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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Resultat 1-5 av 11
Side 12
... 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 ...
... 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 ...
Side 14
... 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 ABC . 10. Similar triangles are to one another in the duplicate ...
... 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 ABC . 10. Similar triangles are to one another in the duplicate ...
Side 15
... bisect the angle CPD , and the segments CO , OD shall bear a given ratio to each other . Take C'O and OD ' ( fig . 7 ) ... bisected by OP and CO : OD in the given ratio . TRIGONOMETRY . SATURDAY , 2ND DECEMBER , 1876. 2 P.M TO 5 P.M. 1 ...
... bisect the angle CPD , and the segments CO , OD shall bear a given ratio to each other . Take C'O and OD ' ( fig . 7 ) ... bisected by OP and CO : OD in the given ratio . TRIGONOMETRY . SATURDAY , 2ND DECEMBER , 1876. 2 P.M TO 5 P.M. 1 ...
Side 18
... bisection of the base , then or Similarly , BD2 = BA2 + AD2 – 2BA.AD cos △ , a2 = c2 + AD2-2c . AD cos A. a = b2 + AD2 — 2b . AD cos§A ; therefore by subtraction we have b2 – c2 — 2 ( b − c ) AD cos§A = 0 ; - therefore bc must 0 or b ...
... bisection of the base , then or Similarly , BD2 = BA2 + AD2 – 2BA.AD cos △ , a2 = c2 + AD2-2c . AD cos A. a = b2 + AD2 — 2b . AD cos§A ; therefore by subtraction we have b2 – c2 — 2 ( b − c ) AD cos§A = 0 ; - therefore bc must 0 or b ...
Side 25
... bisection of the opposite sides is equal to three- fourths of the sum of the squares of the sides of the triangle . Let ABC be the triangle ; D , E , F the middle points of the sides . Then we have similarly and AB2 + BC2 = 2AE2 + 2EB2 ...
... bisection of the opposite sides is equal to three- fourths of the sum of the squares of the sides of the triangle . Let ABC be the triangle ; D , E , F the middle points of the sides . Then we have similarly and AB2 + BC2 = 2AE2 + 2EB2 ...
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Solutions To The Mathematical Examination Papers Set For Admission To The ... D Tierney,Handell Sharratt Ingen forhåndsvisning tilgjengelig - 2023 |
Solutions to the Mathematical Examination Papers Set for Admission to the ... D. Tierney,Handell Sharratt Ingen forhåndsvisning tilgjengelig - 2016 |
Solutions To The Mathematical Examination Papers Set For Admission To The ... D. Tierney,Handell Sharratt Ingen forhåndsvisning tilgjengelig - 2023 |
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 55 - If two triangles have two sides of the one equal to two sides of the...
Side 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.
Side 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Side 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.
Side 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.
Side 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Side 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.
Side 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.
Side 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.
Side 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.