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<br />�t1 ' f S .. >+:.,r" Vie•:
<br />TRIGONOMETRIC FORMULzE
<br />B: n
<br />r1. b CA, b CA. b C
<br />Right riangle Oblique Triangles 4
<br />s
<br />Solution of Right Triangles
<br />.a. b a b c a
<br />4_ For Angle A. sin — — cos = tan = cot = ,sec = , cosec
<br />Given Required z
<br />2_
<br />ti' a,b A,B,a tauA=b=cotB,c= a1+ z=a 1+•as
<br />i a ` az
<br />A, B; b sin A = = cos B, b = V (c+a) (c—a) -- c r' 1— R:.
<br />S4a� e �+ o'
<br />a
<br />}� �, a• c B W 90°—A., b= a cotA, c— a
<br />' �7—
<br />8sin A.
<br />A, b B, a, a =90°=A,a tan _
<br />cos A. ?
<br />A, a. B, a, b B=90°—A,a=osin A,b=ccos A,
<br />1F �I =`� '; <' •° 1 _ Solution of Oblique Triangles
<br />Given Required, a sin B , a sin C _ _''u
<br />A, B,a sind'C=180—{A-� B),e—. sinA
<br />-
<br />b sin d a sin C r
<br />` ..A, a, b . B, c, C sin B =. a .C = 180°—(A f R), c = sin A ti
<br />„ (a—b) tan a (A+B)
<br />q b, C A, B, e A+B=180°— C, tan (A—B)— a + b F
<br />_ a sin C
<br />sin A
<br />r
<br />a + +a
<br />, C s. 2
<br />i3,sin;A-1
<br />sin }B— C=1800—(A.+B)
<br />a+b+c t
<br />_ _ a b, a Area s— 2 , area
<br />b0sin A
<br />u A, b, c Area area = 2
<br />`k
<br />if a= sin B sin C
<br />€_ A, B, 1, a Area area = 2 sin A
<br />1{
<br />REDUCTION TO HORIZONTAL
<br />Y ( Horizontal distance=Slope distance multiplied by the
<br />r I cosine of the vertical angle. Thus: slope distance =319.4 ft. j
<br />:
<br />�11_11;1
<br />Vert. angle— 5' 10'. From Table, Page IX• cos 501(Y=
<br />999.Horizontal distance=319.4X.9959=318.09 ft.ope
<br />distance ttimesdistance
<br />(L ecosine aflvertical angle)tance rWith lthe
<br />VQ same figures as in the preceding example, the follow-
<br />ing Horizontal distance ing result is obtained. Cosine 5° 10 =.9959.1—.8959=.0941.
<br />319.4X.0041=1.91. 319.4-1,31=318.09 ft.
<br />4 ' When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />+ -- aiice less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />slope distance=302.8 ft Horizontal distance=302.6-2 14=3026-0.32=.302.28 ft.X 3024 MA'
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