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�2q,2 - `LL. TRIGONOMETRIC FORMULA . ,
<br /> :8 I3 _. 4a
<br /> as r ? 7 a
<br /> a
<br /> d�—b d C
<br /> _ ' } Right Triangle Oblique Triangles
<br /> Wution of Right Triangles
<br /> a b a b
<br /> _ "For Angle A, sin— o ,cos— c ,tan= h,cot= a,sec=b. 1= a
<br /> Given Required a
<br /> ,y a,b A, B,a tan A=b=cot B,c = a -{-
<br /> s - i a, c A, B, b sand—a—cosB,b=V c+a 1—ea
<br /> A,a B, b, c B=90°—A,b=a cotd.a= sin A----
<br /> A,
<br /> A, b B,a, a B=W—A,a= btanA,c—
<br /> A,c, B,a, b B=90°—A,a=a sin A,b-o cos A,
<br /> ---- Solution of Oblique Triangles G y
<br /> �.
<br /> Given Required
<br /> d, B,a b, c C b=a sin B C_ 180°—(A+B) a a atn C �1
<br /> _
<br /> sin A ' ' sin A
<br /> �.;
<br /> b sin Aa sin C
<br /> c- d, a, b B, e, C sin g= a. ,C= 180°—(A {-B),a= sin d
<br /> a—b tan (A �
<br /> 3 a, b, C A,B,a e J-B=180°—C,tan#(A—B)= a+b
<br /> �.J" 2 f c sin YX2-
<br /> 24 J y a, b, a A'A;C a= 2 sinA—
<br /> stn'-,B= a a ,C-180—(A+B)
<br /> a, ba Area, As=a 2 ,area = a s—a
<br /> 77°47r/i. Of v
<br /> A, b, c Area area = b a sin A
<br /> /
<br /> 2
<br /> arsinBsin C ie
<br /> . ! /q ' A,B,C,a Area area2sinA
<br /> ORIZOREDUCTION TO H
<br /> Horizontal distance=Slope distance multiplied by the
<br /> l
<br /> scale*ofthevertical anale,j��►�sloDedistanes=alA4ft.
<br /> i o� est b°16. lrroai�Dls. .I .oar6°"
<br /> i„ S�o� bp41eoce Sioea manna slo
<br /> pe
<br /> llm-
<br /> T- With the
<br /> f Q satins >s in the p��A .904L
<br /> 1 P f Horizontal distance ing res obtains!.Godes 6.1
<br /> / 4 "� 1 y �19feOt =L9t.a18f—i.si=618 ft
<br /> i / ' When the rise is known.the horisontal distance Is app ros telr:—theslopedist-
<br /> anee less the square of the rims divided b7 twice the slope distance. Thust#se=l4 ft..
<br /> slope distance-302.6 ft. Hositontsl dys�aeas=i f—�Z4I 4js--- fL
<br /> ii E
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