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66e, 22 G3S )9
<br />3 TRIGONOMETRIC FO
<br />/p�9 1 B B�
<br />3S r�� 9�i JS Vl s
<br />g h 0,1 12 h c a c l a�JCC
<br />, 1 / �A b CA
<br />�G$ . 31 15 %� i�i3 �.��Right Triangle —— Oblique Triangles
<br />i5 z 3S ' - Solution of Right Triangles
<br />3 For Angle :±`sin,= c ,cos ,tan= b , cot = a , sec = b, cosec = a
<br />©QpIJ o0.
<br />! 9 a Given Required � ry b2
<br />a, b d, B.;o tan A = b = cot B, c = a1 + bz a 1.+ �:
<br />35.3 2
<br />999 3 9 S b% r%l`fv a; c.' A, B, b sin cos B, i a)(c—a)=c�1—c2
<br />Ff j _ G'p 3 99 j;Z9f' 1 A, a;' .B. -b, c B=90°—A, b = a cotA, e,= sin A. A
<br />j . A, b � . B, a, c B = 90°—A, a = b tan A, c =
<br />fcos A. l
<br />A, c B, a, b I B = 90°—A, a c sin A, b J-1 cos A, g v
<br />n •••,, %i'"/ , Solution of Oblique Triangles .
<br />f i ? ji(� O O. 6 p f� . , Given Required a sin B a sin C
<br />180°—A+B),
<br />sin A c=
<br />' ( sin A
<br />b sin Aa sin C
<br />i A,a b B c,C sinB= a ,C=180—(A+B),c= sin
<br />a, b, C A, B, c A+B=180°— C, tan 2' (A—B)= (a—b) tan z (A+B)�
<br />a + b
<br />. ZD 30 a=asinC
<br />b - fr,� - A
<br />'L D
<br />0
<br />9993. S,
<br />sin
<br />Y? Of,
<br />r'
<br />a, b, e,
<br />A, B, C
<br />a+b+oi — _ I(s— b)(s—c)
<br />s = 2A
<br />2 .,sin --v b c '
<br />UT
<br />sin 2 aB=-) c , C--180 (A+B)
<br />a+b+c.
<br />a, b;- c
<br />Area
<br />s = 2 , area
<br />Z _ _
<br />A, b, c"
<br />Are ab
<br />e sin A '
<br />area = 2'
<br />a2 sin B sin C
<br />Z
<br />A, B, C, a
<br />Area
<br />area = 2 sin A
<br />3•
<br />•REDUCTION TO HORIZONTAL
<br />���•
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the vertical angle. Tb us: slope distance =319.4 ft.
<br />10+=
<br />Stance
<br />Vert. angle =5° 10+. From Table, Page IX. cos 50
<br />e a�
<br />N 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />—
<br />�1p4
<br />riz ttiestane also= slope
<br />CG -co With
<br />Ve
<br />ine ofSlope
<br />angle).
<br />distance (l vertical angle).
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 51 101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31:319.4-1.31=318.09 ft.
<br />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 ft.,
<br />slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =302.6-0.32=302.28 ft
<br />2 X 3026
<br />' MADE ai Y. a. A.
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