TABLE III. TRIGONOMETRIC FORMULAE 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, 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: 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 - 0.32 = 302.28 ft. Rise Slope distance Vert. Angle Horizontal distance