|
�'�6 z7
<br />3
<br />4 0, 6
<br />40, 41
<br />4y�b
<br />I 1ST % Z
<br />oL $679
<br />CC'/ 12 Sz
<br />Oz
<br />jl ts0�(�OOn 9 � )ILdG '�
<br />T t, I k
<br />4°•°�
<br />r
<br />go,ab.
<br />t A, B, C, a Area area =
<br />2 sin A
<br />40.06
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />-
<br />cosine of the vertical angle. Thus: slope distance=319.4ft.
<br />40,06
<br />9 6,
<br />45, a0
<br />aoee
<br />aj5t d Horizontal distance=319.4X.9959=318.09 ft.
<br />9959.
<br />=
<br />r�1CIAg1e Horizontal distance also = Slone distance minus slope
<br />Ar tt
<br />distance times (1—cosine of vertical angle). With the
<br />�1e
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5'101=.9959.1 .9959=.0041.
<br />7�
<br />319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is known, the orizontal distance is approximately:—the slope dist-
<br />h
<br />TRIGONOMETRIC FORMULAE
<br />B B B
<br />a ° a ° a
<br />A ^
<br />A
<br />b C �b Cl C
<br />Right Triangle Oblique Tiiangles —°��
<br />Solution of Right Triangles
<br />'For Angle A. sin= a b a ,cot = b ,sec = a , cosec = c
<br />c c b a b a
<br />- Given Required a x
<br />a,b A,B,c tanA=b=cotB,c= a2+ 2=a 1+ a2
<br />I; a, e A, B, b sin A = c = cos B, b = (c+ a) (c—a) = c 1--
<br />o
<br />A, a B, b, c B=90°—A, b = a cotA, c= a
<br />sin A.
<br />A, b B, a, e B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />A, e B, a, b B = 90°—A, a = e sin A, b= e cos A,
<br />Solution of Oblique Triangles
<br />Given Required _ a sin B ° a sin C
<br />A, B'a b, c, C b sin A'C=180—(A } B),c= sin
<br />b sin Aa sin C
<br />A, a, b., B, c; C sin B= a ,C = 180°—(A + B), c = sin A
<br />o, b, C A, B, c' ' A+B-180°— C, tan z (A—B)=(a—b) tan s (A+B) , a + b -
<br />a sin C
<br />sin A
<br />c
<br />a, b, a A, B, C s= a+b+2 be
<br />be '
<br />sin5BB=,Ni(3—a)( ),C=180°—(A+B)
<br />E "> ac
<br />a+b+c O
<br />C a, b, a Area s`e arta =V/8(8 --T(8— (8—c
<br />bcsin A
<br />A, b, . Area area.==, �Z _ i
<br />i a2 sin B sim'►C
<br />t A, B, C, a Area area =
<br />2 sin A
<br />\ \ 9a
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />-
<br />cosine of the vertical angle. Thus: slope distance=319.4ft.
<br />i Vert. angle=50 101. From Table, Page IX. cos b° 101=
<br />aoee
<br />aj5t d Horizontal distance=319.4X.9959=318.09 ft.
<br />9959.
<br />=
<br />r�1CIAg1e Horizontal distance also = Slone distance minus slope
<br />Ar tt
<br />distance times (1—cosine of vertical angle). With the
<br />�1e
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5'101=.9959.1 .9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is known, the orizontal distance is approximately:—the slope dist-
<br />h
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=3026 ft. Horizontal distance=3026— 14 X 14 _SM6-0.32=302.28 ft.'
<br />2 X 802.6
<br />MADE IN U. S. A.
<br />
|