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<br /> i1oc 7 P 3
<br /> TRIGONOMETRIC FORMULA&
<br /> [)c, B A B
<br /> a a C a a
<br /> '7 r A A
<br /> { C
<br /> t 4 b p b a b
<br /> Right Triangle Oblique Triangles
<br /> - Solution of Right Triangles
<br /> a b a_
<br /> For Angle A. sin = c ,cos a 'tan= b ,cot= a,sec=b, cosec® a
<br /> Given Required ry e
<br /> 9 a, b d, B,o tan d=.b= oot B,e= a► { = a I+s�
<br /> a, a A,B,b ,sin d—a :coo B,b=�/ c}a o—a =a J1—a
<br /> d,a B, b, a B=90°—A,b =a cot A,a= sin A, `1
<br /> l d, b B,a, a B=90°=A,a = b tan A,o=�►--
<br /> coo A.
<br /> 2 A,o B,a, b B=90° A,a =o sia A,b= c cos A,
<br /> Solution of Oblique Triangle
<br /> ' Given Required
<br /> A, B,a b, c, C b= sin A ' C= 180°---(A+B).a =si A
<br /> r J U.gib_ A, a, b B, c, C sin B= b sa A.0= 180°—(A+B),a = � d
<br /> a, b, C A, B,c A+B=180°-C,tan j(A—B)=(a—b)a+bA+B)9
<br /> _a sin C
<br /> sin A
<br /> b, a A, ,-C a=as+2 ,sinJA= Vis- bo—c
<br /> sin fB—_I a—axa—a a c
<br /> 9 ,C=180°--(A+B)
<br /> a+b+a
<br /> a, b, a Area a= 2 , area = s(a—a s—TTF-s
<br /> / A; b, a Area area = bosin A 2
<br /> as sin B sin C
<br /> A,B,a,a Area area = 2 sin A
<br /> REDUCTION TO HORIZONTAX#
<br /> Horizontal distance—Slope distance multiplied by the
<br /> cosine of the vertical angle.Tl+Z slope ce s919.4lt
<br /> ais��ce
<br /> Vert Honzontaladi�tas�ce=81&4XPas 6°.1W=
<br /> Horizontal di also=Slopo distance minas slope'
<br /> i distance times —eosino`of vertical angle). With the
<br /> Yh s.S—o 3 same figures.as a the preceding example,the follow-
<br /> 4
<br /> ine result is obtsiped.Cosine 60 10'=.986®.1= 866=.0011.Hoiiaontal distance 918.4X.0041=1.31.928.4-1.31=91809 ft.
<br /> When the rise is known,tbq hotisuntil distance is approximately:—the slope dist-
<br /> 'Sb'�a`� s once less the square of the rLve divided by twice the slope distance. Thus:9 =l4 ft,
<br /> cT! Y,,y-a 6 slope distance-3026 ft. Horizontal distance=302.6—14 X 14 1 X 0=9020--0.32=308261...
<br /> d�t -# 3'O 7 .. «. r .���.�iirri�iiWeaiiFr..�r.M i�..rr.'1...�. Fmk a►e.s N
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