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T NOMETRIC FORMULIE
<br /> 3 v `�z 9� ; ' Gfia �� B B
<br /> - 3
<br /> o3 �Qgo 7 �
<br /> 7:; ?- 3. da a
<br /> A CJS`S`5
<br /> C
<br /> ✓ g8_: 1 8 Y' Right Triangle Oblique Triangles
<br /> (��-
<br /> .4
<br /> 4W" . Solution of Right Triangles S(�,
<br /> 5 C Ci 4. . r•..
<br /> r '�' M For Angle A.. sin =a cos- b tan- a b a
<br /> fly ' �.ys 2 •s g c = c = b ,cot = -,sec= , cosec=
<br /> c2 a b' -
<br /> _ ✓ a
<br /> r c 1�9 ' �_ Given Retluired
<br /> a, b A, B a a
<br /> ,c tan A= = cot B c = az = 1 —
<br /> as
<br /> q
<br /> T a
<br /> z
<br /> a, c A, B, b `sin A=a a2
<br /> =cos B,b= c�a (c-a =a 1-"2
<br /> A,aB, b, a B=90°-A,b =a cotA,c= a
<br /> sin A. z i.
<br /> %
<br /> r / A, b B a, a B=90°-A a = b tan A c=
<br /> cos A.
<br /> c B, a, b B=90°-A a=c sin A,b=e cos A,
<br /> G y
<br /> 57 :'; S- 5�a l /�� /�rt Solution of Oblique Triangles
<br /> x , Given Required
<br /> 01 �7S - �� __ a sin B °- a sin C
<br /> B B,a b, c, C b sinA ' C= 180 (A+B), c = sinA
<br /> 5I,6r
<br /> yZ �� 4po, b B,c, C. sinB= bsaA,C= 180°-(A+B),a = aeiA
<br /> • sin A
<br /> b, C A, B, a A-�B=180°-C,tan J(A-B)= a-b tan A B
<br /> q2 F� gN3 a _asinC a+ b
<br /> -- A. sin A
<br /> ` ! 0 3 b, a A, B, C 8—a+—2�°,sinjA=�8 be
<br /> r� -
<br /> I yg J 7 3 sin$B=�
<br /> %' ( a ,c-180 (A+B)
<br /> 3, °.
<br /> 3 -7 q b, C Area 8=a+b+8, area = 8(8-a 8- (8-c
<br /> A, b, c Area area = b e sin A
<br /> 2
<br /> A,B,C,a Area area =az sin B sin C 2 sin si
<br /> L ro L REDUCTION TO HORIZONTAL
<br /> Horizontal distance=Slope distance multiplied by the
<br /> 9 • 3 ce cosine of the vertical angle.Thus:slope distance=319.4 ft.
<br /> iy a ytab Vert. angle=50 Id. From Table,Page IX.cos 5°10'=
<br /> L� J' I �' lope a� 9959. Horizontal distance=319.4X.9959=31&09 ft._ 5 Ap¢Xa Horizontal distance also—Slope distance minus slope
<br /> v { 9 Ve
<br /> distance figures
<br /> times(1—cosine of vertical angle). With the
<br /> as in the preceding example,the follow-
<br /> orizogtal distance ing result is obtained.Cosine 510'=.9959,1—.8859=.0041.
<br /> 31&4X.0041=1.31.319.4-1.31=31&O9 fL
<br /> When the rise is known,the horizontal distance is approximately:—the slope dist-
<br /> ce
<br /> Heless the square of the rise divided by twice the slope distance. Thus:rise=l4 fL,
<br /> pe distance=302.6 fL Horizontal disianay.4=6—14X 14—_4M p,32--9022B ft.
<br /> S X 302.6
<br /> . YIIDa IN Y.a.ti
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