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 sin A = a/c = cos B, b = √((c+a)(c-a)) = c√(1 - a²/c²) a, c | A, B, b B = 90° - A, b = a cot A, c = a/sin A. A, a | B, b, c B = 90° - A, a = b tan A, c = b/cos A. A, b | B, a, c B = 90° - A, a = c sin A, b = c cos A, A, c | B, a, b 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 sin B = (b sin A)/a, C = 180° - (A + B), c = (a sin C)/sin A A, a, b | B, c, C A + B = 180° - C, tan ½(A - B) = ((a - b) tan ½(A + B))/(a + b), a, b, C | A, B, c c = (a sin C)/sin A s = (a+b+c)/2, sin ½A = √((s-b)(s-c)/(bc)), a, b, c | A, B, C sin ½B = √((s-a)(s-c)/(ac)), C = 180° - (A + B) area = (b c sin A)/2 a, b, c | Area s = (a+b+c)/2, area = √(s(s-a)(s-b)(s-c)) a, b, c | Area area = (a² sin B sin C)/(2 sin A) A, b, c | Area A, B, C, a | Area 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, Page IX, 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. 7 0 MADE IN U.S.A.