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Transcription
TRIGONOMETRIC FORMULAS
B
c
a
B
c
a
B
c
a
C
b
A
C
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A
C
b
A
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
Horizontal distance=Slope distance multiplied by the cosine of the vertical angle. Thus: slope distance=319.4 ft.
Vert. angle = 5° 10'. Since 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.
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.