An Elementary Treatise on Plane and Spherical Trigonometry and on the Application of Algebra to Geometry: From the Mathematics of Lacrois and BézoutHilliard and Metcalf, sold by W. Hiliard, 1826 - 165 Seiten |
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... and Bézout Silvestre François Lacroix. F B A Ba 7 = 08 DE = FC Grad . BE GF D sin Faciais , bila thomey is Le- Thomas Les . Sila 6 14 } $ * QA 551 41477 E 5 1826 2 A D = a 1 6 208124 A 13 = a 130 = 61 ** G a ) 4D = 1 43 D 骥.
... and Bézout Silvestre François Lacroix. F B A Ba 7 = 08 DE = FC Grad . BE GF D sin Faciais , bila thomey is Le- Thomas Les . Sila 6 14 } $ * QA 551 41477 E 5 1826 2 A D = a 1 6 208124 A 13 = a 130 = 61 ** G a ) 4D = 1 43 D 骥.
Seite
From the Mathematics of Lacrois and Bézout Silvestre François Lacroix. sin Faciais , bila thomey is Le- Thomas Les . "
From the Mathematics of Lacrois and Bézout Silvestre François Lacroix. sin Faciais , bila thomey is Le- Thomas Les . "
Seite 3
... sine of the arc ́ AM , the line P'M ' is also the sine of the arc AM ' , and so of the others . It follows from this ... cosine of an arc is the sine of the complement of this arc , and is equal to that part of the radius comprehended ...
... sine of the arc ́ AM , the line P'M ' is also the sine of the arc AM ' , and so of the others . It follows from this ... cosine of an arc is the sine of the complement of this arc , and is equal to that part of the radius comprehended ...
Seite 4
... sine and cosine . 8. Tangents and secants have with sines and cosines relations , that are very simple , by means of which the one may be found from the other . The triangles CPM and CAN being similar , give CP : PM :: CA : AN ; whence ...
... sine and cosine . 8. Tangents and secants have with sines and cosines relations , that are very simple , by means of which the one may be found from the other . The triangles CPM and CAN being similar , give CP : PM :: CA : AN ; whence ...
Seite 5
... cosine is deduced immediately from the sine , for the right - angled trian- gle CPM , which contains them , and the hypothenuse of which is radius , gives 2 = - R2 , PM + CP CM ( Geom . 186 ) , or ( sin AM ) 2 + ( cos AM ) 2 that is ...
... cosine is deduced immediately from the sine , for the right - angled trian- gle CPM , which contains them , and the hypothenuse of which is radius , gives 2 = - R2 , PM + CP CM ( Geom . 186 ) , or ( sin AM ) 2 + ( cos AM ) 2 that is ...
Häufige Begriffe und Wortgruppen
a² y² abscissas ac² algebraic altitude angle calculate centre circle circumference conjugate axis consequently construction cos a cos cos b sin cosine curve deduce describe determine diameter dividing draw ellipse equal equation y² expression formulas Geom give given line hyperbola hypothenuse known let fall manner mean proportional multiplying obtain opposite ordinates parabola parallel perpendicular PM plane point F question radius right-angled triangle secant side AC similar triangles sin a cos sin a sin sine sine and cosine solution spherical triangles Spherical Trigonometry square straight line substituting subtract supposed tang atang tangent tion transverse axis triangle ABC Trig trigonometry unknown quantity vertex whence α²
Beliebte Passagen
Seite 23 - C' (89) (90) (91) (92) (93) 112. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference.
Seite 26 - ... for this purpose an account is given in a note subjoined to this part. In the solution of this problem we have made use of the theorem, the sines of the angles are to each other as the sides opposite to these angles. We might also apply the rule given for right-angled triangles (Trig. 30), namely, radius is to the tangent of one of the acute angles, as the side adjacent to this angle is to the side opposite ; thus, As radius or sine of 90° . 10,00000 is to 6 c 2,30103 so is tang Abe 47° 30...
Seite 151 - E~JJ and E as centres, and a radius greater than DC or CE, describe two arcs intersecting in F. Then CF is the required perpendicular (I., Proposition XVIII.). 57. Another solution. Take any point O, without the given line, as a centre, and with a radius equal to the distance from O to C, describe a circumference A—V''' intersecting AB in C and in a second ''• •*
Seite 29 - ... others ; thus, 0,17032 2,30103 9,86763 2,33898 the same as before. t Of the manner of measuring the necessary angles and sides and of the instruments that are used for this purpose an account is given in a note subjoined to this part. In the solution of this problem we have made use of the theorem, the sines of the angles are to each other as the sides opposite to these angles.
Seite 85 - I asked whether they were perfectly convinced that in a right-angled triangle the square of the hypothenuse is equal to the squares of the other two sides ? He answered in the affirmative.
Seite 94 - ... both ; that is. their half sum, or their half difference, or a mean proportional between them, or &c., and we shall always arrive at an equation more simple than by employing either the one or the other.
Seite 46 - ABC, the three equations ; cos a = cos 6 cos c + cos A sin b sin c 1 cos b = cos a cos c -j- cos B sin a sin c > .... (B).