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 2
... triangle ABC ( fig . 1 ) , we know the angle B , the angle C , and the side BC , we find in the series of computed triangles that , which has two angles , b and c , respec- tively equal to the angles B and C ; it will necessarily be ...
... triangle ABC ( fig . 1 ) , we know the angle B , the angle C , and the side BC , we find in the series of computed triangles that , which has two angles , b and c , respec- tively equal to the angles B and C ; it will necessarily be ...
Seite 28
... triangles . 32. The principle upon which the resolution of right - angled triangles is founded , leads also to that of triangles of whatever kind . By letting fall from the angle B of the triangle ABC Fig . 14. ( fig . 14 ) a ...
... triangles . 32. The principle upon which the resolution of right - angled triangles is founded , leads also to that of triangles of whatever kind . By letting fall from the angle B of the triangle ABC Fig . 14. ( fig . 14 ) a ...
Seite 29
... triangle , the angle C is not common to the two triangles ABC , BCD ; but the an- gles BCD , BCA , being together equal to two right angles , have the same sine ( 22 ) . The proportion just given admits of a general application , and ...
... triangle , the angle C is not common to the two triangles ABC , BCD ; but the an- gles BCD , BCA , being together equal to two right angles , have the same sine ( 22 ) . The proportion just given admits of a general application , and ...
Seite 30
... triangles AOB and a Ob , that AB : ab :: AO : a 0 , or that AB : 2 sin C :: AO : a 0 ; that is , each side of the triangle ABC , is to double the sine of the opposite angle , as the radius of the circumscribed circle , is to that of the ...
... triangles AOB and a Ob , that AB : ab :: AO : a 0 , or that AB : 2 sin C :: AO : a 0 ; that is , each side of the triangle ABC , is to double the sine of the opposite angle , as the radius of the circumscribed circle , is to that of the ...
Seite 36
... triangles . 1. In the triangle ABC ( fig . 16 ) , the side c , and theangles A and B , being given , to find the side b . Let A 112 ° 30 ′ 21 ′′ , B = 52 ° 54 ′ 40 ′′ , c = 27ch , 348 ; the angle C will be 180 ° — ( A + B ) = 180 ° 165 ...
... triangles . 1. In the triangle ABC ( fig . 16 ) , the side c , and theangles A and B , being given , to find the side b . Let A 112 ° 30 ′ 21 ′′ , B = 52 ° 54 ′ 40 ′′ , c = 27ch , 348 ; the angle C will be 180 ° — ( A + B ) = 180 ° 165 ...
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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).