Mathematical Exercises ...: Examples in Pure Mathematics, Statics, Dynamics, and Hydrostatics. With Tables ... and ReferencesLongmans, Green & Company, 1877 - 413 pages |
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Page 74
... fixed points , and P is a movable point such that PA always bears a fixed ratio to PB . Prove that P lies on a circle , and find the centre of that circle . 5. Define a degree , a grade , and the unit of circular measure . The number ...
... fixed points , and P is a movable point such that PA always bears a fixed ratio to PB . Prove that P lies on a circle , and find the centre of that circle . 5. Define a degree , a grade , and the unit of circular measure . The number ...
Page 115
... fixed line . 12. Investigate an expression for the radius of curvature . in terms of polar co - ordinates . Find a curve in which the radius of curvature is proportional to the radius vector . να 13. Explain the meaning of the ...
... fixed line . 12. Investigate an expression for the radius of curvature . in terms of polar co - ordinates . Find a curve in which the radius of curvature is proportional to the radius vector . να 13. Explain the meaning of the ...
Page 119
... fixed points situated either , both without , or both within , the circumference of a circle whose centre is O , and if R be a point upon that cir- cumference , prove that the sum of the distance PR and QR is the least possible when PR ...
... fixed points situated either , both without , or both within , the circumference of a circle whose centre is O , and if R be a point upon that cir- cumference , prove that the sum of the distance PR and QR is the least possible when PR ...
Page 122
... fixed point . 3. Multiply 5 + 3x3 - x2 + 1 by x2 - 8x + 4 , and divide x6 - 2x3 + 1 by x2 - 2x + 1 ; also resolve into factors the expressions- ( x2 + 3x + 1 ) 2- ( x2 + x + 1 ) 2 and x2 - 23x + 132 . 4. Find the greatest common measure ...
... fixed point . 3. Multiply 5 + 3x3 - x2 + 1 by x2 - 8x + 4 , and divide x6 - 2x3 + 1 by x2 - 2x + 1 ; also resolve into factors the expressions- ( x2 + 3x + 1 ) 2- ( x2 + x + 1 ) 2 and x2 - 23x + 132 . 4. Find the greatest common measure ...
Page 134
... fixed straight lines and PQ a line meeting AB and AC in P and Q , and moving sub- ject to the condition that the area PAQ is constant , prove that when the length of PQ is a minimum AP and AQ will be equal . 9. Prove the following ...
... fixed straight lines and PQ a line meeting AB and AC in P and Q , and moving sub- ject to the condition that the area PAQ is constant , prove that when the length of PQ is a minimum AP and AQ will be equal . 9. Prove the following ...
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Popular passages
Page 123 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Page 10 - A'B'C', and applying the law of cosines, we have cos a' = cos b' cos c' + sin b' sin c' cos A'. Remembering the relations a' = 180° -A, b' = 180° - B, etc. (this expression becomes cos A = — cos B cos C + sin B sin C cos a.
Page 184 - If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other.
Page 78 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Page 184 - To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third (20.
Page 184 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Page 163 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 184 - In right angled triangles the square on the side subtending the right angle is equal to the (sum of the) squares on the sides containing the right angle.
Page 154 - If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D : As AE is to EB, so is CF to FD.