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 páginas |
Dentro del libro
Resultados 1-5 de 96
Página 1
... equal triangles , that we may always construct a triangle , when we know three of the six parts , which constitute it , provided that one at least of these three parts be a side . In order to render the theory of triangles complete , we ...
... equal triangles , that we may always construct a triangle , when we know three of the six parts , which constitute it , provided that one at least of these three parts be a side . In order to render the theory of triangles complete , we ...
Página 2
... equal to the angles B and C ; it will necessarily be similar ( Geom . 203 ) to the proposed triangle ABC ; and since ... equal to the radius of the arc AB . 4. We may form also a series of right - angled triangles , each having one of ...
... equal to the angles B and C ; it will necessarily be similar ( Geom . 203 ) to the proposed triangle ABC ; and since ... equal to the radius of the arc AB . 4. We may form also a series of right - angled triangles , each having one of ...
Página 3
... equals CP , CP ' , CP " , & c . under the name of cosines of the arcs AM , AM ' , AM " , & c . Whence the cosine of an arc is the sine of the complement of this arc , and is equal to that part of the radius comprehended between the ...
... equals CP , CP ' , CP " , & c . under the name of cosines of the arcs AM , AM ' , AM " , & c . Whence the cosine of an arc is the sine of the complement of this arc , and is equal to that part of the radius comprehended between the ...
Página 5
... equal to the sum of the squares of the sine and cosine ; whence cos AM = √R2 ( sin AM ) . The following proposition , which gives the expression of the sine and cosine of the sum and of the difference of two arcs , merits the greatest ...
... equal to the sum of the squares of the sine and cosine ; whence cos AM = √R2 ( sin AM ) . The following proposition , which gives the expression of the sine and cosine of the sum and of the difference of two arcs , merits the greatest ...
Página 6
... equal parts in the point E ( Geom . 106 ) , by the radius CM , which , by construction , bisects the arc NN ' , we infer from the similar triangles NED , NN'G , that NG is also divided into two equal parts in the point D , and that DN ...
... equal parts in the point E ( Geom . 106 ) , by the radius CM , which , by construction , bisects the arc NN ' , we infer from the similar triangles NED , NN'G , that NG is also divided into two equal parts in the point D , and that DN ...
Términos y frases comunes
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
Pasajes populares
Página 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.
Página 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...
Página 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 ''• •*
Página 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.
Página 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.
Página 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.
Página 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).