TRIGONOMETRIC FORMULÆ 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 = √(a² + b²) = a√(1 + b²/a²) a, c A, B, b sin A = a/c = cos B, b = √((c+a)(c-a)) = c√(1 - a²/c²) 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, tan½(A-B) = ((a-b) tan½(A+B))/(a+b) c = (a sin C)/sin A a, b, c A, B, C s = (a+b+c)/2, sin½A = √(((s-b)(s-c))/(bc)), sin½B = √(((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, 403.6x .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= 372.5 - (15x15)/(2x372.5) = 372.5-.30=372.2 ft. 10