Mathematical Exercises ...: Examples in Pure Mathematics, Statics, Dynamics, and Hydrostatics. With Tables ... and ReferencesLongmans, Green & Company, 1877 - 413 pages |
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Page 61
... prove that log.blog , a = 1 . 15. Show that log ( ) = = rlogm ― r logn , and find the logarithms of ' 0625 and 14.4 ... Prove the rule for finding the greatest common measure of two algebraical expressions . 5. Find the greatest common ...
... prove that log.blog , a = 1 . 15. Show that log ( ) = = rlogm ― r logn , and find the logarithms of ' 0625 and 14.4 ... Prove the rule for finding the greatest common measure of two algebraical expressions . 5. Find the greatest common ...
Page 68
... Prove that an + b " is divisible by a + b without remainder , when n is an uneven number . Resolve into their simple factors the expressions ( 1. ) 6 ( x − 1 ) +5 . ( 2. ) ( a2 − 1 ) x2 + ( 3a − 1 ) x + a — a2 . 2. State and prove ...
... Prove that an + b " is divisible by a + b without remainder , when n is an uneven number . Resolve into their simple factors the expressions ( 1. ) 6 ( x − 1 ) +5 . ( 2. ) ( a2 − 1 ) x2 + ( 3a − 1 ) x + a — a2 . 2. State and prove ...
Page 74
... prove that the sum of the two sides containing the right angle is equal to the sum of the diameters of the inscribed and circum- scribed circles . 4. A and B are fixed points , and P is a movable point such that PA always bears a fixed ...
... prove that the sum of the two sides containing the right angle is equal to the sum of the diameters of the inscribed and circum- scribed circles . 4. A and B are fixed points , and P is a movable point such that PA always bears a fixed ...
Page 76
... Prove that cos ( A - B ) = cos A cos B + sin A sin B , and employ this formula to find the value of cos 15 ° . 10. Prove the following equalities : ( 1. ) Sin ( 60 ° + A ) —sin ( 60 −A ) = sin A. ( 2. ) ( Cos A + sin A sin 2A ) 2 ...
... Prove that cos ( A - B ) = cos A cos B + sin A sin B , and employ this formula to find the value of cos 15 ° . 10. Prove the following equalities : ( 1. ) Sin ( 60 ° + A ) —sin ( 60 −A ) = sin A. ( 2. ) ( Cos A + sin A sin 2A ) 2 ...
Page 78
... prove that equiangular parallelograms have to one another the ratio which is com- pounded of the ratios of their ... prove that the three common chords , or these chords produced , pass through one and the same point . Given a circle ...
... prove that equiangular parallelograms have to one another the ratio which is com- pounded of the ratios of their ... prove that the three common chords , or these chords produced , pass through one and the same point . Given a circle ...
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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.