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b J t
<br />1:z�
<br />`� �':c ' .'' r 3S .'TRIGONOMETRI'C =FORMULAE -
<br />GO 25 3 • • '
<br />0 4 ) io 'S
<br />B N B
<br />32
<br />pU 60Z V '� C
<br />7 0 .iZ 6211 3a a a a
<br />� �• �% /76 - S'/5 �--03 0
<br />15 1�.. "
<br />,
<br />Right Triangle Oblique Triangles
<br />0
<br />-z,-7 o. ,' 14- 3,0-D Solution of Right Triangles
<br />F . a b a. b' c e
<br />3O; For Angle A.. •sin = c , cos= c , tan L , cot = a , sec = b, cosec = a
<br />2j — DO '» Given Required — a — — — s
<br />a, b A, I3 ;c tan d b cot B, c a2 -F• 2 a az
<br />ll
<br />¢?/,�� t {� a,, e. A, ,B,. L
<br />sin A = o = coo B, b = (c+a) (tea) = o J 1— p
<br />t. 2tia. 7 h �, 1 - a
<br />G :�Z A; a. B,'b,.4 B=90°—A, b = a cotA, c=
<br />- sin A.
<br />A, B, .a, c B = 90°—A, a b tan A, c = b
<br />.. cos A.,
<br />A,c B, a, b B=90°—A,a=cainA,b=ccos A,
<br />�-200 ,ZG� Solution- of. Oblique Triangles
<br />(e G3T Given Required a'sin Bim. ° a sin C
<br />lD� ; •j vA B L = C-- 180 — A, B .c =
<br />-�,� 1 ,. ;'a •.u, P, -G' sin, 'Li ( 'i- )', sin A 1�
<br />a 4�ZI b`sEn�i aainC
<br />'r n i' �n ( % .� 1 �A,B c, C sin lj `a ,C = 180°—(A { B), c = sin A
<br />(a—b) tan (A+B)
<br />b C* A,`B, c A } B=180o — C, tan j (A=B)= a + b
<br />"-� a sin C
<br />r i c sin A '(
<br />x
<br />pp3g �S j a+b+c
<br />,a� b' c A' B, C s= 2 'aio�A- � .
<br />3 ® b c
<br />r
<br />(s—a)(8_4:=(A B
<br />4, t• (l A� L v. ( ein I,B= C=180 + )
<br />a c
<br />F
<br />a, b, a Area s = a+2+c J area = s(s—a (a— )`(x—c)
<br />\ besin A
<br />111 a7--4- f (70 i A, b, e, Area area = 2
<br />r ��1
<br />+a' sin B sin C
<br />�� �I ���:, ;iA, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL'
<br />X i Horizontal distance=Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance=319.4 ft.
<br />�st'°e Vert. angle=b° 101. From Table, Page IX. cos b° 101=
<br />'•; �%� `b (7' /1.�` oQe ass
<br />9959. Horizontal distance=319.4X.9959=318.09 fL
<br />S ,Q l (£i Ao�1e Horizontal distance also=Slope distance minus slope
<br />iJ v distance times (1—cosine of vertical angle). With the
<br />_ _ e same figures as in the preceding example, the follow-
<br />r G �/l `. n /L Horizontal distance irg result is obtained. Cosine 501o,=.9959.1 .9959=.0041.
<br />1 IV/ 1
<br />319.4X.0041=1.31.319.4-1.31=318.09 ft.
<br />dist-
<br />When the rise is known, the horizontal distance is approximately:—the slope AI / 4 `anee less the square of the rise divided by twice the slope distance. Thus: rise=14 fL.
<br />slope distance=3028 ft Horizontal distance=3028-2 X 302.8 14 X 14 —302. t.
<br />8-0.32=302.28 f `
<br />MAGE IN V. S. A. I
<br />k,�4
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