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400¢9 TRIGONOMETRIC FORMULF, t
<br />B B
<br />o •5 9o4 -z3 2�� �
<br />{o(° 1 8 Z9G�, c^ a a a a
<br />b33 Ul
<br />yC A C
<br />b C A b
<br />3 ,� • �r tRight Triangle Oblique• Triangles "
<br />Solution of Right Triangles
<br />S g a b a b c « c
<br />R , • Z Wit_ 6 For Angle A. sin = ,cos = ,tan = ,cot = —,sec= cosec^' �-
<br />S
<br />p q
<br />"(iven Required
<br />�
<br />a-a�2
<br />, , ,3303
<br />y
<br />v a, a A, _B, b sin A. = C
<br />� cos B, b = �/ (c+a) (c—a) = c 1=•a y
<br />S' t A "ia" B, b, c B=90° -A b= acotA,c= a s
<br />sin A.
<br />Z a
<br />A b B, a, c B= 90°—A a= b tan A c=
<br />t7 ° 1� 1,420 3 _ � _ - , , _ cos A.
<br />0 (p 146 �; (% A, c B, a, b I B = 90°—A, a = c sin A, b = c cos A,
<br />Z ) Jy.�y'�6 3..�5 Solution of Oblique Triangles
<br />13 l p Given Required a sin B a sin C
<br />_ A �B a b, c, C b = C = 180°—(A -I- B), a =
<br />! 9 ? , , sin A ' sin A
<br />17 qq 1 b sin A a sin C
<br />b B, e, C sin B= a ,C= 180° —(A+B),c = sin A
<br />' Q ' (a—b) tan_' (A+B)
<br />a, b, C A, B, a A } B=180°— C, tau , (A=B)= a Z b
<br />J � �� `'�� . i � • a — a sin C .
<br />asin A
<br />b, A
<br />- -}-b -{-c � �t =
<br />I a--b)(s—c
<br />a' , B, C s = 2 sin�A-� be
<br />Ir �3() sin IB=�'s—aa c , C=180°—(A+B)
<br />_ a+} +c
<br />+tai C 5 {� / U t ay..b; ,a " Aiea s ^ 2 , area = s(s—a -)-(8-- ) (s—c 6.
<br />''
<br />1 !t� �� y b c sln.A
<br />b; o Area area =
<br />7 2
<br />.' a sin B sin C
<br />lA, B, C, a' ' Area area = 2 sin A
<br />f
<br />� REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />{ cosine of the vertieal angle. Thus: slope distance=319.4ft.
<br />f� K [) spa Ic Vertangle =5° 101. From Table, Page IX. cos 51 ld=
<br />I Y v 1 e a� d 9959. Horizontal distance=319.4X.9959=31&09 ft.
<br />�N/� 5�op A�.1e a Horizontal mest(lecoslne of'vertcal distance thlthe
<br />y "< -4 same figures as in the preceding example, the follow -
<br />Horizontal Horizontal distance ing result is obtained. Cosine 50 101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is known, the 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=302 6— 14 X 14 =302.6-0.32=302.28 ft.
<br />O 2X3026
<br />0 K"s Ia U. 8. k
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