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S. "T
<br />fly g�t4;% 53- moo_ t� Ory TR1GONaMETRIC FORMULIE
<br />r Fir. 5 B
<br />i,je c c
<br />Awa a
<br />'SL�l ` �f�4'� ' _ S N N.•. ;s Right Triangle Ohlique Triangles -�V, TITi
<br />'Ilk \ Solution -of Right' Triangles
<br />O Oi �� �' a b a b c c
<br />,Y� Far Angle A: cin = c , coe = ,tan = b ,cot = —,.BCC = a, cosec = a
<br />v _
<br />g1 - J� j
<br />.1 �,,.�• ' Given Required a l
<br />C �/ [� X41-Iki a, b A, B ,c tan. A = b = coi B,
<br />1+72 } as
<br />42
<br />r ~ �` �' a, e . A, B, b sin A = a = oos B,
<br />yA
<br />A, a B, b, c B=900—A, b = d cot A, c = a
<br />sin A.
<br />�.b
<br />A, b B, a, c B 90°--A, a = h tanfA, c= .
<br />I _ cos A.
<br />t �A,, c B, a, b B =90°—A, a a sin A, b, c cos A,
<br />�i t��' (�• -
<br />%'' Solution ofOblique Triangles
<br />Given d
<br />a sin Ba sin C
<br />A, B, a b, e, C a ` sin A , O = 180°—(A + B), c = sin A
<br />a sin C
<br />ql ':,'e..• _ c 1 �` A, a, 1i • .B, e, C sin B= bsaip A ,O' — 180°---(A + B), c = sin A
<br />b, C A, B, c A+B�180°— C, tan (A—B)= (a -b) a + bA+B),
<br />/ r
<br />�~.0'3Z� �• =53
<br />aeioC
<br />a
<br />1 j zs sin
<br />- l ,• `5`�� ,., 6+ br •e ..A B, c a_ 2 ,ein'A—_V bo
<br />J
<br />a—a)(s--•e)
<br />•,'� -` � - 1 j\i(1 r sin�B—{
<br />� ,C=180°=(A-{-B)
<br />t {� a+b-i-c
<br />( b, c Areae= 2 ,area s
<br />r 1
<br />A, b, c Area area = b a sin A
<br />2
<br />tlA; B, C,a Area area = o' cin B sin c
<br />2 sin .A Q REDUCTION TO HORIZONTAL
<br />Horizontal distance—Slope distance multiplied by the
<br />cosine of the verticalangle. Thus: slope distance =3119.4ft.
<br />Ce
<br />Vert. angle=b° 101. From Table, Page IX. cos 6 101=
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />�'I ~� r /'� ( (j • ') c��o4e 9g1e a Horizontal distance also=Slope distance minus slope
<br />No
<br />t{f E distance times (1—cosine of vertical angle). With the
<br />�J Re
<br />Paine
<br />u1 figures as in the preceding example, the follow-
<br />'1r1Horizontal distaneeing result Is obtained. Cosine 50 101=.9959.1—.9959=.0941.
<br />- 319.4X.0041=1.31. 919.4-1.91=318.09 ft
<br />�I_, q �. j ' f • When the•rise is known, the horizontal distance is approximately:—the slope dist-
<br />'Ince less the square of the rise divided by twice the slope distance. Thus: rise=14 ft,
<br />slope distance==.8 ft. Horizontal distance=BM6-2X903814 X 14 =3n8�.32
<br />=3012M
<br />ft,
<br />MADE IN V. S. A.
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