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r } TRIGONOMETRIC FORMULlE
<br />B B
<br />--i A a CAA b C A b C
<br />® Right Triangle I Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b o e
<br />For Angle -A., sin = , cos = , tan= , cot = , sec = , cosec =
<br />, t y. ,,d : en7 c c b a b a
<br />�: T 1G►ven• Required a z
<br />,c tan A = b =cot B, c = a2 _+T2_ = a 1 + a
<br />..
<br />.ilgy0 A, B, b• sinA=�=cosB,b=\/(c+a)(e—a)=o,11-02 .
<br />d, a. B, b, c B=90°—A, b= acotA, c= a V
<br />(' 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 B=901—A, a = e sin A, b= c cos A,
<br />Solution of Oblique- Triangles
<br />Given Required
<br />rA, B,a b, c, C b=asinB�C=180°—(A+B),o=asinC
<br />sin A sin A
<br />a
<br />A, a, b B, c, C sin g= b sin A ,C = 180°—(A sin C
<br />a B), c = sin A
<br />•77 �/ �)�(' 0. b, 0 A, B, e A+B=180°— C, tan a (A—B)= (a -b) tan ((A+B)�
<br />a sin C a
<br />sin A
<br />a+b +c I
<br />os, b, o A, B, C s= 2 ,sin,A— be ,
<br />sin 2B= J s a(c ,C=180°—(A+B)
<br />8,' b, a Area s=a+b+c, area
<br />2
<br />A, b, c Area b e sin A
<br />area = 2
<br />a2 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 />taupe Vert. angle =50 101. From Table, Page IX. cos 50 10'=
<br />p40 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />g� ,_,,e Horizontal distance also= Slope distance minus slope
<br />Ve a distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow-
<br />- Horizontal'distance ing result is obtained. Cosine 50 101=.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 />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=3026— 14 X 14 =302.6-0.32=30228 ft.
<br />2 X 302.6
<br />• NAGE W N. S. A:
<br />\1
<br />i
<br />r } TRIGONOMETRIC FORMULlE
<br />B B
<br />--i A a CAA b C A b C
<br />® Right Triangle I Oblique Triangles
<br />Solution of Right Triangles
<br />a b a b o e
<br />For Angle -A., sin = , cos = , tan= , cot = , sec = , cosec =
<br />, t y. ,,d : en7 c c b a b a
<br />�: T 1G►ven• Required a z
<br />,c tan A = b =cot B, c = a2 _+T2_ = a 1 + a
<br />..
<br />.ilgy0 A, B, b• sinA=�=cosB,b=\/(c+a)(e—a)=o,11-02 .
<br />d, a. B, b, c B=90°—A, b= acotA, c= a V
<br />(' 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 B=901—A, a = e sin A, b= c cos A,
<br />Solution of Oblique- Triangles
<br />Given Required
<br />rA, B,a b, c, C b=asinB�C=180°—(A+B),o=asinC
<br />sin A sin A
<br />a
<br />A, a, b B, c, C sin g= b sin A ,C = 180°—(A sin C
<br />a B), c = sin A
<br />•77 �/ �)�(' 0. b, 0 A, B, e A+B=180°— C, tan a (A—B)= (a -b) tan ((A+B)�
<br />a sin C a
<br />sin A
<br />a+b +c I
<br />os, b, o A, B, C s= 2 ,sin,A— be ,
<br />sin 2B= J s a(c ,C=180°—(A+B)
<br />8,' b, a Area s=a+b+c, area
<br />2
<br />A, b, c Area b e sin A
<br />area = 2
<br />a2 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 />taupe Vert. angle =50 101. From Table, Page IX. cos 50 10'=
<br />p40 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />g� ,_,,e Horizontal distance also= Slope distance minus slope
<br />Ve a distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow-
<br />- Horizontal'distance ing result is obtained. Cosine 50 101=.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 />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=3026— 14 X 14 =302.6-0.32=30228 ft.
<br />2 X 302.6
<br />• NAGE W N. S. A:
<br />
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