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39S "q, 7 7
<br /> (,^.2 TRIGONOMETRIC FORMULrE '
<br /> 4.J ?76 B B 14,1 B
<br /> 0"F•�J� 32. .7 4
<br /> A A C
<br /> b C } b C b t
<br /> 7f Right Triangle L-- Oblique Triangles—_-1
<br /> 9 3 , f - Solution of Right Triangles
<br /> For Angle A. sin=S, cos=��,tan=-b°,cot-a,sec=b _.E,Cosec_L
<br /> 24 3 �r Given Required j
<br /> 3v; v 3 a,b A,B,c tan A=b=cot B,c= a'-f b==a
<br /> J ,T a
<br /> c . ?' a,c A,Beb sin A=�=Cos B,b=N/( -a)=c i-°i-,
<br /> 3� .
<br /> 1; V 39 ) `
<br /> A,a B,b,c B=90°-A,b=a cot A,c-sin A.
<br /> j 3
<br /> A.b B,a,c B=90°-A,a=b tan A,c=cos A. 1
<br /> r L Q
<br /> l `� A,c B,a,b B=900-A,a=c sin A,b=c cos A. f
<br /> , Solution of Oblique Triangles
<br /> /1 `: �--�'"' .r►�J Given Required
<br /> 1 �, `1 8 3 7•ff 4 ` A,B,a b,c,C b= sin B
<br /> sin A 'C=180°-(A-i-B),c- sin A
<br /> 39Sin I
<br /> 3 - � 8 A,a,b B,c,C sin B-b sec sin A A
<br /> LC=180°-(A+B),c=
<br /> / �f p 4, .�1 .�
<br /> f 3 -o v a,b,C A,B,c A+B=180°-C,tan*(A-B)=(°-b)tan+b B)
<br /> 3 9 4/o J— c=asin C
<br /> .�•✓ ' '� 3 �'- �"- 3 .P 8 -�1�- sin A
<br /> i j M a 3 3,I 3 y � (2 o, 9 ,C a,b,c A,B,C s=a+2+c,sin*A- (s-b)�s-c'
<br /> 3. �4r c�¢
<br /> ✓; 3--�' 3 .� �Al 1J sin�B= (s-a�s-c>,C=180°-(A I-B)
<br /> ^� 3 3 4. v � a,b,c Area s=°+2+c,area= s(s-a) (s-b) (s-c)
<br /> 0
<br /> 3J .7 9 fi 9 b e sin A 3•/.SF
<br /> q. � A.b,c Area area-
<br /> 10 '-- 3 ,1 y p• A,B,C,a Area area= V it—
<br /> at sin B sin C 31-
<br /> g,
<br /> h J% 3 9 . f 2 sin A
<br /> i ` r ' '� ✓� y REDUCTION TO HORIZONTAL_.'
<br /> r
<br /> 3 '7 ti ��.,,,. Horizontal distance-slope distance multipby the 3•L'
<br /> 31 - _1 cosine of the vertical angle.Thus,for a slope distae oe of est
<br /> 3. � �q . -. 400.6 ft.and a vertical angle of 4'40'-the cosine of Sia
<br /> 3%F
<br /> if 1 ti / 3 �, 4.40'.taken from a table of natural trigommetrical Vim„t+t+g 11
<br /> functions.-.9967.and horizontal distance-400.6:.9967
<br /> _- f6 y. 402.27 ft. Horizontal distance
<br /> .: - y k� - Horizontal distance also-Slope distance minus dope dig-
<br /> 3
<br /> 9,(� s I,.�,,,,�----6�6 "?j •7 tance times(1-cosine of vertical angle).Using the same figams as in the preceding example-
<br /> k p •t '� Cos.4.40'-.9967.1-.9967-.0063.403.6x.0063-1.33 ft.Horizonteldist.-400.6-1.33-40227ft.
<br /> When the rise is known.the horizontal distance may be found by the following approximate
<br /> rule:-the slope distance less the square of the rise divided by twice the dope distance.Thus,
<br /> for a dope distance of 372.5 ft.,and a rise of 15 ft.the horizontal distance=
<br /> 372.5-2X3 372.5
<br /> ,5 -.30=372.2 ft.
<br /> swillm DItTZOM CO.
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