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f TRIGONOMETRIC FORMULtE
<br /> j " r B B B
<br /> a
<br /> �7 /� b CA�b CA
<br /> C
<br /> 13,11'78, Right TrAle Oblique Triangles
<br /> Solution of Right Triangles
<br /> For Angle n a b o
<br /> J 3 t7-73 =6,cos= a,tan= a b d
<br /> b,cot = a,sec= b, cosec= a
<br /> 6, Given Re red a a
<br /> r a, b A, ,d tanA=b= cotB,c = a2-� s = a 1+ E
<br /> A,B, b sin A=e =cos B,b=N/ c+a)(c--a =d J 1 _
<br /> a
<br /> A,a B, b, o B=90°—A,b=a cotA,d= a
<br /> sin A.
<br /> O a
<br /> A,5 B,a, d B=90°—A,a = b tan A,c=
<br /> cos A.
<br /> A,d B, a, b I B=900—A,a =c sin A,,b=c cos A,
<br /> Solution of Oblique Triangles —
<br /> b
<br /> '71 Given Required
<br /> B a R u C b=asinB� C= 180°—A asin C
<br /> sin A — ( �-B). d= sin A�
<br /> b sin A
<br /> B,e, C sin B= a .C= 180°—(A+B),d = -sin C
<br /> �. sin A
<br /> b, 0-A B,d A+B=180°7:V,tan z(A—B)— a—b)tan}(A±B)
<br /> asin C a+ b
<br /> d =
<br /> ^ r sin A
<br /> a, b, o' A. B, C s=a ,sin A= I bo—d
<br /> Bin;B=�s a ,C=180°—(A+B)
<br /> a, b e a=a+a+d,area = s(s—a'
<br /> A, b, d Ake& b e sin A
<br /> (^ area = 2
<br /> a2 sin B sin C
<br /> A,B,C,a Area area = 2 sin A
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance—Slope distance multiplied by the
<br /> cosine of the vettical angle.Thus:slope distance=519.4 R.
<br /> Vert.angle=5°IW. From Table,Page IX.cos 5°1W=
<br /> e a' m 9969 Horizontal diztanoe--319.4X.W*-31&09 iL
<br /> $1°Q Ap41e Horizontal distance also=Slope distance minus slope
<br /> t„ �e distance times(1—cosine of vertical angle). With the
<br /> same figures as in the preceding example,the follow.
<br /> Horizontal distanceIng result is ebtatued.Cosine 5°1 .9969.1—.99b9=.0041.
<br /> 31a4X.004i=L31.91x4-1.31=31$.09 2L
<br /> When the rise is known,the taI eels approximately:—lope dist-
<br /> oe less the square of the rites div%*bg bwtce the aisle distance. Thus:rise=l4 fL,
<br /> pe distance=30U$ HorjwaW 6-WW2=30L26fL
<br /> X 90LS
<br /> ^ K.
<br />
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