TRIGONOMETRIC FORMULAE
B
a
A c C
Right Triangle
b
BC
B
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
a, b A, B, c
tan A = a/b = cot B, c = sqrt(a^2 + b^2) = a sqrt(1 + b^2/a^2)
a, c A, B, b
sin A = a/c = cos B, b = sqrt((c+a)(c-a)) = c sqrt(1 - a^2/c^2)
A, a B, b, c
B = 90° - A, b = a cot A, c = a/sin A.
A, b B, a, c
B = 90° - A, a = b tan A, c = b/cos A.
A, c B, a, b
B = 90° - A, a = c sin A, b = c cos A.
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(1/2(A-B)) = ((a-b) tan(1/2(A+B)))/(a+b)
c = (a sin C)/sin A
a, b, c A, B, C
s = (a+b+c)/2, sin(1/2 A) = sqrt(((s-b)(s-c))/(bc)),
sin(1/2 B) = sqrt(((s-a)(s-c))/(ac)), C = 180° - (A+B)
a, b, c Area
s = (a+b+c)/2, area = sqrt(s(s-a)(s-b)(s-c))
A, b, c Area
area = (bc sin A)/2
A, B, C, a Area
area = (a^2 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 =
Slope distance
Vert. Angle
Rise
Horizontal distance
10