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<br /> n / 3�/ Z 3. TRIGONOMETRIC FORMUL.E
<br /> oG a & 3 ,
<br /> �•� �/ zZ — B B
<br /> a ° a a a
<br /> AA
<br /> �_b a
<br /> a
<br /> Right T�sngle . Oblique Triangles
<br /> Solution of Right Triangles
<br /> For Angle A.iia =a,cos= Q,tan= b,sot=a,sec=b, cosec
<br /> / a
<br /> [given Required a $ �
<br /> a, b A;'B,c tan A= b= cotB,o = az+ s = a
<br /> Z
<br /> q ay c A, , b sin A=!=cos B,b=� o+a o—a =c J 1—a
<br /> r
<br /> A, B=90°—A,b =acotA,c= a
<br /> sin A. --
<br /> b B,a, a B=90°—A,a = b tan A,c= b
<br /> S _ Z - cos A.
<br /> �A,a B,a, bB=900—A,a=e sin A,b=a cos A,
<br /> �51
<br /> Solution of pblique Triangles
<br /> � � "�Given :_ Required B °— asin C
<br /> A, B,a b, e, C b = sin AC (A+B), a =
<br /> l y sin A
<br /> bsin A
<br /> / 1 ca, b B,e, C sin G„
<br /> = S = 180°—(A+B),c = asinC
<br /> _ ? r a sin A
<br /> g'), \div a, b, C A, B,a A+B=180°—C,tan (A—B)= a—b)taB)
<br /> an I(A+
<br /> �J� r
<br /> a sin C +
<br /> sin A
<br /> a
<br /> 3/ a, b, o A, B, Cs'= 2 ,sin;A= b a
<br /> a a ,C=180°—(A+B)
<br /> a; 9�:e , rea s
<br /> , area = a_T8—a s— s—c
<br /> A, b, c Alga area = b a sin A L
<br /> 7 i
<br /> 2 � u
<br /> Wsin Bsin C
<br /> B,C,a Area area = 2 sin A
<br /> REDUCTION TO HORIZONTAL
<br /> Horizontal distance—Slope distance multiplied by the
<br /> fcosine of the vert'co langle-Thus:slope distance=319.4 ft
<br /> ee
<br /> Z) Vert angle=b°10 From Table,Page IX.Cos 60 lot=
<br /> } \ Z �5tlm • .9959. Horizontal distancq=319.4X.9959=31&09 ft
<br /> A¢g1e Horizontal distance also—Slope distance minus slope
<br /> Z(? .9 4e distance times(1—eosine of vertical angle). With the
<br /> rn ) same figures as in the p example the follow-
<br /> /�.}, Zj• N \ Q Horizontal distant ing result is obtained ►afsarmd=.9069.1-.9969=.0041.
<br /> 43 13 319.4X.004141.s1.810.4-1:1q ft,v When the rise is known,the horizontal di is tely:—theslope dist-
<br /> ce lease square of the rise divided by twice.the slo Thuss rice=l4 ft.,
<br /> ope distance=SD8.8 it. Horixontal 32=302.28 ft.
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