TABLE III. TRIGONOMETRIC FORMULAE
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 = e/a
Given | Required | tan A = a/b = cot B, c = sqrt(a^2 + b^2) = a * sqrt(1 + b^2/a^2)
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a, b | A, B, c |
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 | b = (a * sin B) / sin A, C = 180° - (A + B), c = (a * sin C) / sin A
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A, B, a | b, c, C |
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 = (b * c * 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: slope distance = 319.4 ft.
Vert. angle - 5° 10'. From Table. IV. cos 5° 10' = .9959.
Horizontal distance - 319.4 x .9959 = 318.09 ft.
Horizontal distance also - Slope distance minus slope distance times (1-cosine of vertical angle). With the same figures as in the preceding example, the following result is obtained. Cosine 5° 10' = .9959. 1-.9959 = .0041.
319.4 x .0041 = 1.31. 319.4 - 1.31 = 318.09 ft.
When the rise is known, the horizontal distance is approximately: -the slope distance less the square of the rise divided by twice the slope distance. Thus: rise = 14 ft.
slope distance = 302.6 ft. Horizontal distance = 302.6 - (14 x 14) / (2 x 302.6) = 302.6 - .32 = 302.28 ft.