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ALk '40 R
<br />C_ A� ' i , TRIGONOMETRIC FORMIJUE
<br />1f 4'6 B
<br />' Z a f a a'' x•G
<br />I� l°� f
<br />3iy b C Afb C A b C
<br />+" �� ✓ sg R
<br />Right Triangle Oblique. -Triangles
<br />J �4 —*—
<br />' Solution of Right Triangles
<br />0S .
<br />�{ 1 C� a ..1 a b 'ice b
<br />For. Angle A. sin = G , cos = ,fan = fb / icot= a ,sec = b , cosec
<br />� ® Given ' 'Required a �;` z'1
<br />5� 1
<br />p �\ a, b A, B ,c tan A = b = cot'jB c = az -i- z = a 1 + az
<br />Vp i U+. tr
<br />y {� z•
<br />���(t� ���, cc, c A,B,b sinA=� _Ys$b=�(c+a)(c—a)=c.�.1- o
<br />V y 2
<br />j
<br />A, a B, b, -c' B=90°—A, b = a cotA, c= I sin A.
<br />A, b B, a, c B= 90°—A, a= b tan A, c=
<br />r � F � '�✓J - ,�- ° o o � - � cos A. i
<br />3 �
<br />A, c B, a, b B = 90°—A, a = c sin A, b = c cos A,
<br />Solution of Oblique. Triangles '
<br />Given Requireda sin B / ;
<br />61 S- X 9'a A, B, a b, c, C b — sin A ' =.180°�—(A F B) c,= sin A } ri
<br />bsin � i asin
<br />C
<br />A,' a, b B, c, C sin B = a = 180°—(A ' l B); o = sin A
<br />b, C � A B c \ (a—b) tan I (A+B) f Y'
<br />a, A } B=180°— �, tan a (A — 'B)=
<br />a+ b ''
<br />Z _ _ a'sin C +
<br />1 ua2='' 2z-1fp'/ , 7r`' c sin1A
<br />cid b { c t
<br />a, b, c A, B, C s= 2 in A= bc'. Ct,
<br />s
<br />jy`:sin zB= C=180°—(A+B) yryµ
<br />a c N t
<br />b, c.. AreaFp .. t, area = s(8—a) (s—b) (s—c j
<br />bto
<br />sin A
<br />A, b; c Area area t ``2
<br />ai sir1fiB 61n Ci
<br />A,'B, C, a Area area =
<br />1'•C-' 0b' �/ 13
<br />1 � '
<br />U(
<br />j
<br />(2 stn A R
<br />REDUCTIONJ,,HORIZONTAL
<br />Horizontal distance=,Slope distance multiplied by fhe,
<br />cosine of the'wertical angle. Thus: slope distance =319.4'
<br />9959. Horizontal distance-- 19.4X.9959e 318.09 Pt,b°Y- Ops
<br />L 1 1 51oQ ogle Horizontalldistance also=Slope distance minus slo
<br />1� Y, " I ve . p' h�� a 'distance times, (1—cosine of vertical-�t
<br />angle). With
<br />same figures 'as in theipreceding example, the —.follo.
<br />Horizontal distance - ing result is obtained. Cosine 51 101=.9959.19959=.00
<br />l+lt 319.4X.0041=,1.31.319.4-1.31=318.09 ft. l
<br />i When the riseds known, the horizontal distance is approximately:: the slope di
<br />ante less the square of the rise divided by`{twice the slope distance. Thus: rise=141
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =3026-0.32=302.28 ft.
<br />d I !U 17 2X302.6 —
<br />+ ' J MADE Ia V, &.A
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