## Yale University Entrance Examinations in Mathematics: 1884 to 1898 |

### From inside the book

Page 7

... the spherical excess of a spherical triangle . How can you find the area of a spherical triangle when you have its spherical excess given ? Illustrate your answer by a simple example . SEPTEMBER 1886 . 1.

... the spherical excess of a spherical triangle . How can you find the area of a spherical triangle when you have its spherical excess given ? Illustrate your answer by a simple example . SEPTEMBER 1886 . 1.

**Bisect**a given angle and prove 7. Page 8

1884 to 1898. SEPTEMBER 1886 . 1.

1884 to 1898. SEPTEMBER 1886 . 1.

**Bisect**a given angle and prove your construction . 2. Of lines passing through the end of any radius of a circle the perpendicular is a tangent to the circle and every other line is a secant . 3. If from ... Page 17

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**bisected**by , the line joining their centers . 3. To divide a given line in extreme and mean ratio.- What regular inscribed polygons may be constructed by means of this division ? Prove your statement . 4. Two circles are tangent ... Page 20

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**bisected**perpen- dicularly by the line of centers . What does this proposition become when the circles touch ? 3. State and prove the theorems possible when a perpen- dicular is let fall upon the hypothenuse of a right angled triangle ... Page 117

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**bisected**at the point . 3. ( a ) Through a given point P without a given circle ( center O ) draw a tangent to the circle . ( b ) If the distance OP is 15 decimeters and the radius of the circle 9 decimeters , what is the length of the ...### Common terms and phrases

binomial formula binomial theorem bisected chord circle circumference circumscribed common logarithm cone Construct continued fraction cosine Deduce Define Derive the formula Divide drawn 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 hypothenuse inches included angle inscribed JUNE line perpendicular locus loga method of undetermined number of terms parallelopiped perimeter perpendicular polyhedral angle polyhedrons prism pyramid QUADRATICS radians radii radius ratio regular polygons regular polyhedrons Resolve the following right angles secant segments SEPTEMBER series of ascending SHEFFIELD SCIENTIFIC SCHOOL Show similar polygons simplest form Simplify the following simultaneous equations sine SOLID AND SPHERICAL Solve the equation Solve the simultaneous sphere spherical triangle tangent tetrahedron theorem trigonometric functions undetermined coefficients vertex volume

### Popular passages

Page 173 - 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 4 - The sum of any two face angles of a trihedral angle is greater than the third face angle.

Page 190 - In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, minus twice the product of one of these sides and the projection of the other side upon it.

Page 125 - The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.

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 171 - 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 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 186 - It follows that the ratio of the circumference of a circle to its diameter is the same for all circles.