Elements of Geometry and Trigonometry |
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Page 2
... EQUATIONS OF THE POINT AND STRAIGHT LINE - of the CONIC SECTIONS of the LINE AND PLANE IN SPACE - also , the discussion of the GENERAL EQUATION of the second degree , and of SURFACES OF THE SECOND ORDER . DAVIES ' DESCRIPTIVE GEOMETRY ...
... EQUATIONS OF THE POINT AND STRAIGHT LINE - of the CONIC SECTIONS of the LINE AND PLANE IN SPACE - also , the discussion of the GENERAL EQUATION of the second degree , and of SURFACES OF THE SECOND ORDER . DAVIES ' DESCRIPTIVE GEOMETRY ...
Page 38
... equation to , or subtract it from M.P , and we shall have , M.P + N.P - M.P + M.Q ; or ( M ± N ) × P = ( P ± Q ) × M. But M ± N and P , may be considered the two extremes , and P + Q and M , the two means of a proportion : hence , M ± N ...
... equation to , or subtract it from M.P , and we shall have , M.P + N.P - M.P + M.Q ; or ( M ± N ) × P = ( P ± Q ) × M. But M ± N and P , may be considered the two extremes , and P + Q and M , the two means of a proportion : hence , M ± N ...
Page 128
... equation from the second , and observing that the triangles APC , APB , which are both right angled at P , give AC2 - PC2 - AP2 , and AB2 - PB2 = AP2 ; we shall have AP2 + AP2 = 2AQ2 - 2PQ2 . Therefore , by taking the halves of both ...
... equation from the second , and observing that the triangles APC , APB , which are both right angled at P , give AC2 - PC2 - AP2 , and AB2 - PB2 = AP2 ; we shall have AP2 + AP2 = 2AQ2 - 2PQ2 . Therefore , by taking the halves of both ...
Page 207
... equations of the problem . The solu- tion of these equations , when so formed , gives the solution of the problem . No general rule can be given for forming the equations . The equations must be independent of each other , and their ...
... equations of the problem . The solu- tion of these equations , when so formed , gives the solution of the problem . No general rule can be given for forming the equations . The equations must be independent of each other , and their ...
Page 208
... equation x = s - y or Hence , or or x2 = s2 — 2sy + y2 y2 = a2 — s2 + 2sy — y2 2y2 - 2sy = a2- a2 - s ? y2 — sy— 2 By completing the square y2 — sy + 1s2 = 1a2 — ¡ s2 or Hence y = ‡ s ± √ + a2— + s2 = 4 or 3 x = { s = √ { a2 = + s2 ...
... equation x = s - y or Hence , or or x2 = s2 — 2sy + y2 y2 = a2 — s2 + 2sy — y2 2y2 - 2sy = a2- a2 - s ? y2 — sy— 2 By completing the square y2 — sy + 1s2 = 1a2 — ¡ s2 or Hence y = ‡ s ± √ + a2— + s2 = 4 or 3 x = { s = √ { a2 = + s2 ...
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone convex surface Cosine cylinder diagonal diameter dicular distance divided draw drawn equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC logarithm number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular perpendicular let fall plane MN polyedron polygon ABCDE PROBLEM proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABCDE Scholium secant segment similar Sine Cotang slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex