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4, ! TRIGONOMETRIC FORMULAE
<br />B r ` B
<br />`
<br />A'' : bA A C
<br />C b C b J�
<br />Right Triangle Oblique Triangles -
<br />/�/G Solution of Right Triangles
<br />a -b a b` c c
<br />For Angle A. - sin= , cos - — tan= , cot — osec
<br />,sec = —, c =
<br />�• c b a b a
<br />Given Required
<br />q t 21
<br />a,b A,B,'c tan A=b=cotB,c= a +bb?=a 1 az
<br />7. [s — — . — -- -- �7� a, c A, B, b sin A = = cos B, b = \/—(c +a) (c—a) a�
<br />o
<br />A, a B, b, c B=90°—A, b = acotA,c= a
<br />J sin A.
<br />p0�' l 001 �Lo A, b B, a, e. B=90'—A, = btanA,c= cos A. .
<br />•l zQ7s b.� D �i ..zOC-�a�3% 5 `t A, c B, a, b B=90°—A, a = csinA, b= e cos A;
<br />3 n 5^ 6 x Solution of Oblique -Triangles
<br />-Xff' Given Required _ a sin B —
<br />4r7 ��53� . ��0 A' B'a b' C b sinA'C=180°—(A+I3)�c=asinC
<br />sin A
<br />Z 3. 5 a (7 56 �r �T 3 ,A, d," b B, c> .L' sin B= b Sian A,C = 180°—(A ¢-B),,, = sin A
<br />t�// 0 1 (a -b) tan � (A+B)
<br />Z ia, b, C �, B, c A+B-180= C, tan a (A—B)—
<br />(� a b '
<br />} . 7, T3 sin A
<br />b c A, B, C s=a+b+c �(s_ b)(s—c
<br />7 ,sin ZA-
<br />2 b c
<br />j s—a s—c
<br />M. sin'B=�( )( ),C_180°—(A B)
<br />i / ' ,� ac
<br />Ub F Hca, b, c Area s= +2+ ,area = s(s—a s—b)(s=c
<br />)A, b, c Area area, = b c sin A
<br />��. 2
<br />Cq0 CoJ 1 ,�� 9z41 'h_ az sin B sin C
<br />A, B, C, a Area area 2 sin A
<br />Y. REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied. by the
<br />/ r —' 11+ cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />(o ' tante Vert. angle =5° 101. From Table, Page IX. cos 5° 10'=
<br />�A� j,�✓ © 2 r� 6 �\i e ass _ y 9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />!lIJJJ $fop ogle Horizontal distance also=Slope distance minus slope
<br />}� P a distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />i.. b .• r� : Horizontal distance ing result is obtained. Cosine 51 101=.9959.1=.9959=.0041.
<br />td 319.4X.0041=1.31. 319.4-1.31=318.09 ft.
<br />r� When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />q 1 \ ante less the square of the rise divided by twice the slope distance. Thus: rise=14 ft.,
<br />�1g2l 1; 3 k, slope distance=302.6 ft. Horizontal distance=302.6— 14 X 14 =302.6--0.32=302.28 fL
<br />2 X 302.6
<br />K
<br />woe u, U. S.A.s. ti
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