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<br />TRIGONOMETRIC FORMULjE
<br />B B
<br />a
<br />a a
<br />C
<br />A A
<br />C b C b.
<br />4ight Triangle Oblique Triangles
<br />• Solution of Right Triangles
<br />a I
<br />for Angled. sin = , coo= b tan = a cot = b see cosec
<br />Given Required
<br />a, b A, B,c tan A cot B, c = NAW _+b2
<br />a2
<br />a, a 2, B, b sin A=2�=cos B,b=VTi+_a)(c—a)=c a2
<br />. : 0 —;72
<br />A A b, c -B 900—A
<br />a b= a cotA, 0=
<br />sin A.
<br />A, b B, a, e B 90--A, 4 = b tan A, -c = b
<br />co. A.
<br />A, c B, a, -b B=90*—A, a = c sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />�Given Required a sin B asin C
<br />'A, B, a b, c, C b = , C 1800—(A + B), c = —
<br />sin A sin A
<br />asin C
<br />'A, a, b B, c, C sin B 180*—(A + B), e
<br />a sin A
<br />b, C A, A 0 A+B=180*— C, tan 1, (A --P) —(a—b) tan _�_(A+-B)
<br />asinc
<br />-
<br />a + b
<br />sin A
<br />b +
<br />)av b, A, B, C 8=a+ 2 sinjA=C_C,
<br />sin jB=
<br />C=1800--(A+B)
<br />a+b+o ac
<br />Area s = 2,area __N1W((s—a
<br />b c sin A
<br />A, h, a Area area = — 2
<br />a2 sin B sin C
<br />1. A, B, C, a A . rea area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Ijorizontji *distance = Slope distance . muftiplie . d'by the
<br />cosine of the vertical angle pe'disfancc=319.4ft.
<br />Vert. angle=50101._ Fr6, NiT.bsi,!IoPage IX. cos 51 I1d=J
<br />9959. Horizontal distance=3193X-9959=318.09 ft.
<br />(D .
<br />swoe, Nc.9Horizontal distance. also= Sloe6 distance minus slope
<br />ve 17' --distance times (1—cosine'of *yertical,angle).- With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 60101=.9959. 1---�-.9959=.0041.
<br />3l9.4X.0041=1.31.319.4=1.31=318.09 ft.
<br />i When the rise is known,,tbe horizontal distance is'd —the slope dist-
<br />ance less the square of the rise divided y twice the slope distance- Thus: riie=14ft,
<br />slope distance=302:6h. Horizontal dls.tance=302:6— 14 X 14 =302.6-432=302.28%
<br />2X3026_
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