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