--,MIGONOMETRIC FORMULA ,
<br /> B - jr� B
<br /> a a
<br /> 1 J'�
<br /> _ Lc
<br /> ��J �� '�v+ ?.•� / i b C A b C A b C
<br /> ,Right Triangle Oblique Triangles---j
<br /> (� L �, Solution' ght Triangles
<br /> 7-4
<br /> f \ `� 7i 0 For Angle A. sin =c tan— a,cot = a,sec=b,coedc= 4
<br /> V9
<br /> `X C� }� .. ,( Given Required
<br /> �l a, b A, B,a . tan Ab t B,c = a� } = a 1 + az
<br /> _ _
<br /> a, a A, B, b sin ma =coo B, (c-)-a =c 1-03
<br /> A,a B, — 0°— = a cot A,o a
<br /> to A. 2-
<br /> A, b
<br /> -A' b B, B= — to c= os A. l5f z / 7
<br /> 1 D B, —90°— a b— cos A
<br /> io of Ob ' iangles
<br /> Requirn a sin C
<br /> 1 -g ( A, B,a c, C — 1 =7 ° " B), a = sin A
<br /> C 1 -(A B) a — a sin C'
<br /> s
<br /> A,,a, b B, e, C sin A
<br /> Gr
<br /> a—b tan A B l a1, b, C A, B,o A+B= °—C,tan z -B)=�) z( )
<br /> a sirs C a+ b '
<br /> 95 c c sin A
<br /> b, a A, B, C a=a+b+c,sinjA=Vla-�(s—c
<br /> ori V o
<br /> > > ? 6 sin B— a c C=1800—(A+B)
<br /> `p ag b, a Area , area
<br /> A /
<br /> A b a Area �Z�b c sin A
<br /> area —
<br /> b`� �3 �-� p �yy a'sin Bsin C
<br /> A,B,-C a Area area = 2 sin,A i
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance=Slope distance multiplied by the
<br /> wl a cosine of the vertical angle.Thus:slope distance=319.4 M
<br /> Vert..angle=50 101. From Table,Pace IX..cos 6°101=
<br /> � 9959. Horizontal distance=3l9.4X.9969=31&09 ft
<br /> v Arg1 Horizontal distaince 41so=Slope distance mints slope
<br /> e a distance times (1--cosine of vertical angle). With the
<br /> rrA �f same figures as in the preceding daample,the follow-
<br /> / J Horizontal distanceing result is obtained.Cosine 5°10'=.9959.1—.9M=.0041.
<br /> {t k X519.4X.0041=1.51.519.4-1.91=318.09 ft.When the rise is known,the horizontal distance is approximately:-the slope dist-
<br /> ance
<br /> V less the square of the rise divided by twice the slope distance. Thus:rise=14 ft.,
<br /> slope distanoe=3028 ft. Horizontal distaaeo=302e-1-�4_302.8-0.32=30226 ft.
<br /> 2 X 302.6
<br /> sADa IN Y.&/.
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