The Elements of Analytical Geometry ... |
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Common terms and phrases
AB² abscissa AC² asymptotes axis of x becomes bisect centre circumference coefficients conical surface conjugate diameters consequently construction coordinates cos.² cosine curve denote determine the triangle difference directrix distance draw ellipse equa equal equilateral expression find the sides formulas Geom geometrical given point given straight line gles hence hyperbola hypothenuse indeterminate equation inscribed circle length Let ABC locus major diameter negative oblique ordinate origin parabola perpendicular plane of xy plane triangle point of contact positive primitive equation principal diameters PROBLEM proposed line radii radius vector rectangle rectangular axes referred represent right angle right-angled triangle square substituting subtangent suppose surface system of conjugate tangent three sides tion transformed triangle ABC vertex vertical angle whence
Popular passages
Page 31 - To find the side of an equilateral triangle inscribed in a circle, multiply the diameter of the circle by .866.
Page 30 - Having given the side of a regular decagon inscribed in a circle whose radius is known, to find the side of a regular pentagon inscribed in the same circle.
Page 200 - Given the base and the sum of the sides of a triangle, to find the locus of the point of intersection of lines from the angles bisecting the opposite sides.
Page 109 - They may cut each other, having two points common, when the distance between the centers is less than the sum and greater than the difference of the radii.
Page 153 - ... does. From the general expression for the subtangent, just given, it follows that that is, as in the ellipse, the rectangle of the subtangent and abscissa of the point of contact is equal to the rectangle of the sum and difference of the same abscissa and semi-transverse axis. Thus OM-MR = A'M-MB...
Page 196 - Given the base of a triangle and the difference of the angles at the base, to determine the locus of the vertex. Taking the same axes as before, and putting a, a', for the tangents of the angles at the base...
Page 37 - In an oblique-angled plane triangle ; there is given the difference of the sides which includes the angle of 71° 10,' equal to 11, and the line that bisects the said angle is equal to 24 ; from whence is required a theorem that will determine the base and sides of the said triangle. The figure can be supplied by the reader.
Page 67 - In a triangle, having given the ratio of the two* sides, together with both the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle.
Page 104 - ... y", and there results for the tangent the equation y — y' = ; (x — a/), as before found. which is obviously that of a perpendicular to this, through the point (a/, y',) that the tangent through any point is perpendicular to the radius at that point, a property which was assumed in the preceding investigation. PROBLEM II. (26.) To draw a tangent to a circle from a given point without it. Let (a...
Page 37 - In an isosceles triangle, the square of a line drawn from the vertex to any point in the base, together with the rectangle of the segments of the base, is equal to the square of one of the equal sides of the triangle.