|
.•fit• ��• r 1 •1, ` �') E
<br />• TRIGONOMETRIC FORM1JL;2E
<br />t1 B B B
<br />a d a d a
<br />z,
<br />C
<br />Right Triangle Oblique Triangles
<br />Solution of Right _Triangles
<br />For Angle A. sin= c , cos = , tan = b , cot= , sec = b , cosec = a
<br />Given Required --72-
<br />I a, b -4, B ,c tan A = b cot B, c = V az -{ = a y , T
<br />a, d A, B, b sin A = a = cos B, b = %/ (c+cc) (c—a) = e 1— a x
<br />z
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />A, b B, a, c B=90`—A, a= b tan A, c= b
<br />cus A.
<br />A, c B, a, b B = 90°—A, ¢ = c sin A, b -- c cos A,
<br />Solution of Oblique Triangles
<br />Given Required
<br />A, B, a b, c, C b = s sl n 4 C = 180°— (A -F B) , d ` s n 4
<br />b sin A ¢ in C
<br />A, a, b B, c, C sin B= sin
<br />a, b, C A, B, c A -{-B-180°— C, tan HA-1(a—b) tan 'a (A+B)*
<br />c=
<br />a sin C
<br />sin A
<br />a, i5, c A, B, C s= 2.
<br />sin 3B=yf ( a d C=180°—( A+B)
<br />a, b, c Area s= 2 area a(a—cc) {s—b) (s—d
<br />bcsinA
<br />i A, b, c Area area =
<br />i 2
<br />a" sin B sin C
<br />A, B, C, a Area area =
<br />L sin 4
<br />REDUCTION TO HORIZONTAL,
<br />Horizontal distance= Slope distance muItiplied by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />ll i�'ce Vert. angle= 5" 101. From Table, Page IX: cos 51 10Y=
<br />e iS 9959• Horizontal distance=319.4X.9959=318.69 ft.
<br />c�1ot' Y,g1e Horizontal distance also=Slope distance minus slope
<br />1e P4 distance times (i—cosine of vertical angle). With the '
<br />same figures as in the preceding example; the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 10'=.9959.1—.9959=.0041.
<br />319.4 X.06 -11=1.3L 319.4-1.31=318.09 St.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 %,
<br />slope distance =302.6 ft. Horizontal distance --m6— 14 X 14 =302C- 0.32=802.28 ft.
<br />2X302.6
<br />MAOI IN U, 6. A.
<br />
|