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<br /> - ti �til�}Ori #i�- f31a$ •�
<br /> G� � ,,3 7 Y� B B B
<br /> .-- �f' ,
<br /> �- A
<br /> , o �b a b
<br /> Oblique Triangles
<br /> Solution of Right Triangles
<br /> a b a b o
<br /> For Acle : do = o ,cos— o ,tan= b ,cot = ,sec s—b, cosec a
<br /> (liven Required 1 �
<br /> a, b A, B,o tan A=b=cotB,a a } '=a 1 I !
<br /> \ a
<br /> a, 0 A, B, b sin A=o =cos B,b=%/ e+a o---a =c I—!'-
<br /> i' o
<br /> B=90°—A b =acotA a= a
<br /> sin A.
<br /> c� AAb. J 3t, • B=90°—A,a = b tan A,c= b
<br /> cos A.
<br /> b I B-94°—A,a =e sin A,b=c cod A,
<br /> S6lution of Oblique Triangles
<br /> sin A ' C= 180°—(A h B), ., a sin C
<br /> J T u sin A
<br /> ,r L ) c� C sin B= b sin A,C=180°—(A-(-B),a = a
<br /> sin C
<br /> re J sin A
<br /> Ab, ,B,o A+B=180°—C,tan (A—B)= a—b tan bA}8),
<br /> o —
<br /> aeinC
<br /> }...�.'� � sin A
<br /> t _
<br /> a b, o A,B, C .=aa t,sin
<br /> ,
<br /> ( -r
<br /> sin J1B_�a a o ,C=180°—(A+B)
<br /> �t v
<br /> i
<br /> a b, e s= 2 , area =Vq
<br /> b c sin A.
<br /> A, b, crca area = 2
<br /> _a 2 sin B sin a
<br /> d,B,C,a Area area 2 sin A
<br /> r
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance—Slope distance multiplied by the
<br /> cosine of the vertical angle.Tb ks:slope distance-319.4 ft.
<br /> 1 �gts4°e 'Vert. angle=8°IW. From Table,Page IX,eftb°1W=
<br /> .6966. Horizontal 1llstanoe-31W.9VA-313.06 ft
<br /> 1 i gl bAg1e Horizontal distance also=Slope distarticalnce minus slops
<br /> distanfiQnns as i the
<br /> tp�(1--cosine
<br /> preco of
<br /> ipg i amVt t erffoilow--
<br /> Horizontal distanos. ing result is obtained.Uine b�IW!= 8 ,1—.6689=.mL
<br /> 919.4X.0041=1.31.819.4-1.91=51&0 ftWhen the rise is known,the horizontal distance is apprortmately: the slope diab.
<br /> less the square of the rise divided by twice the slope distance. Thus:riise=14 ft.,
<br /> distanee=302.6It. 'Horizontal distance-305L1 =SOZ6-0
<br /> 2x302.6
<br /> f
<br /> V - s
<br />
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