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TRIGONOMETRIC FORMUL)E
<br /> 4, B I3 B
<br /> a C
<br /> 11 q a a
<br /> J a
<br /> A b,. C A�b C A b , C
<br /> Right Te Oblique Triangles
<br /> Solution of Right _Triangles
<br /> a b a b a a
<br /> For Angle i1n = cos=— tan=
<br /> o a , e , b ,cot= a,sec= b. cosec = a
<br /> Y (liven
<br /> a, b A6o tan A=b= cot B,e = a$+ 2 =a
<br /> 1�I��s
<br /> `sin A=o =cos B,b=/ +-a (a—a
<br /> G
<br /> _ S4 -7 `� A,a all B=90°—A,b=a cotA,e=
<br /> sin A.
<br /> y ,a B=90°—A,a = b tan A,a= cos A.
<br /> 3 9 l 3 `� �_. A,o i I B=90°—A,a=e sin A,b=o cos A,
<br /> Z l
<br /> 'ZSolution of Oblique Triangles
<br /> Reagired a sin BC= 1800—(A+B),a =Ba,in C
<br /> sinA '
<br /> sin A
<br /> bsinA
<br /> d, e, bte, C sing= a ,�'= 180°—(A-{ B), Bin sin
<br /> F �)_3 / �� a, b, C A,B,a A+B=180°—C,tan j( —A—B) a—b)tan (A+B)
<br /> a+ b
<br /> a =asin C
<br /> sin--A
<br /> 0 1 Z 3 ? �� � a ds ,BC
<br /> 3 ,a 2
<br /> sin B=�( a,, ,C-180°—(A+B)
<br /> a+b+a
<br /> 6
<br /> a,.b, c2 2 , area =N/8(8 as— e—a
<br /> A, b, c Tea area = baBin A 2
<br /> as sin B sin C
<br /> 1 A,B,C,a fArea area = 2 im A
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance=Slope distance multiplied by the
<br /> cosine of the vertical angle.Thus:slope distance=319.41t.
<br /> ge Vert.angle—so Id. From Table,Page IX,cos b°
<br /> i�
<br /> slop. g 9960. Horizontal distance=318.4X.9068=31&OB ft.
<br /> $� 1e Horizontal distance also=Slope distance minus s�Ope
<br /> ve distance times(1—cosine of vertical angle). With the
<br /> same figures as in the preceding example,the follow-
<br /> Horizontal distanee ing result is obtained_Cosine 60 10'=.9060.1—9869=.0041. -
<br /> 31 .4X.9041=1.91.31&4-1.31=318.00 ft.
<br /> When the rise is(mown,the pori�distance is appnoxlmately-the slope dist-
<br /> anceless the square of the rises lb tQWD the elope wee• Thus:rise=l4%,
<br /> slope distance—Mle ft. HorizoaW dist mee..WLA—l4)e 1A-*M—M-30.20 f4
<br /> 2X302.6
<br />
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