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TRIGONOMETRIC FORMUL/E
<br />f� e�
<br />B B B
<br />t a a o a
<br />C' A
<br />L'
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />ka b a b c o
<br />For Angle A, sin = ,cos = —, , ten = 3 cot=—,sec=.b , cosec = a
<br />Given. Required
<br />a,b A,B,o tanA=b=cotB,c=
<br />a
<br />a, a A, B, b sin A = ='c& B, b = V c—+a) (c—a) =C
<br />A. a •- B, b, 0 B=90°—A, b =_a cotA,
<br />sin A.
<br />A, b B, a, c B'= 90°—A, a = "b tan A, e = b
<br />- cos A.
<br />A, o B, a, b B = 90°—A, a--•=. o sin A, b = c cos A.,
<br />Solution of Oblique Triangles
<br />(liven Required a sin B
<br />B)
<br />A, B, a b, e, C b = 180°—(A -}- � o = a sin C
<br />� sin A sin A
<br />' A, a, b B, c, C sin B= a b sin A ,C = 180°=(A -{- B), c = sin A
<br />a, b, C A, B, c A+B-180°— C, tan 7 (A—B)= (a—b) Can ; b { B)
<br />a } b
<br />_ a sin C
<br />o _ sin A
<br />a, b, c A, B, C,sin A-
<br />2 # VVl b e
<br />3.
<br />sin }B=.�l a a 0 ),C -180`—(2+B)
<br />Areas—a+2+c, area
<br />A, b, e. Area area = b.e sin A
<br />2
<br />A, B, C, a Area area = a= sin B sin O 2 sin A
<br />REDUCTION TO HORIZONTAL .
<br />Horizontal distance—Slope distance multiplied by the
<br />ce cosine of the vertical angle. Thus:slope distance =319.4ft.
<br />5tafi Vert angle=5 td. From Table, Page DL cos 50S11W=
<br />S1O�° paa�e a H9orizontal distancesalsoe Slope distanceminusslope
<br />yc 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 5010=.9959.1—.9959=•o041.
<br />Wherithe rise is known, the9horiiontal distan a is approxirsmately:—tbe slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise= 141t.,
<br />slope distance=302,6 fL Horizontal distance=sme— 14 X 14 =3M"32=90`L2SfL
<br />2 X 902.6
<br />MADE In U. 9, A.
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