Elements of Geometry and Trigonometry |
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Page 55
... right angles . PROPOSITION XIX . THEOREM . The angle formed by two chords , which intersect each other , is measured by half the sum of the arcs included between its sides . Let AB , CD , be two chords intersecting each BOOK III . 55.
... right angles . PROPOSITION XIX . THEOREM . The angle formed by two chords , which intersect each other , is measured by half the sum of the arcs included between its sides . Let AB , CD , be two chords intersecting each BOOK III . 55.
Page 56
... formed by two secants , is measured by half the diffe- rence of the arcs included between its sides . Let AB , AC , be two secants : then will the angle BAC be measured by half the difference of the arcs BEČ and DF . Draw DE parallel to ...
... formed by two secants , is measured by half the diffe- rence of the arcs included between its sides . Let AB , AC , be two secants : then will the angle BAC be measured by half the difference of the arcs BEČ and DF . Draw DE parallel to ...
Page 63
... formed with the given sides and the given angle . Cor . If the given angle is a right angle , the figure will be a rectangle ; if , in addition to this , the sides are equal , it will be a square . PROBLEM XIII . To find the centre of a ...
... formed with the given sides and the given angle . Cor . If the given angle is a right angle , the figure will be a rectangle ; if , in addition to this , the sides are equal , it will be a square . PROBLEM XIII . To find the centre of a ...
Page 64
... formed by two tangents , must pass through the centre of the circle . PROBLEM XV . To inscribe a circle in a given triangle . Let ABC be the given triangle . Bisect the angles A and B , by the lines AO and BO , meeting in the point 0 ...
... formed by two tangents , must pass through the centre of the circle . PROBLEM XV . To inscribe a circle in a given triangle . Let ABC be the given triangle . Bisect the angles A and B , by the lines AO and BO , meeting in the point 0 ...
Page 71
... formed , all equal to each other , because all have the same base and altitude . The rectangle ABCD will contain seven partial rectangles , while AEFD will contain four : hence the rectangle ABCD is to AEFD as 7 is to 4 , or as AB is to ...
... formed , all equal to each other , because all have the same base and altitude . The rectangle ABCD will contain seven partial rectangles , while AEFD will contain four : hence the rectangle ABCD is to AEFD as 7 is to 4 , or as AB is to ...
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Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book centre chord circ circumference circumscribed common cone consequently convex surface cylinder diagonal diameter dicular distance draw drawn equal angles equally distant equation equiangular equivalent figure formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC let fall logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE prism proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant secant line segment side BC similar solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex