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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Página 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 ...
Página 15
... bisected by OP and CO : OD in the given ratio . TRIGONOMETRY . SATURDAY , 2ND DECEMBER , 1876. 2 P.M TO 5 P.M. 1. Define the degree and unit of circular measure , and shew how to pass from one of these units to the other . Tod . Trig ...
... bisected by OP and CO : OD in the given ratio . TRIGONOMETRY . SATURDAY , 2ND DECEMBER , 1876. 2 P.M TO 5 P.M. 1. Define the degree and unit of circular measure , and shew how to pass from one of these units to the other . Tod . Trig ...
Página 18
... bisection of the base , then or Similarly , BD2 = BA2 + AD2 - 2BA.AD cos1⁄2A , 4a2 = c * + AD2 - 2c . AD cos A. - a2 = b2 + AD2 − 2b . AD cos§A ; therefore by subtraction we have b2 — c2 — 2 ( b − c ) AD cos1⁄2A = 0 ; = - therefore bc ...
... bisection of the base , then or Similarly , BD2 = BA2 + AD2 - 2BA.AD cos1⁄2A , 4a2 = c * + AD2 - 2c . AD cos A. - a2 = b2 + AD2 − 2b . AD cos§A ; therefore by subtraction we have b2 — c2 — 2 ( b − c ) AD cos1⁄2A = 0 ; = - therefore bc ...
Página 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 AB + 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 AB + BC2 = 2AE2 + 2EB2 ...
Página 35
... bisection of the line joining p and q , then this point has no motion . Now if g is the C of G of p and 2 , therefore 99 pg :: p q ; and hence we may readily prove that Og Opp - q : P + L , and therefore that accel . of g : accel ...
... bisection of the line joining p and q , then this point has no motion . Now if g is the C of G of p and 2 , therefore 99 pg :: p q ; and hence we may readily prove that Og Opp - q : P + L , and therefore that accel . of g : accel ...
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Solutions to the Mathematical Examination Papers Set for Admission to the ... D. Tierney,Handell Sharratt Sin vista previa disponible - 2016 |
Solutions To The Mathematical Examination Papers Set For Admission To The ... D. Tierney,Handell Sharratt Sin vista previa disponible - 2023 |
Términos y frases comunes
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
Pasajes populares
Página 55 - If two triangles have two sides of the one equal to two sides of the...
Página 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.
Página 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Página 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.
Página 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.
Página 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Página 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.
Página 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.
Página 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.
Página 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.