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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Seite 1
... that if they could , by any means whatever , calculate a series of triangles , the angles of which should be of all possible values , this series would necessarily Trig . 1 Fig . 1 . contain a triangle similar to the Plane trigonometry.
... that if they could , by any means whatever , calculate a series of triangles , the angles of which should be of all possible values , this series would necessarily Trig . 1 Fig . 1 . contain a triangle similar to the Plane trigonometry.
Seite 2
... calculated , we are led to inquire into the method of constructing such a series . To take the most simple case first , I will suppose that the triangles to be determined are right - angled ; it is evident that they may all be formed in ...
... calculated , we are led to inquire into the method of constructing such a series . To take the most simple case first , I will suppose that the triangles to be determined are right - angled ; it is evident that they may all be formed in ...
Seite 5
... calculating the sines and cosines ; and even the 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 ...
... calculating the sines and cosines ; and even the 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 ...
Seite 7
... calculated . 12. The equation sin 2a = 2 sin a cos a R leads also from the sine of an arc a to the expression of the sine of half of this arc . If we substitute for the cos a its value R2 sin a2 * , the equa- I would apprise the learner ...
... calculated . 12. The equation sin 2a = 2 sin a cos a R leads also from the sine of an arc a to the expression of the sine of half of this arc . If we substitute for the cos a its value R2 sin a2 * , the equa- I would apprise the learner ...
Seite 9
... calculate , but only the ratio they have to radius ; since it is sufficient to know in the triangles CPM , CP'M ' , & c . ( fig . 2 ) , the ratio which the sides have amongst themselves . We Fig . 2 : may , therefore , on account of the ...
... calculate , but only the ratio they have to radius ; since it is sufficient to know in the triangles CPM , CP'M ' , & c . ( fig . 2 ) , the ratio which the sides have amongst themselves . We Fig . 2 : may , therefore , on account of the ...
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).