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TRIGONOMETRIC FORMULIE
<br />7 3 i B B B
<br />71
<br />71 '/7 9
<br />r y° b C b C A• b C 34�•i�
<br />7g ' ,o�oa e 3 z © __� , �j
<br />q- .Right Triangle Oblique Triangles
<br />0 5 t' _ Solution of Right Triangles �G
<br />a b c c
<br />(q. �� �. - For Angle A. sin= a , cos= —,-tan= b , cot ¢ , sec = b; cosec
<br />equired a
<br />Given R 2
<br />{ 62 a }
<br />A,,B,c tangy=—=cocB,c= a2 = 1 z
<br />b ! a
<br />(� b, /�t�
<br />0 . �/ u; ' "a' ;� 'A, B, b sinal = ! = cos B, b = \1 (o+a) (a—a) = o J 1— o
<br />A' ` B`/, M x` �� 5 A,`a. B, b, o B=90°—A, b = z cotA, c= a
<br />sin A.
<br />A, b.B,a,,c B=90°—A,a=btanA,c=cos A. - <
<br />g;c- B, a, b. I B=90°—A,a=0sin A,b=ccos A, Ze-) • „'
<br />Solution of Oblique Triangles
<br />J _ Given Required a sin B , a sin C
<br />A, B; ,a b, c, C b = sin A ' C = 180 —(A + B), c = sin A
<br />G • ; 6 {{
<br />Il Q r- ? 1.l' i C) b sin A a sin C 1
<br />A, a, b Ac,•C sinB= ,0=180o—(A+B),c=
<br />G a sin A Sim
<br />i
<br />rf�+} ; !J , t4w $ • �1 ~J 7 �r�. b . ir< n®�� o, b, C A, B, o A+B=180°—C,tan ''J(A—B)=(a_b)•an bA+B),� t:..,.
<br />4. J.�4 _asin C
<br />sin
<br />pL j �� )� ' 07 - a+b+c 'A= •� ls—'b)(&—c
<br />Com. ®3a�fl a b; o d, B, .0 s= 2 ,ein7V be ,
<br />--s=a s -
<br />Al
<br />sin I( )1 ) C=180'—(A B)
<br />/.►lS: 1 0'3 1 as +
<br />(� t l , �' J • ( 63� 31 r'. = a+b+o i r
<br />= - -- I .`s9'b;. c`• Area s= , areas—b (s,
<br />"'•_� A, b,.c Area area —1 b.a sin A
<br />0 ' S6 aR sin B sin C %J
<br />1 S A, B,' C, a 'Area area =
<br />2 sin A
<br />.REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by D the _ 1 cosine of the vertical angle. Thus: slope distance=319.4ft
<br />.4% Vert. Vert. angle = 51 101. From Table, Page IX. cos 50 101=
<br />; 3 o aie H 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />� ?; — T P A��1e a
<br />distanceHorizonttimest(lccosi a oflvertcal angle).istance rnus Withlthe
<br />same figures as in the preceding example, the follow-
<br />'Horizontal distance ing result is obtained. Cosine 50 �101=.9959.1—.9959=.0041. �}
<br />j 319.4X.0041=1.31.319.4-1,31=318.09 it. , ✓ _ - - When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />i • ' ry , G 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 =3026-0.32=302.28 ft.
<br />3 f 2 X 302.6
<br />wtae IN U. a.&
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