|
7-
<br />12 2
<br />0
<br />TRIGONOMETRIC FORMULfE
<br />B. B B
<br />c a r.�c a c a
<br />iAb CA b C�� C
<br />fRight Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i s b a b c c
<br />For Angle A. sin= e , cos = 0 , tan= 3 , cot = a , sec = b , cosec = a
<br />Given Required a
<br />tan A=L= cot B,c= a'=i z=a 1 } d,
<br />a, c A, B, b sinA=G=cosB,b=V/ o+a)(c—a)=c�1—Q=
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />b
<br />b B, a, c 8=90°—A,a = btanll, r,= cosA.
<br />I
<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
<br />A, B, a b, c, C b = a sin B C = 180°—(A -)- B), c = a sin C
<br />sin A sin A
<br />b sin A Ctt sin C
<br />A, a, b B; c, C sin B=. a ,=.180°—(A1 B),c= sin
<br />f a, b, C A, B, c A+B=180°= C, tan ' (A—B)= (a—b) tan
<br />a
<br />- a sin C +
<br />=sin A
<br />a, b, c A., B, C s=a+2+c,sin-2A=I—
<br />be
<br />sin 4B=_N1(.r
<br />- �
<br />a, b, c Area s = 2
<br />bcsin A'
<br />Area area = 2
<br />ii
<br />a'- sin 13 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 />y taoce Vert. angle=5° 101. From Table. Page IX. cos 50 10'=
<br />` e bis 0 9959. Horizontal distance=319.4X.9959=318.09 ft..
<br />S1o0 ope
<br />distan ettimesal t(1—cosine ofance �veaminustance
<br />rtoe cl angle)1With lthe
<br />l ye same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 511 101=.9959.1—,9059=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 It.
<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=14 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3020— 14 X 14 =302.0-0.32=302.28 ft.
<br />2 X 302.6
<br />- — UAD2 1j U. S. A.
<br />11 v
<br />Cj -% Ll
<br />99 1
<br />7-
<br />12 2
<br />0
<br />TRIGONOMETRIC FORMULfE
<br />B. B B
<br />c a r.�c a c a
<br />iAb CA b C�� C
<br />fRight Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i s b a b c c
<br />For Angle A. sin= e , cos = 0 , tan= 3 , cot = a , sec = b , cosec = a
<br />Given Required a
<br />tan A=L= cot B,c= a'=i z=a 1 } d,
<br />a, c A, B, b sinA=G=cosB,b=V/ o+a)(c—a)=c�1—Q=
<br />A, a B, b, c B=90°—A, b= a cotA, c= a
<br />sin A.
<br />b
<br />b B, a, c 8=90°—A,a = btanll, r,= cosA.
<br />I
<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
<br />A, B, a b, c, C b = a sin B C = 180°—(A -)- B), c = a sin C
<br />sin A sin A
<br />b sin A Ctt sin C
<br />A, a, b B; c, C sin B=. a ,=.180°—(A1 B),c= sin
<br />f a, b, C A, B, c A+B=180°= C, tan ' (A—B)= (a—b) tan
<br />a
<br />- a sin C +
<br />=sin A
<br />a, b, c A., B, C s=a+2+c,sin-2A=I—
<br />be
<br />sin 4B=_N1(.r
<br />- �
<br />a, b, c Area s = 2
<br />bcsin A'
<br />Area area = 2
<br />ii
<br />a'- sin 13 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 />y taoce Vert. angle=5° 101. From Table. Page IX. cos 50 10'=
<br />` e bis 0 9959. Horizontal distance=319.4X.9959=318.09 ft..
<br />S1o0 ope
<br />distan ettimesal t(1—cosine ofance �veaminustance
<br />rtoe cl angle)1With lthe
<br />l ye same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 511 101=.9959.1—,9059=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 It.
<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=14 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3020— 14 X 14 =302.0-0.32=302.28 ft.
<br />2 X 302.6
<br />- — UAD2 1j U. S. A.
<br />
|