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<br />/pJ , 73
<br />3, i-
<br />!oB �
<br />�T GONOMETRIC 'FORMULA
<br />B
<br />c
<br />a
<br />A A
<br />b C
<br />rCAA b C �
<br />Right'Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For'Angle •A, sin = c , cos = o , tan= b , cot = a , sec = cosec = a
<br />Given Required a
<br />i
<br />¢,b A,B,c tan A=b=cot B,c= a2-47 b -=a`1 } a +
<br />— — z
<br />r ¢' c .4,B,b sinA=0a—cosB,b—�(c4a)(c—¢).= J1—ap oot
<br />li. A, ¢ B, b, c B=90'—A, b = a cotA, c= a
<br />sin A.
<br />i A, b B, a, c B = 90'—A, a = b tan A, c = b 25
<br />cos A.
<br />A, c . B, a, b I B = 90'—A, a = c sin A, b = c cos A,
<br />I Solution of Oblique Triangles
<br />t Given Required a sin B
<br />4 A, B, a b, c, C b = sin A ' C = 180'—(A + B)., c = n A
<br />b sin A a sin C
<br />A, a, b B, c, C sin B.=a , C = 180°—(A -i- B), c = sin A
<br />(
<br />• ¢, b„ G A, B, c A -{-8=180o — C, tan _ (A—B)— tan z A+B)a + b ,
<br />a sin C
<br />c sin A
<br />v 2 a, b, c A., B,'C s_ry+2+c,.Sin'A
<br />�•
<br />oG
<br />),C-180°—(A+B)
<br />'
<br />sin•'_.B=\I(•
<br />ac
<br />a+b+c
<br />¢, b, a
<br />Areas=
<br />2 ,area = •� x(,�—a) s— 1 (d—c)
<br />1
<br />Area
<br />b c sin A
<br />area = .
<br />a2 sin B sin C
<br />A; B; C, a
<br />Area
<br />area = 2 sin A
<br />REDUCTION
<br />TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />_
<br />t
<br />cosineoftheverticalanale.Thus: slopedistance=3l9.4tt.
<br />aty �3n°e
<br />Vert. angle= 5' 10'. From Table, Page IX. cos 5' 101=
<br />ea\epe
<br />c 9959. Horizontal distance=319.4X.0959=318.09 ft.
<br />" Horizontal distance also=Slope distance minus slope
<br />t. hn�\e
<br />x distance limes (1—cosine of vertical ankle). With the
<br />4e
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance
<br />ing result is obtained. Cosine 5°l0'=.9959.1—.9'J59=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=31309 ft.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=302.dft..IIorizontal
<br />distance=302.6-14X14 =302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />.` '-
<br />MACH IN V.B.A.
<br />—�•
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