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TRIGONOMETRIC FORMULIE
<br /> .� B B B +` ;
<br /> a c a c a
<br /> b C �,� b C A 6 ` C
<br /> �i Right Triangle Oblique Triangles f
<br /> Solution of Right Triangles
<br /> p C For Angle A. sin = o ,cos= a ,tan= b ,cot = a,sec= b, cosec= a
<br /> 3 2. O f 7 3 S Given Required a
<br /> a, b A, B,o tan A=b= cot B,c =Vra + 2 ='a i a=
<br /> o 'A,B, b 'sin A=—=cos B,b— c+a c—a a
<br /> Y
<br /> 6 b as —c —��) =c J1-02
<br /> e� av B, b, c B=900—A,b =a cotA,c= sin A.
<br /> j� b
<br /> b B, a, c B=90'—A,a = b tan A,c= cos A.
<br /> /�•7 v' ' CE — c B, a, b B=90°—A,a=c sia A,b= c cos A,
<br /> Solution of Oblique Triangles
<br /> ✓ �' � iven Required
<br /> a b c C b = a sin B C= 180°—(A+B); c a sin C
<br /> sin A ' sin A
<br /> �✓ ` A, a, b B, c, C sin B= b sin ASC= 180'—(A-i-B),c = a sin C
<br /> a sin A
<br /> ;7, a, b, C A,B, o A+B-1800—C,tan I(A—B)=(a—b)taan.1(A+B
<br /> c —
<br /> asinC +
<br /> 416 sin A
<br /> B, a+b+o 1 (a—c
<br /> O b, o A, B, C a— 2 ,sin:3 be '
<br /> f sin?B—�(a—sea ,1=180°--(A+B)
<br /> a, b, o Area 8=a+b+c
<br /> 2 ,area
<br /> z G 3 �, b, o Area area = b o sin A
<br /> 2
<br /> A B C s Area area=az sin B sin C
<br /> s
<br /> ' 2 sin A
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance—Slow distance multiplied by the
<br /> cosine of the verbvd=gle.Thus:slo distance 519.4 ft '
<br /> Vert.angle-6°10f., Fro:!W9 ,I as^e IX.an b°10'=•
<br /> _,,�s • 9W Horizontal.dLianoe�319,4X.996 318,
<br /> 061t
<br /> !AP° Apg1Q HodIzontal distpnos also—Slope distance minas slope
<br /> distance times El--cosine of vertical an With the
<br /> same fispres as in the p=VS ezample,the follow-
<br /> Horizontal diata3m i result is obtained.Cosies 6°1N .9Y68 1—.9669=.0011.
<br /> 51&4X.0011=1+51.519.4-1.51—Si&�B 1�L
<br /> `Whoa the rise is known.the horizontal distance fe aP; ozlmately—the slope dist-
<br /> an"less the square of the rise divided by twice the slope distanoe. Thus.rise=l4 fL,
<br /> elope disd►aa 8 f6 2S!et3 tat R 3Oe.2li ft°
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