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TRIGONOMETRIC FORMULAE
<br />�i 1 33 B B
<br />} d a c a a
<br />a
<br />b ° a�b
<br />Right Triangle Oblique Triangles
<br />"Solution, of Right Triangles
<br />a b a b c a
<br />i, Tor Angle A, sin = a , coe = C , tan= 3 , cot= a —,see = b , cosec =
<br />(liven Required a
<br />I� a,b A,B,o tan A=b= cot B,e= a' }=a 1 } s
<br />9
<br />74
<br />t{ a� a A, B; b gin = = COB B, b (d +__T(` --a) ='c
<br />A, a B, b, a B=90*=A, b= a eotA, e= a YY
<br />I
<br />gin A.
<br />A, b B, a, a, B = 90°—A, a = b tan A, c = cos A.
<br />!' A, e B, a;! b B = 90°—A, a = c sin A, 6= e, cos A,
<br />Solution of an Oblique Triangles
<br />9 g
<br />Given Required _ a sin B in C
<br />' A, B, a 'b, e, iC b ein A ' C = 180°—(A } B), a = sin A
<br />b min A a ein C
<br />c, C sin B =a . C = 180°—(A + B) , n = sin A
<br />q b, C A, B,'6 A+B-180°— C, tan (A—B)= b) tan (A +B
<br />I' k _asin C a+b
<br />• a sin A
<br />I
<br />bi a A, B; C s=o' 2 iniA= g b
<br />. Bm }B— a d ,C-180° lA+B)
<br />d Area s— 2
<br />l I b- a Bin A
<br />j, .el, b, a Area area = 2
<br />t
<br />of sin B sin
<br />.A, B, Ca Ares aArea = C
<br />2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance —Slope distance multiplied by the
<br />cosintepee Vert Bof ang angle 50 10'. cal From Tablus;e,Page IX cos b°81
<br />all m 9969. Horizontal distance=918.4X.9959=318.09 ft
<br />C,% Ante Horizontal distance also=Slone distance minus slope
<br />Ve distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance Ing result is obtained. Cosine 5° 10'=.9959. 1--.8859=.0041.
<br />319.4X.0041=1.31.319.4-1.31=318,09 ft.
<br />Whe¢ 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=l4 ft..
<br />',slope distance=502.8 ft.,Horizontal distance=9628— 14 X 14 3028-0.32=30228 ft-
<br /><.� 2XS28
<br />-'� MADE IN U. S. A.
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