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 = c/a Given | Required tan A = a/b = cot B, c = √(a²+b²) = a√(1 + b²/a²) a, b | A, B, c 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 b = (a sin B)/sin A, C = 180°-(A+B), c = (a sin C)/sin A 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 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 Slope distance Horizontal distance 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 × 14)/(2 × 302.6) = 302.6-0.32=302.28 ft. 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×.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×.0041=1.31. 319.4-1.31=318.09 ft.