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 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 deg - A, b = a cot A, c = a/sin A. A, b | B, a, c | B = 90 deg - A, a = b tan A, c = b/cos A. A, c | B, a, b | B = 90 deg - 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 deg - (A + B), c = (a sin C)/sin A A, a, b | B, c, C | sin B = (b sin A)/a , C = 180 deg - (A + B), c = (a sin C)/sin A a, b, C | A, B, c | A+B=180 deg-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 deg-(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: slope distance = 319.4 ft. Vert. angle - 5 deg 10'. From Table. IV. cos 5 deg 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 deg 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.