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19
<br />� J
<br />7k0 7
<br />t
<br />i
<br />- TRIGONOMETRIC FO'RMIJUE
<br />B• 1s B.
<br />0. b� C A G ti
<br />Right Triangle Oblique Triangles
<br />Solution_ of Right Triangles
<br />a b a b c a
<br />For Angle A. sin = c , cos =' a ., tan = b , cot. = � , see = b, cosec = a
<br />Given' Required _ a 2
<br />( a, b A, B ,c tan A = = cot B, c = al + b'= _. a 1 { as
<br />a, a A, 03
<br />I1.. A, a B„b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />h
<br />j A, b B, a, c B=�O°—A,a = bran A, a'='
<br /># , cos 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 Ba sin C
<br />: A, B, a b, c, C h = sin A ' C = 130°—(A + B), 0 sin A
<br />b'sin A a sin C
<br />A, a, b 73, c, d sin B= a C = 180p—(4 --t- B), a — sin A
<br />a, b, 0 A, B, c A-FB=180°—C, tan z (A—B) (a—b) ten z (A,+B) ..
<br />a +.b
<br />_ a sin`l>'''''
<br />c sin A
<br />a+b+e ' kr
<br />a, b, c .A, B, C s — 2 a A.=-�h a
<br />1, r, sin g=1f (—alai. ).,C 184 (A+B)
<br />a+h i C
<br />Area s 2 ,area s(a—u) s- b) (s—a)
<br />A, b, c Area h e sin A
<br />area = 2
<br />a"- sin P sin C _
<br />j' A, B, C,a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slone distance multiplied by the
<br />-�"
<br />ine ofth ] 1 Th 1 d' t —319 —
<br />cosa Ve lea a g e. Thus
<br />ape is ante— i
<br />$�afiooVert. anffle � 5° 10', From Table, Face 1X. cos 5' 101=
<br />e 3 N 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />glop g1e Horizontat distance also=Slope distance minus slope
<br />Ve . An LK distance times (t—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distanceing result is obtained. Cosine 5°
<br />10'=.99G9.1—.9959=.0041.
<br />3t9.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise is ]mourn, the horizontal distance is aoproximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. '1`has: 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 302.6
<br />MAOE IN U.S. A.
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