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TRIGONOMETRIC FORMUL S,,
<br />B B
<br />17
<br />_9 a c a
<br />` a a a
<br />70
<br />7 A�b C A C
<br />knight Triangle Oblique Triangles — o--�� .
<br />} 2 '] � Solution of Right Triangles
<br />I�� z For Angle A. sin = a , cos = a , tan= b , cot = a , sec = b- cosec = a
<br />V Given " Required a
<br />A=I
<br />�S a,b A,B,c tan. b=cot B4O a2+ z=a 1+a-=
<br />` .4a1 4YI J i �� z
<br />). A, B, b sin A = a = cos B� b = (a i a) (a—a) = a 1= a' / . .
<br />c o
<br />5 Y d I 7a -2- �, a B, b, c B= 90°—d, b= a cotA, c-
<br />7.sin A.
<br />b'
<br />qqyy� �Z,� A, b B, a, c B = 90°—A, a = b tan A, c =
<br />5
<br />.n.i Z,cOs A.
<br />3-,A, c B, a, b . B = 90°—A, a = c sin A, b = c cos A, -
<br />Solution of Oblique Triangles
<br />Given Required
<br />_ asinB , asin C 1
<br />A' B'a -b' c' C b sin'A'C=180—(A+ B),c= sin A'
<br />+ /r�
<br />A,a,b B,c,C sin B=bsaA,C=180°—(A+B),�= sin
<br />a, b, C A, B, c A+B=180°— C, tan ; (A—B)= (a—b) tan z (A+B)
<br />a+b
<br />v a sin C
<br />sin A
<br />a, b, o A, B, C s= 2 ,sin A= be
<br />_
<br />9 1 ;. cl ° g D % f �• sin 1B= I(s—a)(s—c) C=180°—(A+B)
<br />ac
<br />c6 5G ✓may -tea_ -. 1S o: a+b+c
<br />ay b, a Area s= 2 , area = s(s—a) (s— (s—c
<br />\� S
<br />/7 9 G d A, b, c Area aiea = b o sin A
<br />-►
<br />2
<br />Scl 4 o z _ 0 7 z
<br />?.d - P-- ` r /�!Zr'�' Ste' I A,B, C,a Area area = a s2 sin
<br />B C
<br />%11 2��1 03 J tr /�, zZ v REDUCTION TO HORIZONTAL p� 2 Horizontal distance= Slope distance multiplied by the
<br />§'i 1 \ % ✓' /-- Icosine of the vertical angle. Thus: slope distance=319.4 ft.
<br />t�nce Vert. angle =50 101. From Table, Page IX. cos 50 10'=
<br />o ass d 9959. Horizontal distance=319.4X.9959=31&09 ft.
<br />51oQ Angle ix Horizontal distance also=Slone distance minus slope'
<br />}�
<br />4e distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow-
<br />a! F rl ` } p , r� , j Horizontal distance ing result is obtained. Cosine 5° 101=.9959.1—.9959=.0041.
<br />0 rf 319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />U When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise =14 ft.,
<br />l / slope distance=302.6 ft. Horizontal distance=302 6— 14 > 14 =302.6-0.32=302.28 ft.
<br />® 2 X 302.6
<br />qtr
<br />encs is U. e. a
<br />.i
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