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<br />TRIGONOMETRIC FORMULA'
<br />' B B
<br />c a c a c a
<br />. d b A�b C A C
<br />a,1 Right Triangle Oblique Triangles
<br />ISolution of Right Triangles
<br />For Angle A. i n = , cos = a , tan = b , cot = a , sec = b , cosec= a
<br />jGiven Required
<br />a,b A,BI,c tan A=b= cot B,e= al z=a 1+as
<br />a A, B , b ein A = Q - = cos B, b = V (e+a) (c—a) = c 1— a 2
<br />A a' B 6 c B= b=a = a'
<br />.a. 90 A, cotA,c—
<br />�� BID A.
<br />A, b B„a, a B=90°—A,a = btan A, c= b
<br />cos A.
<br />A, a B,�a, b- B'= 90°—A, a = c sin A, b = e coa A,
<br />Solution of Oblique Triangles
<br />Given Regaired a sin B , a sin C
<br />A, B, a b, c; C b = sin A ' C = 180 —(A +-B), c = sin A
<br />( b sin A a sin C
<br />A, a, b B, c; C sin B = a , C = 180°—(A -1 B) , c = sin A
<br />a, b, C A, B, c A+B-180°— C, tan 3 (A—B)= (a—b) tan 11a -f- b } B)
<br />b '
<br />a sin C
<br />c sin A .
<br />� �• a, b, a A, B; -\C a = a+2 , Sin 3A=.\'(s— be
<br />sin 3B=ti�(4 aa c— c),C=180°—(A+B)
<br />! .
<br />�a, b, c Area s=a+2+c, area
<br />A, b, c Area area = b e sin A
<br />2
<br />a= sin B sin C
<br />'A, B, C, a Area area = 2 sin A
<br />REDUCTION TO. HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />ale Vert. angle= 6' 104 From Table, Page IX. cos 50 lor=
<br />e 655 9959. Horizontal distance=319.4X.9959=31&09 ft. pp
<br />S,00 Aog,( a distaHorizontal
<br />ce timest(iccosine ofverticalangle).lWithlthe
<br />Qe same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 51 101=.9959.1—.9959=.0041.
<br />i 1 319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />When the rise Is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the squaee of the rise divided by, twice the slope distance. Thus: rise =14 ft.,
<br />slope distance=3020 ft. Horizontal distance=3026— 14 X 14 =302.6-0 32=302.28 4t.
<br />2 X 302.6
<br />MADE IN U. e. A.
<br />A
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