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TRIGONOMETRIC FORMULIE
<br />B B B
<br />e a a c
<br />a
<br />:A 6 C A G C
<br />c
<br />RightTriangle Oblique Triangles
<br />Solution of -Right Triangles
<br />For Angle A. sin = C , cos= tan= b cot= a , sec = b, cosec =
<br />Given-, I Re4uired ¢
<br />B, + —
<br />;�.. b as
<br />sin A = 6 =Cosa:b=�{c l a { j aa.
<br />C �� Qa
<br />� r' -A, a.::•B3' b, o B=90°—A,fb _ ¢cotA,o= a
<br />sin A.
<br />y. A, b B, a, c B=90°—A, a = b tan A, c=b
<br />_ eosA:-
<br />A, c B, a, b B=90°—A, a= c sin A T r c cos 11,
<br />Solution of 'Oblique Triangles
<br />Given Renuired a sin.V
<br />!A, B,a b, C, C b= C=180°—(A+B),c=¢sin(
<br />sin A sin A
<br />b -sin A a sin C
<br />A, a,, b B, c, C ia ,C 180°—(A 1 B), c = sin A
<br />M a, b„ C.. A, $, e A -{-B=1$0°— C, tan -'2 A+B}
<br />(A—B) ¢—b} tan � (
<br />a -)- b
<br />' c =
<br />a sin C
<br />sin A
<br />1 .a, b, c A, B, C s=¢+2+C,sinaA= i .
<br />�
<br />-A`
<br />a c ,C=180°—(A+13� •
<br />.�
<br />a, b, a Area s 2 , area = s(s—a (s—b j (s—c
<br />b, e Areab C sin A
<br />a area = 2
<br />a'- sin B sin U
<br />q :A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />i
<br />s cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />tarpVert. angle=5" 101. From Table, Page IX. cos 611a'-
<br />m 9959. Horizontal distance=319.4X.9959=316.09 ft.
<br />5104�gle a"^ Horizontal distance also=Slope distance minus slope
<br />Vest. distance times (1—eoslne of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance inu result is obtained. Cosine 5° 10'=.0959.1—.9959=.0041.
<br />319.4X.0041=1.3!..119.4-1.31=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 distanee=302.6 ft. Horizontal distance=302.6— 14 X. 14 =302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />MADE IN U. 8, A.
<br />'19
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