TRIGONOMETRIC FORMULÆ
B
A c B
a C b A c
Right Triangle Oblique Triangles
Solution of Right Triangles
For Angle A. sin = a/c, cos = b/c, tan = a/b, cot = b/a, sec = c/b, cosec = c/a
Given | Required
tan A = a/b = cot B, c = √(a²+b²) = a√(1+b²/a²)
a, b | A, B, c
sin A = a/c = cos B, b = √((c+a)(c-a)) = c√(1-a²/c²)
a, c | A, B, b
B = 90°-A, b = a cot A, c = a/sin A.
A, a | B, b, c
B = 90°-A, a = b tan A, c = b/cos A.
A, b | B, a, c
B = 90°-A, a = c sin A, b = c cos A.
A, c | B, a, b
Solution of Oblique Triangles
Given | Required
A, B, a | b, c, C
b = (a sin B)/sin A, C = 180°-(A+B), c = (a sin C)/sin A
A, a, b | B, c, C
sin B = (b sin A)/a, C = 180°-(A+B), c = (a sin C)/sin A
a, b, C | A, B, c
A+B = 180°-C, tan(½(A-B)) = ((a-b) tan(½(A+b)))/(a+b)
c = (a sin C)/sin A
a, b, c | A, B, C
s = (a+b+c)/2, sin(½A) = √(((s-b)(s-c))/(bc)),
sin(½B) = √(((s-a)(s-c))/(ac)), C = 180°-(A+B)
a, b, c | Area
s = (a+b+c)/2, area = √(s(s-a)(s-b)(s-c))
A, b, c | Area
area = (bc sin A)/2
A, B, C, a | Area
area = (a² sin B sin C)/(2 sin A)
REDUCTION TO HORIZONTAL
Horizontal distance = slope distance multiplied by the cosine of the vertical angle. Thus, for a slope distance of 403.6 ft. and a vertical angle of 4° 40'—the cosine of 4° 40', taken from a table of natural trigonometrical functions, = .9967, and horizontal distance = 403.6 x .9967 = 402.27 ft.
Horizontal distance also = Slope distance minus slope distance times (1—cosine of vertical angle). Using the same figures as in the preceding example—Cos. 4° 40'=.9967, 1-.9967=.0033, 403.6 x .0033=1.33 ft. Horizontal dist. = 403.6-1.33 = 402.27 ft.
When the rise is known, the horizontal distance may be found by the following approximate rule:—the slope distance less the square of the rise divided by twice the slope distance. Thus, for a slope distance of 372.5 ft., and a rise of 15 ft. the horizontal distance =
(15x15)/(2x372.5) = 372.5-.30=372.2 ft.
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