
<br />Given Required a• sin •B a sin C �r
<br />'B, .B, a. b, c, C b = G = 180°—(-3 } B) c =
<br />sin A s:n A 15 4 3 r
<br />:y fe .. -- + b sin A ° a sin C
<br />41�-�� ' a, b B, c, C sin I3= a C = 180'-(A I B),
<br />.. 1 r •J -D - `' ' a-b) tan ' (A+B)
<br />jg' (•.70 , 7' a, b, G A, B, c A �B=180 tang 4-13)= a + b
<br />q v ,
<br />1ST 06�a
<br />—_2 U / 1 Xr�f 2-a.? c r,�
<br />sin 3 _�•,rY3 4Z.C5�
<br />2 ` Q ,
<br />b c A, B, C s = . 2 qn''-•_4=1 b c `>' 3(•sfig= j("-all''-c) C=180°-(A+B)
<br />—'��•Z 10 `t 1 �S� ' C C� a c
<br />iy •�, ii'- a, b, c Areas-rrT2+c area
<br />l- !L9,bcsinA
<br />A,b,c Area
<br />+ _ a a•'-sinBsinC 4z•o
<br />3 ; 3 � 1.72 1 % 5 � • Z • 3 , 9, B, C, a Area arca =
<br />17 S -5-�o 4 l 2sin A 5F �
<br />REDUCTION TO HORIZONTAL
<br />�• ht: �1 Horizontal distance= Slope distance multiplied by the
<br />I d � ' )-f , ce cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />Z r~ I { •s ,,�' r % baa Vert. anglb=5° IV.. From Table, Page IX. cos 50 101=
<br />'7 4 c i1 ti 9959. Horizontal distance=319.4X.9959=318-09 ft.
<br />3 c�,o a i --�_�- • slop ��1e Horizontal distance also=Slope distance minus slope
<br />7d y0, o ' �` �. A distance times (I-c ine of vertical angle)• With the
<br />I- - { 3 •,,y 1` - ', c ( �� 1 q, /� yep same figures as in the preceding example, the follow-
<br />��
<br />Horizontal distance
<br />z7 a C Z- ing result is obtained. Cosine 50 101=.9959.1-.9969=.0041.
<br />l - 1
<br />319.4X.0011=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is known, the horizontal distance is approximately:-the slope dist-*
<br />y ante less the square of the rise divided by twice the slope distance. Thus: rase=14 ft.,
<br />_
<br />• � y a Z; � • � - _ - `'� .. / J l� slope distance=302.6'ft. Horizontal distance=3026- 74 X 14 -302.6--0.32=302.28
<br />ft-2 X 302.6
<br />n MADE IN U. B.A.
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