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<br />TRIGONOMETRIC F'ORMULIE i
<br />TP / �'� 9 . s "jam•
<br />B
<br />G a G a G a z
<br />p Z �t 4r �4 : b d d b Q A G
<br />S, 37 :saar�t �e
<br />Right Triangle Oblique Triangles r
<br />TPI- s; Solution .of Right Trian lea
<br />Tf' ! 9'. y :sem, I a g
<br />a b a b c
<br />cy j p -ten= cot = - ,sec = cosec
<br />A- =
<br />Given Required a z
<br />$;�(3a; b h A, B ,c tan A = b =cotB, G = 1/ar + s = a 1 -}-�1
<br />C
<br />Ts< ti
<br />Pock-
<br />A,
<br />tea a, c ' A, B, b sin A - = cos B, b = � (c-�a) (�-a) mm e
<br />• f .
<br />. 12i1 A, a B, b G B=90°-A, b- acc c=
<br />9 a sin A. -
<br />b
<br />S 1cs�r ! A, b B, a, ' e. B = 90°-A, a = b tan A, e = i
<br />X00 cos A.
<br />2 Gs`a 487. o $ ?' t A, c B, z,'b B = 90°-A, a = c.sin A, b = e cos A.,
<br />Solution of Oblique Triangles
<br />Given Required
<br />�oct.p asin B asin C
<br />9.P 7. o •�B, B,-a b, c,. C b sin A , C =` 180°-(A } T3}, c - sin A
<br />-7 ��.c 7
<br />b sin.A , a sin C
<br />Jf�i,ur r; .Bcb d, a, b B, e, C sin B = a , C = 180 -(A -f' B), e - kin A
<br />9 E3`t: a2 i 4 (a-b) `(A+B)
<br />q b, C A, B, G A+B=180°- C, tan j (A-B)- a + b
<br />Y i a sin C
<br />COL
<br />cv ti 1
<br />I G —
<br />� � sin A
<br />N �
<br />b,. C A, B, C' s=a+2+GYY
<br />in
<br />.
<br />- -.
<br />kin iB=',1 a G ,Ci=180 —(A'F'B)
<br />Gf.- f��4�1!/C3t ,'..ilO�'Tjl: Asa-.. • � Y
<br />Alve4-yl G/.c/L. �c�d./ALp/. �,?r•+,o 6"a.i' - a+b+o
<br />995—g ! a, b, •o Area s= 2 ,area = s(s-a s- ) (s-c)
<br />A, b, c Area b e sin A
<br />'/'V_av area =
<br />G•fi2�, ..at�,yov.5i.47� 2
<br />�ac6 cam,
<br />a2 sin B sin C
<br />A, B, C, a Area area = 2 sin A ;
<br />RArGP�yca REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e cosine of th e vertic al angle. Th us: slope distance =319.4 ft.11Pn
<br />e a15' y V99brt9. Hor== 51 101i izontal distance 319.4X.9959= 3 8.09 ftIX. cos . 10 —
<br />S%o �pgle Horizontal distance also=Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />Ve same figures as in the preceding example, the follow- §
<br />Horizontal distance irg result is obtained. Cosine 5° 10'=.9959.1—.9959=.0641.
<br />919.4X.6041 1.31, 319.4-1.31= 318.09 ft.
<br />1 , - ); When the rise is known, the horizontal distance is approximately;—the slope dist-
<br />once less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft,
<br />slope distance=302.8 ft. Horizontal distance=3026— 14X 14 =3M6-0.32=30228 ft.
<br />2 X 902.5
<br />MADE In U. 5. A. ,!�
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