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 = √(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(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) = √((s-b)(s-c)/(bc)),
sin(1/2 B) = √((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
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 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) = 302.6-0.32 = 302.28 ft.