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<br />TRIGONOMETRIC"FO:Ril"fUL
<br />R'�F - ,�i`i+.�Y�2r•- �.�j� _ � q.� �. �J • /,e} 7 i � t.� Y � � a LE 6
<br />e"s . C ` --- Ri hi Triangle Obliyu.e= Triangles ��1 3-
<br />�: Solution of Right Triangles f
<br />?� µ $��..• j tJ rr' ��/ �f . 4� s % i� �6". = ,: �, Y` For Angle A _ ;sin a i, cos = b ; tan = a ,'coat = , sec = cosec =
<br />a'
<br />/.. ty "'' - .: '� % l Given ' 7ec iuired�i a
<br />f%3 37 i, a f L'
<br />Q' b tf• .13 ,c tan A =cot T; c = y/a,s 1 - F-
<br />Z?�_ry 'A; B, b sink. �=cos B,b—�✓[c+a.)(c—a) =c �1—ate
<br />_.. „ _� '��. dam,/ . �•� .- _ �. � � v _ ; - ..
<br />'- Ai a ;13;`;b,.o •-B=90°—A, b = acof4,c= rz
<br />• 1 , y•� sin A.
<br />�.—G•r r -- - �: A, b 73,•a, c 8=90°—A,a btanA,
<br />� . cos A. v
<br />A;a B, a, b- 8.-90°—�,a=csin l b=ccosA, `� f?�:_•
<br />Solution of Oblique Triangles
<br />fcr, Given Required
<br />a sin JV , a sin C s
<br />A, B, a : b c, C b. = stn A G = 18D —(A -� B) c = sin A
<br />.-� Z b sin .1 a sin C r �r
<br />/y A... 7 k t d, a, L . ]3, o, .0 4 sin B= C—:1;$0°—(A t B), c =
<br /># a sin A L r
<br />3.3 - �� _ i a b C 1; B, o A+B=180°= C, tan. (A--i3)= a—b) tan (A+B}/ ,l `
<br />/ �✓ q p r- 6g # '
<br />a sin C a b '
<br />f
<br />sin A �.
<br />�r Y •.^ " �o� i?`-'.^� .- - '� Sr �' a, b; C A, 2
<br />sin '.'L_,1
<br />.:� F;
<br />sin'I3—_
<br />Area s=a+h+c, area
<br />:
<br />b c sin 1
<br />j,5r S cy % �— �, . b c Area —
<br />• / G, / fe - _- "' area
<br />� 2
<br />a'sin Z; sin {'f
<br />"-.
<br />d,B,C,a Area area= 2sinA'7..3
<br />REDUCTION TO 'HORIZONTAL x 7
<br />�i! ..:x•: t .," ��� -, �
<br />' '3oriaontal distance= Slope distance multiplied by the
<br />cosine oftheverticalangle.Thus: slope distance -319.4 ft.
<br />i, .- t y YI/ e atst�rce. Vert.
<br />e69aHarizonYal From
<br />Table,319.499� efLIX. cos�� la
<br />7 �fig1e Horizontal distance .also='lope distance minus slope
<br />t
<br />. . / r x _ A� V �Y distance times (1-cosine of vertical angle). With the
<br />�,` -gsame figures as iu the preceding example, the follow-
<br />/ r / � f ing result is obtained- Cosine 5° 10'=.9959.1-.9959=.0041.
<br />t .� Horizontal distance
<br />--'' � ' '� ' 319.4X.0041=1,31. 319. tl-1.31=318.03 ft.
<br />-
<br />When the rise is known, the horizontal distance is approximately:-the slope dist-
<br />4
<br />ist-
<br />ance less the square of the rise divided by t)viee the slope distance. Thus: rise=l4 it.,
<br />slope distance=302.8 ft, Horizontal distance=302:4'x-
<br />14X 14 =302.0-0.32=302.28 ft.
<br />2 X 3026•
<br />MADE IWU. S.A.
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