|
DIA'
<br />HOW
<br />FR
<br />Enter on;
<br />-of the s
<br />stag aco
<br />run ve(ti
<br />line repr
<br />location
<br />the dott
<br />gives th`
<br />from th
<br />be add(
<br />Oistancl p
<br />scale r
<br />S. of v( _
<br />is the
<br />TRIGONOMETRIC FORMULIE
<br />B B B
<br />C c c'
<br />a a a
<br />A b C .A�b C A b C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = c ; cos = , tan = b , cot = a , sec = 6 , cosec = a .
<br />Given Required t
<br />a,'b A, B, c tan A = b = cot B, c = a' } b' = a 1 } ar
<br />a, c A, B, b sin A = =cos B,b = (c+a) c—a) = c I—
<br />A, a B, b, c B = 90°—A, b = a cot A, c =sin A.
<br />A, b B, a, c B = 90°—A, a = b tanA, c = b
<br />cos A.
<br />A, c B, a, b B = 90°—A, a = c sin A, b = c cos A.
<br />Solution of Oblique Triangles
<br />Given Required a sin B
<br />A, B, a b, c, C b= sin A ' C =180° — (A +B), c = sin A
<br />A, a, b B, c, C sin B=b sin A, C =180°—(A }B), c = a sin'C
<br />a sin A
<br />a, b, C A, B, c A}B=180°-C,tan#(A—B) =(o -b) tangy (A}B)
<br />a -{- b
<br />a sin C
<br />c = sin A-
<br />a, b, c A, B, C s= a+2 c, sin J- A = (S— b(S—c)'
<br />sin I B= (S -a ((cs—c) C=180°—(A+B)
<br />a, b, c Area s=a+2+c, area= s(s—a) (s—b) (s—c)
<br />A, b, c Area area= b c sin A
<br />2
<br />A, B, C, a Area area = at sin B sin C
<br />2sin A
<br />REDUCTION TO HORIZONTAL.
<br />Horizontal distance -slope distance multiplied by the
<br />cosine of the vertical angle. Thus, for a slope distance of
<br />400.6 ft. and a vertical argle of 4' 40' -the cosine of lolx•
<br />4' 40', taken from a table of natural trigonometrical S Vel• An'�1a C
<br />functions, 9967, and horizontal distance -4M.6 x .9%7
<br />=402.27 ft. llori2ontal distance
<br />Horizontal distance also -Slope distance minus elope dis-
<br />tance times a -cosine of vertical angle). Using the same figures as in the preceding example -
<br />Cos.4'40'=.9967.1-.9967=.0033.403:6x.0033-1.33 ft. Horizontal dist.-403.6-1.33-402.27 it.
<br />When the rise is (mown, the horizontal distance may be found by the following approximate
<br />rule: -the slope distance less the square of the rise divided by twice the slope distance. Thus,
<br />for a slope distance of 3722 ft., and a rise of 15 ft. the horizontal distance=
<br />372.5- 15x15 =372.5-.30=372.2 ft.
<br />2X372.5
<br />EUGENE DIETZGEN CO. MADiM°'" i+
<br />
|