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TRIGONOMETRIC FORMULAE
<br /> B" Q B
<br /> rl 'l l d a
<br /> a o
<br /> a
<br /> A
<br /> �ba A C
<br /> S W7Right Triangle Oblique Triangles ^- r
<br /> q /o° Solution of Right Triangles
<br /> • 1 L f" Z�' F°r AngF°a on =a,cos—A,t.— b,cot = a.sec=b. cosec_ °—a
<br /> Given Required
<br /> d a,b. .A,B,o. tanA=b= cotAa = a + s = ta 1 +
<br /> a
<br /> 5
<br /> 00
<br /> a, e. A,B, b min =cosB,bo�-a a—a) =a`�1—
<br /> - V
<br /> B=9tl°—d,b=a cotA,a= a
<br /> sin A.
<br /> :
<br /> rl d,b B,a, a . B=90°—A,a 777erfr
<br /> b tan A,o= cos / 6 a t v
<br /> A,
<br /> a'-. B,a, b I B=900—A,a=e sin A,b=o cos A, �9 3%.A.1
<br /> Solution of Oblique Triyangle&
<br /> Given� Repaired _a sin B
<br /> 3 7 i o 7 d' A a b, o, �' b sin A ' C= 180°—(A+B), a _ a sin C
<br /> 3 sin A ,.
<br /> X4e. L/ L � bsinA
<br /> A, a, b B,a, C sinB== a ,a= 180°—(A 1-B),a _ a sin C
<br /> 3. 'f / sin A
<br /> a b, C A, B,a A+B-1800—C,tan j(A-B)_ a—b)tan#(A+B)
<br /> a + b ,
<br /> a =
<br /> a-Sha C
<br /> sin A
<br /> B, C a—aa,sin A—
<br /> 2 be ,
<br /> sing=J(a—aa ,C-180°—(A+B)
<br /> Area a=a 2 area =V7(8—.TF— J
<br /> (a—a a— ) .__
<br /> A, b, o Area Brea = b a sin A
<br /> 2
<br /> VO A,B,C,a Area area =a=Sin B sin C
<br /> 2 sin A
<br /> X REDUCTION TO HOtRIZONTAL
<br /> Horizontal distince=Slope distance multiplied by the
<br /> cosine of the vertical angle.Thus:slope distance=919.4&
<br /> Vert.angle=51 101. From Table,Page IX.cos b°itr=
<br /> 9869. Horizontal distance=81&4X.9959=81&o8 ft.
<br /> 5�0 Apa1e Horizontal distance also=Slope distance minus slope
<br /> Qe distance times (1—cosine of vertical angle). With the
<br /> Horizontal distance ing resusame lt figures btained Cosine 6 101==Vii=the ell �
<br /> 819.4X.0941=1.81.819.4-1.81=81&09 it
<br /> When the rise is]mown,the horizontal distance is approaimat the slope dist
<br /> aaee less the square of the rise divided by twice the slope'd=nce us:rise=14 ft.,
<br /> slope d3 me!1820& Horizontal dlstaace=80¢0--1 4.
<br /> t4s-o SS02 28 ft. dt
<br /> '- S XIOLB• �� ,,
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