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Entrance Examinations in Mathematics, 1884 to 1898 [With Supplements to 1900]
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binomial formula binomial theorem bisects chord circle circumference circumscribed cone Construct continued fraction cosine Define Derive the formula dihedral angle divided drawn equal angles Expand Express an angle expressions into factors feet Find the area Find the number Find the value following expressions fraction geometric geometric progression given points inches inscribed JUNE line perpendicular locus loga method of undetermined middle points parallel planes parallelopiped perimeter polyhedral angle polyhedrons prism pyramid QUADRATICS radii radius ratio regular polygons regular polyhedrons Resolve the following right angle secant segments SEPTEMBER 1891 series of ascending SHEFFIELD SCIENTIFIC SCHOOL Show similar polygons simplest equivalents Simplify the following simultaneous equations sine SOLID AND SPHERICAL Solve the equation Solve the simultaneous sphere spherical triangle tangent tetrahedron theorem triangle ABC trigonometrical functions undetermined coefficients vertex volume
Page 175 - Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.
Page 113 - IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.
Page 173 - The area of a circle is equal to one-half the product of its circumference and radius.
Page 33 - If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.
Page 35 - In two polar triangles each angle of the one is the supplement of the opposite side in the other. Let ABC, A'B'C
Page 125 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Page 188 - It follows that the ratio of the circumference of a circle to its diameter is the same for all circles.
Page 147 - An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Hyp. OM is a right section of oblique prism AD', and OM ' a right prism whose altitude is equal to a lateral edge of AD'. To prove AD' =0= GM' . Proof. The lateral edges of GM