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TRIGONOMETRIC FORMULIE
<br />B B B
<br />a c a c a
<br />Ig b C Ab G A C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i For Angle A. sin = a , cos = b , tan= a , cot = b , sec o , cosec = o'
<br />c c b a b a
<br />Given Required a
<br />a,b. A,B,c tanA=b= cot B,c= a2+T2 a 1T as
<br />a,c A,B,'b sinA=a=cos B,b=�/(C+a)(o—a)=c�1—off
<br />A, a B, b, c B=90°—A, b = a cotA, c= sin A.
<br />A, b B, a, e B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />A, c B, a, b I B = 90°—A, a =.c sin A, b = e, cos A,
<br />Solution of. Oblique Triangles
<br />Given Required a sin Ba sin C
<br />A, B, a. b, c, C b = sin A , C = 180o—(A + B), c = sin A
<br />b sin Aa sin C
<br />A, a, b B, e, C sin B = a , C = 180°—(A + B) , c = sin A
<br />a, b, C A, B, c A+B=180°— C, tan 1,b
<br />(A—B)= (a—b) tan '-, B)�
<br />t asin C a+
<br />e,=
<br />sin A
<br />b, c A, B, C s=a+2+c'sin'A= V(3— a(C—c
<br />sin aB= J(s—aa C=180°--(A+B)
<br />a, b, c Area 8= a+2 +c , area = s (s—a (s—) (s— c
<br />b c sin A
<br />A, b, c -Area area = 2
<br />az sin B sin C .
<br />B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e
<br />1 cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />�9opVert. angle= 51 101. From Table, Page IX. cos 51 10'=
<br />040
<br />6
<br />1s y 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />5, Anse a distancet timest(lecosine oflvertcalance ngle).11Withlthe
<br />.v0 same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 10'=.9959.1—.9959=.0041.
<br />319.4X.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 />ante less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3026— 14 X 14 _302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />xAne w u.e.A-
<br />I
<br />A
<br />yJ
<br />f�
<br />�i
<br />�s o
<br />TRIGONOMETRIC FORMULIE
<br />B B B
<br />a c a c a
<br />Ig b C Ab G A C
<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />i For Angle A. sin = a , cos = b , tan= a , cot = b , sec o , cosec = o'
<br />c c b a b a
<br />Given Required a
<br />a,b. A,B,c tanA=b= cot B,c= a2+T2 a 1T as
<br />a,c A,B,'b sinA=a=cos B,b=�/(C+a)(o—a)=c�1—off
<br />A, a B, b, c B=90°—A, b = a cotA, c= sin A.
<br />A, b B, a, e B = 90°—A, a = b tan A, c = b
<br />cos A.
<br />A, c B, a, b I B = 90°—A, a =.c sin A, b = e, cos A,
<br />Solution of. Oblique Triangles
<br />Given Required a sin Ba sin C
<br />A, B, a. b, c, C b = sin A , C = 180o—(A + B), c = sin A
<br />b sin Aa sin C
<br />A, a, b B, e, C sin B = a , C = 180°—(A + B) , c = sin A
<br />a, b, C A, B, c A+B=180°— C, tan 1,b
<br />(A—B)= (a—b) tan '-, B)�
<br />t asin C a+
<br />e,=
<br />sin A
<br />b, c A, B, C s=a+2+c'sin'A= V(3— a(C—c
<br />sin aB= J(s—aa C=180°--(A+B)
<br />a, b, c Area 8= a+2 +c , area = s (s—a (s—) (s— c
<br />b c sin A
<br />A, b, c -Area area = 2
<br />az sin B sin C .
<br />B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e
<br />1 cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />�9opVert. angle= 51 101. From Table, Page IX. cos 51 10'=
<br />040
<br />6
<br />1s y 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />5, Anse a distancet timest(lecosine oflvertcalance ngle).11Withlthe
<br />.v0 same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Cosine 5° 10'=.9959.1—.9959=.0041.
<br />319.4X.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 />ante less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope distance=302.6 ft. Horizontal distance=3026— 14 X 14 _302.6-0.32=302.28 ft.
<br />2 X 302.6
<br />xAne w u.e.A-
<br />I
<br />A
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