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TRIGONOMETRIC FORMUL/E
<br />B 8
<br />i a a
<br />?J, 1. y 35a y s Z� Right Triangle Oblique Triangles
<br />Solution of Bight Triangles
<br />D
<br />p�__r` For An le A. sin = a a b e a
<br />,I '31 {y g/ @ ep,/ �, g e , coo = ,tap = b' .cot = a , Bec = b, cosec =
<br />1'o' y r' �'+a"`%. C7 I Given Required a
<br />/% ® ;j � 1�a _, 3� c?� ( _ a, b.. A, B ,c tan = =._cot B; c = ax+ s = a 1 x
<br />t� ABdO,BinA coo Bb e a'
<br />3le
<br />02
<br />d 39� �f / f f'' %'I ) A, a�� '=B, b, c B=90°—A, b = a cotA, c= a.
<br />6� � sin A.
<br />L
<br />_ A.b •11,a, c ��=90°—A.a=btan A,c=
<br />7! _
<br />� Cos A.
<br />[ f J .y d :, a :_c ` A, ° B. a, L B=90°—A, a = c sin A, b'= cos A,
<br />f .a a a
<br />r�I 4$. !l
<br />�I/0",5 ! t�¢4 q $ �� �tj'" ` ! Solution of Oblique' Triangles
<br />ft Giren Required
<br />4 r� > �1/� 1� � � :` { A, B; a b, c, C b = a sin B C = 180°—(A + B) a Bin C -
<br />L i Bing ' o — sin,A
<br />=flz a .0=180'—(A+e),c Y
<br />��f� A, a, b. B, c, C Bin B— Bin A train G
<br />„!i '��• G jj�.�'l Y "�' a sin A
<br />1,
<br />q, b, r' A, B, c A+B-180°-0, tsn: I B)— a—b) tan (A+B)
<br />B)
<br />I'I !S~ y"jl.p l ra r3�-— 4!= o =Jain C a + b
<br />a.�:�/ }�✓ �'� lain A
<br />.77
<br />�i - �� �!'S ��� r 6, `br G _'Ar , �i S—a+&+
<br />Tb �, Bin JA— (S (g_0)
<br />�N C
<br />Ar / '� i
<br />sin ill $—c .0=180°--(A-} B)
<br />y�{�! ' : .
<br />�- 4 r ^/ } ( f t' % o 3 0 ; tt' b, c Area s= 2 ,area = s (s—a s—) r—c)
<br />A; b, a Area area = b sin A tj
<br />Bin S sin C
<br />r-3 =' A. Br C,a Area a[ea =
<br />2 sin A
<br />73,
<br />•I 7� — .03 �. _3r 3' REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the -
<br />cosine of the ver tical angle. Thus: slope distance =319.4 ft. +
<br />iifl tt,, p Q( �J ✓ ` _ yta� Vert. angle=b° 10'. From Table, Page 7X. cos 51 NY=
<br />1 yam'e6 �' M. Horizontal distance=319.9X-9959=318.00 fL
<br />bugle � Horizontal distance also=Slope distance minus slope
<br />. '� ' �e distance times (1—cosine of vertical angle). With the .
<br />r..I I crS ,y U# (y. / same figures as in the preceding example, the follow -
<br />Horizontal distance rng result is obtained. Cosine 50 10'=.9959.1—.9959=.0041. -
<br />ii 319.4X.0041=1.31. 319.4-1.31 =318.09 ft.
<br />�- 3� •r , i�- f i f.. O 3l �' . When the -rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=14 fL,
<br />slope distance=3x2.6 St. Horizontal distance=3026— 14 X 14 =3028-0.32= 30228 fL
<br />2 X 802.6
<br />L s'MA°t
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