"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 = √(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 Slope distance Vert. Angle Rise Horizontal distance When the rise is known, the horizontal distance is approximately:—the slope dist- ace 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, Page IX. 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 follow- ing result is obtained. Cosine 5° 10'= .9959. 1-.9959=.0041. 319.4×.0041=1.31. 319.4-1.31=318.09 ft. MADE IN U. S. A.