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3�• �z _
<br /> 37� TR�GONCtMETRIC .FORDJUL1E
<br /> 32.f3 ?. "9 Z2Ge
<br /> z f �•°d - - a.sy z8�6 B B B
<br /> ss o
<br /> � . a
<br /> zf, A a
<br /> C
<br /> ¢,�zss q tea/ _ _.;.4 - `Right Triangle Oblique Triangles
<br /> —✓ - ' ' •y z6 �h✓- o Solution of
<br /> zy,a z ✓ 1Y '/ 7 ,� L�. Right Triangles
<br /> a b a b c
<br /> 2 ✓---- �+'' .r ;- 9G� 7 For Awe A. an = coo=- tan= o
<br /> _-- V c ' e ' b ,cot =a,sec=b, cosec= -
<br /> -s- 13 j3�'f, �"0 3 2•` c Given, Required a
<br /> ✓ 7 ,J/` 3 ;2799 t/ _ �,GJ A B,e tanA=b= cotB,c= a2+ z =a 1+—
<br /> w , as
<br /> 71 ��3 J
<br /> 2J 77 iz " 7✓ �`• A B°b Siad=� a
<br /> f of - �.Z¢ / 3✓ 1 / Cab �/ � 9-a = — Of
<br /> 4
<br /> � __� Z�-- -✓ --- f / - B, b, c =90°-A b=a cotA c=a ma
<br /> _ .q, , c
<br /> r sin A.
<br /> 2Z.S ✓/i3 J..�¢✓ -z-2'%1 j22,1
<br /> Ili& _ B, a, c B-90°Ada = btanA,c=
<br /> �/�-- ----y _ ✓ - cos A.
<br /> `
<br /> 17' S 4 9z —t f .l�,s� 1s.f , a, b B=90*-A,a= csinA,b=ccosA,
<br /> 4, /. _- �� f a• Solution of Oblique Triangles
<br /> 2,3 .L 6 33•o // ZJ.a equired a in B
<br /> ZJ 3Y
<br /> _ 3 P 3 t G-9y _ �,3 z c, C b = r✓'= 180°_(A+B), o =asin C
<br /> ' sin A, ,, sin A
<br /> Z� i✓ "�9.1� + .3-77 Zy,B z z ,06✓'rw b sin A
<br /> >'3•2 J7 '7-3 s'3 ° s� - B,0, C sinB= "--„-,C= 180°_(A +B)�c - asin C
<br /> / ---•9 At, a .. sin A
<br /> 3 t L 9 __ o - ° x (a-bl`tan (A+B)
<br /> 2¢.G 3 •moo ✓ o:� s� 1d
<br /> ---- 3 _L A,B, c A+B-180 -�C,tan-a(A-B)=
<br /> 27 ✓, z ✓ .s, 30•l` 3 / 15.J j asin C
<br /> b-F o f
<br /> 13 60 _ 3 .13 �,r t 7•'r� 1 B, ri 8= 2 ,sinjA= ,
<br /> 2, - / �� / ^ - b c
<br /> - z��'� �� s 1 1.8,
<br /> �9
<br /> .3^ sinaB=,•1(8-a)(8--c ,C=18a°-(A+B) •. ,
<br /> 23.3 G ✓ G ✓ - _3_75^ a,r6✓ 33, V ac
<br /> 3 , a+b+c
<br /> V 8 2 a a s a 8 u-c
<br /> 2 3 7,9X. ✓ t o' area = b e sin A
<br /> 21.%3 ✓ _ S ✓ , N �� 2 9l4( � La
<br /> r
<br /> ✓ p as sin B sin C 4
<br /> f� � area = q
<br /> 2s
<br /> in
<br /> 2�•3�, -__ -y� ��. ✓ j 9 4fJ� A DUCTION TO HORIZONTAL 9 Q
<br /> 5-✓ Horizontal distance=Slope distance multiplied by f$e
<br /> •�+i?ter...
<br /> /� ✓ 3 ! r f cosine oftbevertical angle.Thus:slope distance=! 4ft.
<br /> t
<br /> J.�3 ' _ + a Vert. angle=6°10,. From Table,Page IX.cos 610,= -'
<br /> �/ SG•y '✓ s"f - •Q ;; 9969. Horizontal distance=319.4X.89 =31&09 ft,
<br /> L"_ s/ --- - ✓ �1 a Horizontal distance also=Slope di tance minus slope
<br /> .3, y J z. ?/ e I Via•. distance times(1-cosine of ver
<br /> tic angle). With the
<br /> - -- -
<br /> ,/ 2 Y.; ✓ 3. f same figures as in the:preeedl'AC example,the follow-
<br /> ,1 Ho da nce ; ing result is obtained.Cosine 51lW=.9960,]-.8869=.0041.1
<br /> F•°_9 - 1 9 33.1 t 819.4X.0041=1.81.319.4-1.91=31&09 ft.
<br /> - Z ✓ �� a is knodyb the horizontal distance is approximately:-the slope dist-
<br /> '� 2 ✓ Hare of We rlee divided by twice the distance. Thus:rise=14 ft..
<br /> 3 %/ —�9'
<br /> L,5. �✓ _ ✓ 80t idWance=3026.- g,
<br /> C
<br /> 5
<br /> 22 ¢3 ✓ 2 t ✓ 2.9C 004a
<br /> - �• _ MADE flFti.�,A. ,,,,
<br /> i
<br /> R /
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