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le oda S .G 3,, is . 'TRiFORMUL)E
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<br /> Triitpgie Oblique
<br /> Solution aof Rlthta TrIanglos
<br /> For Angle A. sin = o ,cos— ,tan— b,cot= a,sec*■b, cosecIle
<br /> Given Required
<br /> a, b A, B,c tan A m b= cot B,c a `— a 1-1- a,
<br /> I a v u Z Kms— o A,B, b sinA.=c—cos B,b—�/ c-1 a a—a =a J l—02
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<br /> c 3 a i V
<br /> T✓ 113 8 a B, b, o B=90 A,b =awtA,a= sin A.
<br /> 3 f o •v b B,a,-,, B_90°—A,a,= b tan A,c= cos A. L y
<br /> )4 & W N n ,r a B, a, b B—90°—A,a=o sin A,b=e cos A, 7 i
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<br /> 1 g g l 'D * A,1p I 1 I a 18 / /L• 7 0 en t Solutiona sin B Oblique TrianglesJ-
<br /> j,o p } , lcq. b _ C= 180o—(A+B),a =
<br /> �' 7x e, C ein A ' sin A +f i
<br /> L4 9 S � y' b-e'n A a sin C
<br /> J 7 - �-�- 1�5� 3 ;mv b .I�,a; C sing ,C= 180°—(A+B),a =
<br /> min
<br /> 7 /I .° I/ti 9. ". B a A B=180°—'C,tan' A—B— a_b tan (A+B
<br /> 3.7 .r Z •y � y, a � a sin C1
<br /> sin A 'at 4-7 �` tf
<br /> 1 )3,9 $`. A.
<br /> �' 8= 2 ,sin�A=.`� b c
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<br /> Bln
<br /> o 3'Sr JJ__a.r #B=�N a a �C=1g8 F(A B�
<br /> b, o Area 8=a+ ,area = s u—a s— ) 8--c
<br /> 2
<br /> 3• !i �" ' y� b o sin A OL ,cf-d
<br /> I I J a y .y �.b, a Area area = 2 1 j
<br /> 8 a'sin B sin'C'
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<br /> y ,..M: `� /�•�, 9� �!—�' �. 4• A.$,�`�a Area area = 2siah,
<br /> 4
<br /> YU � "REDUCTION TO HORIZONTAL
<br /> Horizontal distance distance IDnZMDIled by the
<br /> Vert.angle
<br /> e+sestleslaaail Thus:s m �66 10 s
<br /> .9m Hors=• t l d Fmm TOWNUX Psis
<br /> 98b8. Horizontal dSslaaoe-1i9.4X�8.68 it.
<br /> Hodzonal distance also—910PO ee minus dye
<br /> distance tlmss t1-oQsiae of vertlral With the
<br /> same ftgum as in the pro
<br /> the lolbw-
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<br /> Horizontal tame s 4x`e'u ° lam. ued.CCO'slate l-
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<br /> P ;; the rise is known.the tsl anion,+►iistnee �sil4ft•.
<br /> -t 7 thr agaare of the ries dtrided b�tse
<br /> F"... / 7 L al 8028 iw Horizontal distaaoe-702.8—14 X 14
<br /> —�t8-8> 7Elt.
<br /> leiJ (. 2 X 802.8
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