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TRIGONOMETRIC.FORMULA,
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<br /> I b 3 i 1 ��l 5� r� b C d� b C 4 b L C
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<br /> Right Triangle Obligee Triangles
<br /> •/ -G 4 y 3oluttan of Right Triangles
<br /> 10 9 For Angle d. .b a b' a b o
<br /> 6,71 1 9 17 �:�[� 2 o 3 J.8 �� Q 3 9 Given � ° ,cos e ° ,tea=
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<br /> zi? � a,b ,�f,a t�nd�becotB,o= �7= a Y 1T
<br /> ��3 -� _LSU ✓ -D , "' a
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<br /> /15-73 � a A b sin d=a=cos B,b=%/ o a a-a =o I 1_a''
<br /> 71 1� i 3 vim. d,a .o B=90°—A,b=a cotA.,o= sin A.
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<br /> a, a B=90°--,d,a = b tan A,a= b
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<br /> doe A.
<br /> b B s 90°-d a=o sin b= e coo A
<br /> 5 .- _ 7-L .. A�.a, , , ,
<br /> U ✓. �/ g �' ri- Solution of Oblique Triangle*
<br /> Required a Bin B ,-( �-�.a a sAn Ain C
<br /> a b, o> �' b= , C=
<br /> 180*_(A =
<br /> sin d ,
<br /> B,e, C sin B= b°a A,0-� 180°-(A+B),a = a din C
<br /> din d
<br /> _ A,B,a A+B=1800-C,tan (A-B)- a-8 tan�(d_}B)
<br /> 7J � c �yq7 g , o _asin C a+ b
<br /> sinb ad3. 7 ° -
<br /> ��L7b d,B, C a= .in}d-1, 3s2, �7
<br /> ein}B= sa e-c 'C=180°-
<br /> -(A+B
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<br /> a, b, a a s—a
<br /> a= 2
<br /> C) ,area U:' a-- e—a
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<br /> ✓� �9Si G,z-5� A, b, c °Area bo sin
<br /> d
<br /> area = f
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<br /> o % i d,B,C,a Area area -ae 2 tBn sin
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<br /> i s 7.6 REDUCTION TO HORIZONTAL 6 ,0
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<br /> 9 , (p✓ HorIsoatal alsta>iee—Dope distance multiplied by the
<br /> y 7 oe Vat of tthhe vertical 51 1W. From Ta�bk� °°°�i'10,
<br /> �L4 ��O o0e �e Honba ts�ditbwea dtstartea os slope
<br /> �� v oe viae ofaI fie) the
<br /> samet a e exam 'Ibe follow.
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<br /> Horizontal distance i is 1�
<br /> Cwine 6 3—.9th=.0011.
<br /> / ? � Ut �4--l.bl
<br /> Cb �� When the rise is known,the horisauttal
<br /> Ion diats:♦s is app tett'—tis alopedist
<br /> aace lethe sanare of the alae divided 1»twos tf5ie dope 3 hn�e: =14 1L.
<br /> N/h dope a ft. Horiwettal dis4Nbe�DD26- X fi.
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