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
tan A = a/b = cot B, c = √(a²+b²) = a√(1 + b²/a²)
a, b | A, B, c
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
b = (a sin B)/sin A, C = 180°-(A+B), c = (a sin C)/sin A
A, B, a | b, c, C
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
Horizontal distance
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 × 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, IV. 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 following result is obtained. Cosine 5° 10' = .9959. 1-.9959=.0041.
319.4×.0041=1.31. 319.4-1.31=318.09 ft.