Solutions to the mathematical examination papers set for admission to the Royal military academy, Woolwich, and for the Royal military college [&c.] by D. Tierney and H. Sharratt1877 |
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Page 15
... distances from C ' and D ' are in this constant ratio . Let this circle intersect OP in P ' . Join C'P ' and P'D ' . Draw PC , PD parallel to P'C ' and P'D ' respec- tively , then C and D are the points required , since C'P ' : P'D ...
... distances from C ' and D ' are in this constant ratio . Let this circle intersect OP in P ' . Join C'P ' and P'D ' . Draw PC , PD parallel to P'C ' and P'D ' respec- tively , then C and D are the points required , since C'P ' : P'D ...
Page 25
... distance of any point of the curve from the centre is a mean proportional between the distances of the same point from the foci . Salmon , Conic Sections , Art . 174 . If in any hyperbola CD be conjugate to CP , we have SP.HP - CD , but ...
... distance of any point of the curve from the centre is a mean proportional between the distances of the same point from the foci . Salmon , Conic Sections , Art . 174 . If in any hyperbola CD be conjugate to CP , we have SP.HP - CD , but ...
Page 26
... distance equal to the difference between the segments of the line given , describe a circle cutting latter segment in D. Join AD , and produce it to meet the outer segment in C. Join CB . Then ACB is the triangle required . Join DB . We ...
... distance equal to the difference between the segments of the line given , describe a circle cutting latter segment in D. Join AD , and produce it to meet the outer segment in C. Join CB . Then ACB is the triangle required . Join DB . We ...
Page 34
... Distance described = 289800 feet . 2. Establish the relations between space and time , time and velocity , velocity and space in the case of a heavy body falling freely in vacuo . Parkinson's Mechanics , Arts . 67 , 68 , 69 , 70 . If s ...
... Distance described = 289800 feet . 2. Establish the relations between space and time , time and velocity , velocity and space in the case of a heavy body falling freely in vacuo . Parkinson's Mechanics , Arts . 67 , 68 , 69 , 70 . If s ...
Page 44
... distance from the wall , W , the weight of the rod , W , weight attached to the end B , the 2 inclination of the rod to the horizon when it is in the position of equilibrium . 1 Resolving vertically we have R , cos 0 = W1 + W1 . Taking ...
... distance from the wall , W , the weight of the rod , W , weight attached to the end B , the 2 inclination of the rod to the horizon when it is in the position of equilibrium . 1 Resolving vertically we have R , cos 0 = W1 + W1 . Taking ...
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Solutions to the Mathematical Examination Papers Set for Admission to the ... D. Tierney,Handell Sharratt No preview available - 2016 |
Solutions To The Mathematical Examination Papers Set For Admission To The ... D. Tierney,Handell Sharratt No preview available - 2023 |
Common terms and phrases
acceleration axes axis ball base bisection body cent centre circle coefficient common condition Conic contained cose curve Define described determine diameter difference direction distance Divide divisible draw drawn equal equation equilibrium expression feet focus forces four fraction gallons give given gravity half Hence inches inclined increases integration intersect Join least length limiting means Multiply obtain opposite original parabola parallel parallelogram Parkinson's Mechanics particle passing perpendicular places plane positive produced Prop prove radius ratio rectangle remainder represented respectively rest Result right angles roots seconds segments shew sides signs Similarly space square straight line string Subtract tangent triangle Trig unit velocity vertex vertical virtual weight whence yards
Popular passages
Page 55 - If two triangles have two sides of the one equal to two sides of the...
Page 71 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Page 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Page 12 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Page 13 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Page 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Page 13 - BAC is cut off from the given circle ABC containing an angle equal to the given angle D : Which was to be done. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.
Page 62 - In every triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle.
Page 13 - PROP. X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Page 70 - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD • (ax.