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r.
<br />+ ` ' TRIGONOMETRIC FORMULAE d•
<br />B B
<br />75
<br />� r
<br />jA C
<br />Right Tiie�gle Oblique Triangles
<br />Solution of Right Triangles
<br />\ � r b_ a b c
<br />s
<br />For Angle;, cos ,tan = -b ,cot = — ,sec = —, cosec =
<br />c a b a
<br />1 - Given
<br />Z -
<br />f
<br />4
<br />a,b
<br />A,B,c'
<br />tsnA=b=cotB,c= aE+ 2=a 1 E -
<br />az
<br />3�9 �. J(9
<br />a, c
<br />A, B, b
<br />sin A = G = cos B, b = \/ — a (c+a) (ca) = c 1= 2
<br />L
<br />A, a
<br />,-
<br />B„b, c
<br />a
<br />B=90°—A, b= acotA,"c=
<br />'
<br />sin A.
<br />6 ¢
<br />A, b
<br />B, a, c
<br />B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />G
<br />}
<br />- A, c
<br />B, a, b I
<br />B = 900—A, a = c sin A, b = c cos A,
<br />b ` Solution of Oblique Triangles
<br />o f
<br />Given
<br />Required
<br />B a a sin C
<br />A, B,a
<br />b, c, C
<br />_sin °
<br />b
<br />A'C=180—(A+B),c=
<br />l7 d�
<br />A,a,b
<br />B,c,C
<br />sin sin
<br />b sin Aa sin C
<br />sin B= a ,C=180—(A+B),c=
<br />sin
<br />A B— a (a—b) tan (A+B)
<br />+ -180 — C, tan , (A—B)=
<br />to+b ,
<br />a C
<br />sin
<br />e—
<br />sin A
<br />-
<br />a+b+c
<br />a, b, c
<br />A, B, C
<br />;A_ �(a)
<br />s— 2 ,sin \ b c '
<br />(s—a)(s--o)
<br />sin jB= `� a c C=180°—(A+B)
<br />C� 7
<br />.r
<br />'a, b, o
<br />Area
<br />a+b+c
<br />s= , area = 1/s(s—a (s— ) (s—c
<br />2
<br />, b e.
<br />-
<br />Area
<br />b0Rio A
<br />area = 2 r�
<br />as sin B sin C
<br />-
<br />A,B, C,a
<br />Area
<br />area = 2 sin A
<br />REDUCTION
<br />TO HORIZONTAL
<br />tt
<br />Horizontal 'distance= Slope distance multiplied by the
<br />^^
<br />taoce
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />Vert. angle =5° 10+. From Table, Page IX. cos 50 10+=
<br />5� 000
<br />ass
<br />q 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />a Horizontal distance also=Slope distance minus slope
<br />MILO
<br />distance times (1—cosine of vertical angle). With the
<br />ve
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance
<br />ing result is obtained. Cosine 50 10+=.9959.1—.9959=.0041.
<br />Y 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
<br />ft.
<br />Horizontal distance=302.6— 14 X 14 ==6--0.32=302.28 ft.
<br />2 X 3026
<br />v
<br />MADE IN U.S.A.
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