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GONOMETR IC FORMUL&
<br />GY- `
<br />a r a c
<br />a
<br />Right Triangle C ` —� Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin- _ , cos = c , tan= b , cat = a , see = L , cosec =
<br />Given Required az a4
<br />a,b A;B,e tauA=b= cot B,c=-V—a-} a—a 1 7�_
<br />06 a A, B, b sin A= 6 = cos B, b =1✓ (c { a (c—a) = c V 1- 42
<br />A, a- B, b, c B=90o—A, b = acot A,c= a vv
<br />sin A.
<br />b
<br />A, b -9, a, a IB = 90'—A, a = b tan A, c =
<br />I cos A.
<br />A, a B, a, b B= 90°—A, a = c sin A, b = c cos A,
<br />j Solution of Oblique Triangles
<br />Given Required asin B
<br />I A, Aa L, c, C b= sin A'C=180°—(A+B),c= snA
<br />b sin A
<br />A,a,•b B,c,C sin B= a ,G''=180'—(A-pB),o= sinA
<br />sin A
<br />a, b _C.,- A, B c A 8-180"— C, tang"(A—B�= (a—b) tan a (A+�
<br />., b,_
<br />o
<br />a sin C a-+ b
<br />sin A
<br />o, b, c A, B, C s.sin IA—
<br />2
<br />ti
<br />sin 5B= V (d ac
<br />a, b, c Area s=a�2+c, area = s(s—a s— (s—c)
<br />A, b, c Area ares = b o sin A
<br />2
<br />a' sin B sin C
<br />A, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e cosine of the vertical angle. Thus: slope distance=319.4 ft.
<br />�aoc Vert. angle=b° 101. From Table, Page IX. cos 5, id=
<br />ass m 995.9. Horizontal distance=319.4X.9959=316.09 ft
<br />51° ple a°'y Horizontal distance also= Slope distance minus slope
<br />4e 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 5° 10'=.9959. 1--.9959=.0041.
<br />819.4X.0041=1.31.319.4-1.31=316.09 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 />sltiope distance=81126 ft. Horizontal distance^SQ'L,6— 14 X 14 =8M" 32=3M23fL
<br />2 X 8026
<br />MADE 14 V. S. A.
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<br />GONOMETR IC FORMUL&
<br />GY- `
<br />a r a c
<br />a
<br />Right Triangle C ` —� Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin- _ , cos = c , tan= b , cat = a , see = L , cosec =
<br />Given Required az a4
<br />a,b A;B,e tauA=b= cot B,c=-V—a-} a—a 1 7�_
<br />06 a A, B, b sin A= 6 = cos B, b =1✓ (c { a (c—a) = c V 1- 42
<br />A, a- B, b, c B=90o—A, b = acot A,c= a vv
<br />sin A.
<br />b
<br />A, b -9, a, a IB = 90'—A, a = b tan A, c =
<br />I cos A.
<br />A, a B, a, b B= 90°—A, a = c sin A, b = c cos A,
<br />j Solution of Oblique Triangles
<br />Given Required asin B
<br />I A, Aa L, c, C b= sin A'C=180°—(A+B),c= snA
<br />b sin A
<br />A,a,•b B,c,C sin B= a ,G''=180'—(A-pB),o= sinA
<br />sin A
<br />a, b _C.,- A, B c A 8-180"— C, tang"(A—B�= (a—b) tan a (A+�
<br />., b,_
<br />o
<br />a sin C a-+ b
<br />sin A
<br />o, b, c A, B, C s.sin IA—
<br />2
<br />ti
<br />sin 5B= V (d ac
<br />a, b, c Area s=a�2+c, area = s(s—a s— (s—c)
<br />A, b, c Area ares = b o sin A
<br />2
<br />a' sin B sin C
<br />A, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e cosine of the vertical angle. Thus: slope distance=319.4 ft.
<br />�aoc Vert. angle=b° 101. From Table, Page IX. cos 5, id=
<br />ass m 995.9. Horizontal distance=319.4X.9959=316.09 ft
<br />51° ple a°'y Horizontal distance also= Slope distance minus slope
<br />4e 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 5° 10'=.9959. 1--.9959=.0041.
<br />819.4X.0041=1.31.319.4-1.31=316.09 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 />sltiope distance=81126 ft. Horizontal distance^SQ'L,6— 14 X 14 =8M" 32=3M23fL
<br />2 X 8026
<br />MADE 14 V. S. A.
<br />t
<br />I
<br />-
<br />GONOMETR IC FORMUL&
<br />GY- `
<br />a r a c
<br />a
<br />Right Triangle C ` —� Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin- _ , cos = c , tan= b , cat = a , see = L , cosec =
<br />Given Required az a4
<br />a,b A;B,e tauA=b= cot B,c=-V—a-} a—a 1 7�_
<br />06 a A, B, b sin A= 6 = cos B, b =1✓ (c { a (c—a) = c V 1- 42
<br />A, a- B, b, c B=90o—A, b = acot A,c= a vv
<br />sin A.
<br />b
<br />A, b -9, a, a IB = 90'—A, a = b tan A, c =
<br />I cos A.
<br />A, a B, a, b B= 90°—A, a = c sin A, b = c cos A,
<br />j Solution of Oblique Triangles
<br />Given Required asin B
<br />I A, Aa L, c, C b= sin A'C=180°—(A+B),c= snA
<br />b sin A
<br />A,a,•b B,c,C sin B= a ,G''=180'—(A-pB),o= sinA
<br />sin A
<br />a, b _C.,- A, B c A 8-180"— C, tang"(A—B�= (a—b) tan a (A+�
<br />., b,_
<br />o
<br />a sin C a-+ b
<br />sin A
<br />o, b, c A, B, C s.sin IA—
<br />2
<br />ti
<br />sin 5B= V (d ac
<br />a, b, c Area s=a�2+c, area = s(s—a s— (s—c)
<br />A, b, c Area ares = b o sin A
<br />2
<br />a' sin B sin C
<br />A, B, G; a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied by the
<br />e cosine of the vertical angle. Thus: slope distance=319.4 ft.
<br />�aoc Vert. angle=b° 101. From Table, Page IX. cos 5, id=
<br />ass m 995.9. Horizontal distance=319.4X.9959=316.09 ft
<br />51° ple a°'y Horizontal distance also= Slope distance minus slope
<br />4e 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 5° 10'=.9959. 1--.9959=.0041.
<br />819.4X.0041=1.31.319.4-1.31=316.09 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 />sltiope distance=81126 ft. Horizontal distance^SQ'L,6— 14 X 14 =8M" 32=3M23fL
<br />2 X 8026
<br />MADE 14 V. S. A.
<br />
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