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XII Natural Tr f !� ,l c� 4';6 TRIGONOMETF4C i�3ltl�[UL� S'
<br /> ._i 1
<br /> Angle.Sin. Tan. Soo �- S 1,1G`- /J ..� .^ 7 7 3 J Z <1 i
<br /> j
<br /> - 3 a
<br /> y L� -3) 3 I. a o ` ► b 3���
<br /> _ Right Triangle
<br /> .SS C
<br /> ��O61i ue Triangles
<br /> Solution of Right Triangles }I
<br /> i s / iy, i For Angle sin = ,cos= o ,tea= b ,cot = b,sec= b, aosie e
<br /> �- (liven Required a
<br /> 7 7 a, b A, B + 3 .+
<br /> ,e tanA=b= cotB,c = aE = a } �_
<br /> a' c A,B, bsinA_—=conA b=\/ o a e—a =o�Y—a
<br /> e
<br /> A,a B, b, c B=90°—A,b=acot A,c= a 3/, /
<br /> ��- sin A, d
<br /> A, b B,a, c B—90°—A,a = b tan A,c= b 3. 4(7 `1
<br /> m ----- cos A�-7 ,7 of Oblique Triangles
<br /> 9 AG'
<br /> 35 .b7s A.,�_ lc �. A,c B, a, b B=90°—A,a=e stn A,b=c cos A, �:�� 4- l
<br /> 10.5766 1 Solution' G,
<br /> Given Required
<br /> 3 .S3o7�4 �� _ 7 _ SJ ...Z A, B, a b, e, C b = sin A ' C= 180°—(A-I B), c = 81i a� ;r
<br /> 583F; !r ` � --
<br /> .5s:. ,! j .1 A, a, b B,e, C sin B= b ar30 ia Ari = 180°—(A {-B),c = sin A
<br /> a, b, C A, B, c A+B=180°—C,tan z(A—B)— a—b tan A B)
<br /> 20.592, ZU — ,
<br /> 30.5948.��/ F- •l •3 _ a sin C a+ b
<br /> 40.5972.'
<br /> 50.5995.1,
<br /> JC�ff
<br /> 0.6041: �, J / «rya, b c A, B, C s=a+2+c sinjA=_Ils
<br /> T V
<br /> ✓.y be
<br /> 8 .6088.76 . 2 i � v 3y <, b6. / sin'B=�(1 a ,C-180°—{A+B)
<br /> 40.6111.772 1.263a % —
<br /> .6134.778 1.2661 1.0 /"/ l,'3 3 7 a b
<br /> 3 a, b, a Area a= + +°, area = s(e—a s— )(s—e
<br /> 86 .6157.78131.26901.62, ;, r 2
<br /> 10.8180.78 1.27191.618; b c sin A
<br /> 20.6202.7907 1.2748 1.612] / f ' ' A, b, c Area wren = 2
<br /> 30.8225.79541.27781.606' r-i g r_.
<br /> az sin B sin C
<br /> 40.6248.80021.28081.801 i._- A,B,C,a Area area = --
<br /> 50.8271.80501.28381.595 1.242 6 ' 2 sin A
<br /> REDUCTION TO HORIZONTAL
<br /> Cosh.Coto.Cosec. See. Tan. ShL Mole Cosh.Cote,Con. Z11-5 Horizontal distance=Slope distance multiplied by the
<br /> ' cosine of the ve�ticalangle.Thum alopedistance X919 41 G
<br /> ,/8 a�gte�pe Vert angle=6 IW. From Table,Page IX cos b°10,=
<br /> 3 o- ° 1 3-i _ loge 1e Horizontal d1stance=819.4X.9959-31800 ft.
<br /> Y4—
<br /> $
<br /> Aro a Horizontal distance also=Slope distance minus slope
<br /> distance times(1—cosine of vertical angle). With the
<br /> same figures as in the preceding example,the follow-
<br /> /61-34' Horizontal distance ing result is obtained.Cosine 50 10'=.8960.1—.go69=.004L
<br /> 3 y s19.4X.60t1=t.$L 319.4-1.31=31809 tt
<br /> 7mt
<br /> _�.�� When the rise is known,the hodzentai dhslax is RPMwimatel ••—the slope dist.
<br /> 1 ' ante less the square of the rise divided by tries the slope distance. Thus:rise=14 ft.,
<br /> slope distance-300.8 ft. Horizonhd distauee M s-14 X 14� 8-0.3Z=30S 98 ft.
<br /> 2 X 3426
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