|
b�` c %i TRIGONOMETRIC FORMUL/E '
<br />B
<br />ir_ 15
<br />c C C
<br />/2. 30: �� - y /-� a o a.
<br />G. r 7�- r ,:
<br />9 ii7 l �; �• -- -= 4tit3106 CAA ti t o b C
<br />` Z' �{p� /06• oa _ _ _ Right Triangle Oblique 'Triangles
<br />Solution of Right Triangles
<br />'
<br />-Fe
<br />or Angle A. sin = , 'os e , tan= b , cot = a , sec = b, cosec =
<br />a
<br />Given Required a - z
<br />a, b -d, B ,c [at) A = =cot B, c = ti.`cc + o- = rc 1 + 2
<br />- vq
<br />a, c A, B, sinA=�=Cos B,b=TicJ1-02
<br />A, ra B, b, c B=90°—A, b = acotA, c= a
<br />sin A.
<br />1, _'. �._-.•' % Z , o / `5� i't b
<br />��✓� A, b B, ra, c B = 90°—A, a = b tan A ,, c =
<br />cos A.
<br />r/. .. �!�y U' -A, c B, a, b B=90°—A, a = csin A,,b= c cos A,
<br />I „G z r--�'
<br />Solution of Oblique Tiangles
<br />O J ✓ -Given Required
<br />3_j /i/�Z3—• '7G asinB asinC
<br />%2?_- A, B,a b, c, C b= sinA G'=150°—,A+B),c=
<br />sin A
<br />/2. o b sin A - ctsin G'
<br />A, a, b B, c, G' sing= a C = 1S4`—(A = B), c = sin A
<br />13 �79
<br />a, b, C A, B, c A+B=180°— G', tan NAI—B)= a—b) tan! �A , B)
<br />.G8 yc +
<br />•e sin C a b
<br />�� / rin A.
<br />a, b, c A, B, C s=a+2+c,sin?.A= `�'.`•^b a—c)
<br />-5 77^
<br />�T /¢D /•t 9B `�!S ,22 j b_ sin?B=1j(a—a}ts—c) G'=150°—(A+B) .
<br />a+b,+c �.
<br />a, b; 'c• 'Area s = 2
<br />J , urea = 1.. ,(s—a (.3—b) (s—c
<br />f i b e sin A
<br />C1 g 7 A, b, c Area area = 2
<br />y •� ;� l !o: {•y
<br />a2 sin B sin C
<br />6 t'A 1 B, C,a Area area = 2 sin A
<br />�arr 'fT e I' REDUCTION TO HORIZONTAL
<br />73 r 1 �� 1 Horizontal distance=SGape distance multiplied by the
<br />cosine of the verticalan ie. Thus: slope distance=319.4ft.
<br />e t __ staff Vert. angle=51 10r. Fnom Table, Page IX. cos 51 lor=
<br />(Q yi 0 h 9959. Horizonlal distance=319.4X.9959=315.09 ft.
<br />�i :?� .. e ea�oP n¢le tz Horizontal distance' a1so=Slope distance minus slope
<br />distance times (1—cosine of vertical angle). With the
<br />same figures as in the Freceding example, the follow -
<br />Horizontal distance ing result is obtained. Ccsine 50 101=.9959.1—.9959=.0041.
<br />0 319.4X.0011=1.31. 319.4-1..31=318.119 ft.
<br />When the rise is known; the horizontal distance fx approximately:—the slope dist-
<br />- ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 it.,
<br />- - slope distance=_302.6 ft. Horizontal distance=302.6— 114 X 14 =302.6-0.3.2=302.28 ft.
<br />f tti P'X 302 6
<br />NAGE In -U. B. A. V
<br />
|