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q �, r/v haw TRIGONOMETRIC FORMUL/E
<br /> v, I�� B B B
<br /> {
<br /> c a e a a a
<br /> CA b QA b C
<br /> Right Triangle Oblique Triangles
<br /> Solution of Right Triangles
<br /> ' I For Angle A. cin =a,cos= b tan= a,cot = b,sec= o, cosec =o a b a b
<br /> a
<br /> (liven Required
<br /> a,b. A, B,o tan A=b= cotB,c =V--2—+T2-= a 1 { ��.
<br /> a, c A, B, b ` sin d=a =cos B,b= a a a--a
<br /> a � S=c.Jl-`os
<br /> A,a B, b, a B=90°-A,b =a cotA,a= a
<br /> sin A.
<br /> A, b B,a, a B=90°-A:,a =b tan A,c=cos A.
<br /> j\
<br /> _jC r' A,a B,a, b B=90°-A:,a=c sin A,b=c coo A,
<br /> v Solution of Oblique Triangles
<br /> Given
<br /> tl, B,a b, a, C b = snA ' C= 180°-(d 1-B),a= sin
<br /> b sin A
<br /> a i A, a, b B,c, C sin B= a ,C= 180°-(A {-B),o =asin A
<br /> / a, b, C A, B, a A+B=1.80°-C,tan (A-B)= -b tan
<br /> aU4±M
<br /> ✓'� a+ b . '
<br /> c =
<br /> a sin C
<br /> sin A
<br /> a, b, o A, B, C s=a+2,sin jA=
<br /> in'B—
<br /> a(� ,C=180°—(A+B)
<br /> 7 v a, b, a Area 8=a 2, area = a(s—a a— (8—a
<br /> A, b, c Area area = b o sin A
<br /> $ ,e e ----- 2
<br /> 4 3 z r a2 sin B sin C
<br /> A,B,C,a Area area = 2 sin A
<br /> REDUCTION TO HORIZONTAL
<br /> — Horizontal distance—Slope distance multiplied by the
<br /> u cosine of the vetticalangle.Thus:slope distance=319.4ft.
<br /> 'i. bce Vert. angle=5*id. From Table,Page IX.cos 5110-
<br /> AM Horizontal distance=319.4X.6969=31&09 ft
<br /> 51oQ Arg1e Horizontal distance also=Slope distance minus slope
<br /> v f 4e
<br /> distance
<br /> ame fiegntrreesea(1--cosine
<br /> i the P eeedingC am the ifollow�
<br /> Horizontal distance ing result is obtained.Cosine 60
<br /> 10,=.9669.1—.9959=.0041.
<br /> 3I&4X.0041=1.31.319.4-1.31=31&09 ft
<br /> When the rise is known,the horizontal distance is anprozimately:—the slope dist-
<br /> ance less the square of the rise divided by tVfte the abspe distance. Thus:rise=14 ft.,
<br /> slope distance=3026 ft. Horizontal distana r—M16—14 X14=302.6-0.32=30¢.28 ft.
<br /> 2X 3026
<br /> MADE IN D.LA.
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