|
I.Wlq.flee,y-
<br />1ZCl
<br />"c
<br />1 . TRIGONOMETRIC FORMULtE 1 6 , o
<br />L ''f{CAA b\ C A b C
<br />Right TfiiiEiAbe Oblique Triangles I
<br />Solution of Right Triangles
<br />For Angle A. 'sin — a , cos= —,tan= b , cot= a, see = b, cosec = a
<br />Given Re Qirire'd
<br />a, b -11 B ,c tan Ab' = cot B. c = y/a= + b= = a ; Y
<br />a
<br />a, c. sinA=�-c �I,g=4� G+ )(--a)=c�1—E
<br />a
<br />f i A, a B, b, c B=90°—A, b= a cotA, e=
<br />gin A.
<br />Lt
<br />A, b B, a, c B=90'—A, a = b tan A, c = b
<br />Cos A. - -I
<br />A, c B, a, b B=90°—A, a = csinA, b= ccosA,
<br />Solution of Oblique Triangles
<br />Given Required a sin B a sin C
<br />A, B, a ,h, o, Cb — sin A ' C = 180°—(A + B), c = sin A
<br />b sin A a sin C
<br />B, c, C sin B=a , C = 180°—(A -l• B), e, = sin A ' .
<br />(a—b}
<br />a, b, C A; B, c. �1-f-B=180°-C, tan :', tan '- (A+B•)(A—B)= a + b ,
<br />II asin C
<br />o=
<br />sin A
<br />( a, b, e A, B, C a=a-f-a+c,sin'A—lg~
<br />2 be
<br />g.I sin
<br />t I,
<br />a+b+c
<br />ccs, 6, a Area, B = 2
<br />f1' bcsin A
<br />A, -h c= .Area' area = 1
<br />a'- sin B sin C
<br />A, B,C,a .,Area area =
<br />, 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosineofthevertical angle. Thus: slope distance=319.4ft.
<br />w°e Vert. angle =6' 101. From Table, Page IX. cos 5° lW=
<br />e''� �9959. Horizontal distance=319.4X.9959=318.08 ft.
<br />1
<br />1 'De Pnq\e . , Horizontal.distance also= Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5°10'=.9959. 1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 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=302.6 ft, Horizontal distance=3026= 14 X 14 =3026-0==3a28ft.
<br />2 X 502.6
<br />4 - MAGE IN U. B•. Ib r
<br />i
<br />15 33��
<br />e: _
<br />7 �
<br />X17
<br />r
<br />HI
<br />J.
<br />VV
<br />IV
<br />I.Wlq.flee,y-
<br />1ZCl
<br />"c
<br />1 . TRIGONOMETRIC FORMULtE 1 6 , o
<br />L ''f{CAA b\ C A b C
<br />Right TfiiiEiAbe Oblique Triangles I
<br />Solution of Right Triangles
<br />For Angle A. 'sin — a , cos= —,tan= b , cot= a, see = b, cosec = a
<br />Given Re Qirire'd
<br />a, b -11 B ,c tan Ab' = cot B. c = y/a= + b= = a ; Y
<br />a
<br />a, c. sinA=�-c �I,g=4� G+ )(--a)=c�1—E
<br />a
<br />f i A, a B, b, c B=90°—A, b= a cotA, e=
<br />gin A.
<br />Lt
<br />A, b B, a, c B=90'—A, a = b tan A, c = b
<br />Cos A. - -I
<br />A, c B, a, b B=90°—A, a = csinA, b= ccosA,
<br />Solution of Oblique Triangles
<br />Given Required a sin B a sin C
<br />A, B, a ,h, o, Cb — sin A ' C = 180°—(A + B), c = sin A
<br />b sin A a sin C
<br />B, c, C sin B=a , C = 180°—(A -l• B), e, = sin A ' .
<br />(a—b}
<br />a, b, C A; B, c. �1-f-B=180°-C, tan :', tan '- (A+B•)(A—B)= a + b ,
<br />II asin C
<br />o=
<br />sin A
<br />( a, b, e A, B, C a=a-f-a+c,sin'A—lg~
<br />2 be
<br />g.I sin
<br />t I,
<br />a+b+c
<br />ccs, 6, a Area, B = 2
<br />f1' bcsin A
<br />A, -h c= .Area' area = 1
<br />a'- sin B sin C
<br />A, B,C,a .,Area area =
<br />, 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosineofthevertical angle. Thus: slope distance=319.4ft.
<br />w°e Vert. angle =6' 101. From Table, Page IX. cos 5° lW=
<br />e''� �9959. Horizontal distance=319.4X.9959=318.08 ft.
<br />1
<br />1 'De Pnq\e . , Horizontal.distance also= Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5°10'=.9959. 1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 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=302.6 ft, Horizontal distance=3026= 14 X 14 =3026-0==3a28ft.
<br />2 X 502.6
<br />4 - MAGE IN U. B•. Ib r
<br />
|