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.