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-7
<br />TRIGONOMETRIC FORMULtE
<br />B B B
<br />e U c c a
<br />_ - b C �� b C A b C
<br />Right Triangle' Oblique Triangles
<br />Solution of Right Trianglis
<br />_ For Angle A. sin = cos = b ,tan — ¢ ,cot = U ,sec = cosec = e
<br />¢
<br />Given Required a z
<br />I' '¢,b _A,B,c tanA=—cotB,c= ¢= 1 be=¢ 1•-1-=
<br />g a'
<br />a,, c _4, B, b a a
<br />A, a B, b, c B=90—A, b = ¢ cotA, a= a
<br />jj sin A.
<br />4' A, b 11, a, c B=90°—A, a= b inn A, c= b
<br />cos A.
<br />A, c B, ¢, b B=90° "A, a = c sin A; b = e. cos A,
<br />Solution of Oblique Triangles
<br />Given Required _ a sin B
<br />�.A B, a b, e, C 8 sin A ,•O = 180°—(.A -+ B), c =asin C
<br />sin A
<br />b sin A a sin C
<br />'A a b B, c C_ sing= a ,C= 180°—(A f B7,c sink r/ .—
<br />• ¢, b, C A, B, c fi-}-B-180 — C, tan
<br />c =
<br />a. sin C 1`
<br />sin A
<br />Y
<br />a, b, e A B, C s —a+2 °"in zA=alis bbta—c)
<br />sinl,D— �a�i� c),C=ISV—(A+B)
<br />a, 'b, c Area s=¢+b+C at
<br />A, b e Arca b e sin A
<br />area =
<br />A, B, C, a Area area =
<br />a' sin B sin C
<br />2 sin A
<br />4
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />5 - e Cos ineOftile vertical angle, Thus: slope distance=319.4ft.
<br />Vemc .9959. aHarizontal distance=910.4X.99a�5110% From Table. 9e 5IX, Cos ° 1tN-
<br />3 8.09 ft.
<br />,1Q Horizontal distance also =Slope distance minus slope -
<br />111distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained, Cosine 51 l0'=.9959. 1—.9959=.0041.
<br />3L9.4X.0041-1.31, 319,4-1.31=316.09 ft.
<br />When the rise isoknown, the horizontal distance is approximately:—tbe slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise =14 ft.,
<br />slope distance=302.G ft, Horizontal distance =302,6— 14 X 14 =302.&-0,32=302.28 ft.
<br />' 2 X 3026
<br />&AUE IN V, B.A.
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