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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Αποτελέσματα 1 - 5 από τα 11.
Σελίδα 11
... join FE ' . Then , since the angle E'DF equals the angle BAC , and BA and AC are equal to E'D and DF respectively , the triangle FDE is equal to the triangle BAC . But the triangle FDE is equal to the triangle FDE ' , " since DE is ...
... join FE ' . Then , since the angle E'DF equals the angle BAC , and BA and AC are equal to E'D and DF respectively , the triangle FDE is equal to the triangle BAC . But the triangle FDE is equal to the triangle FDE ' , " since DE is ...
Σελίδα 12
... Join AD . Then the square on AD is equal to twice the square on AB . From D draw DE at right angles to AD and equal to AB . Join AE . Then the square on AE , being equal to the squares on DE and AD , is equal to three times the square ...
... Join AD . Then the square on AD is equal to twice the square on AB . From D draw DE at right angles to AD and equal to AB . Join AE . Then the square on AE , being equal to the squares on DE and AD , is equal to three times the square ...
Σελίδα 14
... Join FG . Then the triangle EFG is equal to the triangle ABC . Bisect EF in K. Draw KH perpendicular to EF and GH parallel to EF . Join HE , HF . Then EHF is the triangle required , for it is isosceles and equal to EFG , that is , to ...
... Join FG . Then the triangle EFG is equal to the triangle ABC . Bisect EF in K. Draw KH perpendicular to EF and GH parallel to EF . Join HE , HF . Then EHF is the triangle required , for it is isosceles and equal to EFG , that is , to ...
Σελίδα 15
... 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 ' :: C'O : OD ' ; therefore by similar triangles CP : PD : CO : OD ; therefore the angle CPD is ...
... 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 ' :: C'O : OD ' ; therefore by similar triangles CP : PD : CO : OD ; therefore the angle CPD is ...
Σελίδα 26
... Join AD , and produce it to meet the outer segment in C. Join CB . Then ACB is the triangle required . Join DB . We shall find CD = CB , and therefore difference between AC and CB = difference of segments of base , and angle ACB ...
... Join AD , and produce it to meet the outer segment in C. Join CB . Then ACB is the triangle required . Join DB . We shall find CD = CB , and therefore difference between AC and CB = difference of segments of base , and angle ACB ...
Άλλες εκδόσεις - Προβολή όλων
Solutions To The Mathematical Examination Papers Set For Admission To The ... D Tierney,Handell Sharratt Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2023 |
Solutions to the Mathematical Examination Papers Set for Admission to the ... D. Tierney,Handell Sharratt Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2016 |
Solutions To The Mathematical Examination Papers Set For Admission To The ... D. Tierney,Handell Sharratt Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2023 |
Συχνά εμφανιζόμενοι όροι και φράσεις
AB² ABCD acceleration AD² ARITHMETIC axis BC² cent centre of gravity circle coefficient of friction Conic Sections cose cubic curve decimal described diameter Differential Calculus directrix Divide dy dx equal and parallel Euclid expression feet Find the equation forces fraction given straight line hyperbola inches inclined integration intersect Join latus rectum least common multiple logarithmic mechanical advantage METCALFE AND SON moment of inertia Multiply opposite angles parabola parallelogram Parkinson's Mechanics particle perpendicular plane point of bisection Prop prove quadrilateral radius ratio rectangle contained Result right angles roots seconds segments semicircle shew sides Similarly sin² sine square string subtending Subtract tangent Todhunter Todhunter's Trigonometry triangle ABC Trig vertex vertical virtual velocities weight whence whole number yards
Δημοφιλή αποσπάσματα
Σελίδα 55 - If two triangles have two sides of the one equal to two sides of the...
Σελίδα 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.
Σελίδα 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Σελίδα 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.
Σελίδα 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.
Σελίδα 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.