Elements of Geometry and Trigonometry |
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Page 212
Adrien Marie Legendre Charles Davies. PROBLEM XVII In a right angled triangle , having given the perimeter and the perpendicular let fall from the right angle on the hypothe- nuse , to determine the triangle . PROBLEM XVIII . To determine a ...
Adrien Marie Legendre Charles Davies. PROBLEM XVII In a right angled triangle , having given the perimeter and the perpendicular let fall from the right angle on the hypothe- nuse , to determine the triangle . PROBLEM XVIII . To determine a ...
Page 253
... angles of a spherical triangle are to each other as the sines of their opposite sides . IV . From K draw KE perpendicular to OB , and from D draw DF parallel to OB . Then will the angle DKF = COB = a , since each is the complement of the ...
... angles of a spherical triangle are to each other as the sines of their opposite sides . IV . From K draw KE perpendicular to OB , and from D draw DF parallel to OB . Then will the angle DKF = COB = a , since each is the complement of the ...
Page 264
Adrien Marie Legendre Charles Davies. right angled at A ' , and hence every case may be referred to a right angled triangle . But we can solve the quadrantal triangle by means of the right angled triangle in a manner still more simple ...
Adrien Marie Legendre Charles Davies. right angled at A ' , and hence every case may be referred to a right angled triangle . But we can solve the quadrantal triangle by means of the right angled triangle in a manner still more simple ...
Contents
BOOK | 7 |
Problems relating to the First and Third Books 57 | 57 |
BOOK IV | 68 |
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book VII centre chord circ circumference circumscribed common cone consequently convex surface cosine Cotang cylinder diagonal diameter dicular distance divided draw drawn equally distant equations equivalent feet figure find the area formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE PROBLEM Prop proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant segment similar sine slant height solid angle solid described sphere spherical polygon spherical triangle square described straight line TABLE OF LOGARITHMIC tang tangent THEOREM triangle ABC triangular prism vertex