TRIGONOMETRIC FORMULÆ B A B C A B C A C 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) s=(a+b+c)/2, area = √(s(s-a)(s-b)(s-c)) a, b, c Area area = (b c sin A)/2 A, b, c Area area = (a² sin B sin C)/(2 sin A) A, B, C, a Area REDUCTION TO HORIZONTAL Slope distance Horizontal distance Vert. Angle Rise 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. When the rise is known, the horizontal distance is approximately:-the slope distan- ce 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. MADE IN U. S. A.