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<br />fi� - -*h TRIGONOMETRIC FORMULAE
<br />B
<br />id A A C
<br />—b C
<br />t -s Right Triangle Oblique Triangles
<br />'• '' Solution of Right Triangles
<br />'3f
<br />cFar Angle A. Bin = a ,cos= b , tan = a , cot = b ,sec = cosec =
<br />e a b a b a
<br />[liven Required ax
<br />:a,b A,B,o tanA=b=cotB,e=N/-a2-+-T2 aI+ap
<br />a, c A, B, b sinA= =cosB,b=✓ a -I -a (c-a)=c�1-oe
<br />lir A, a B, b, 0 B=90°—A, b = a cotA, c=
<br />in A.
<br />A, b B, a, c B = 90°-A, a = b tan A, c = b
<br />X08 A.
<br />A,c B, a, b B=90°----A,a=asin A,6=ccoo .9,
<br />a Solution of Oblique Triangles
<br />Given Required a sin.B
<br />A, B, a b, c, C b -
<br />in A ' C = 1800-(A + B), c sin A
<br />bsin A asin C
<br />`=1 A, a, b B, c, C sin B-= a ,G' = 1801—(A -(- B), c = sin A
<br />g; l 'a, b, C A, B, is A+B=180°- O, tan J (A -B)- fib, tan 1,,(A+B)�
<br />a + b
<br />a=
<br />a sin C
<br />in A
<br />A, B, C s=a+2+c,sinA-�$ ba-°
<br />v
<br />? I sinjB=�(s-aa
<br />r Y
<br />a+b+c
<br />a, b, o Area s= 2
<br />I
<br />l ,) A, b, a Area area = b a e2 "A
<br />k W sin B sin C
<br />4,B4O,a Area area = 2sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance —Slope distance multiplied by the
<br />a cosine of the vertical ansle.Thus: slope distanee =319.4 ft.
<br />L StgPIlm Vert. angle=6° 101. From Table, Page lK. cos 6'1[Y=
<br />9959. Honzantal distance=819.4K.9959=318.09 ft.
<br />^ gloms ApINe a Horizontal distance olso=Slope distance minus slope
<br />Ve distance times (1 --cosine of vertical. angle). With the
<br />same fit urea as in the preceding example, the follow -
<br />H
<br />orizontal distance Ing result is obtained. Cosine 51 pY=.9959.'1—.9959=.0041.
<br />819.4X.0041=1.91.819.4-1.31=318.09 ft.
<br />rise is known, the horizontal distance is approximately:—the slope dist-
<br />quare of the rise divided by twice the slope distance. Thus: rise=l4 It.,
<br />=902.9ft Horizontal distance=302.8— 1-4 Y14 =9029-0.32=30Z7Bft.
<br />2 X 8026
<br />MAOE/k U. B. K.
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