What people are saying - Write a review
We haven't found any reviews in the usual places.
Other editions - View all
base and altitude binomial formula binomial theorem bisected chord circumference cone Construct continued fraction cosine Define Derive the formula dihedral angle divided drawn equation x2 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 number of sides number of terms parallel planes parallelopiped perimeter polar triangle pyramid Quadrat1cs radians in degrees radii ratio regular polygons regular polyhedrons Resolve the following right angles secant segments September 1891 series of ascending SHEFFIELD SCIENTIFIC SCHOOL Show similar polygons simplest equivalents Simplify the following simultaneous equations sine Sol1d and Spher1cal Solve the equat1on Solve the equation Solve the simultaneous spherical excess spherical triangle square root tangent tetrahedron theorem triangle ABC trigonometric functions undetermined coefficients vertex
Page 177 - 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 115 - 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 175 - The area of a circle is equal to one-half the product of its circumference and radius.
Page 35 - If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.
Page 37 - 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 127 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Page 190 - It follows that the ratio of the circumference of a circle to its diameter is the same for all circles.
Page 149 - 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