Mathematical Questions and Solutions in Continuation of the Mathematical Columns of the Educational Times Volume 39 (Paperback)


This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ... (By Professor Hudson, M.A.)--Prove (1) that, if the curve r = a + b cos fl have a point of inflexion, a must lie between b and lb; and (2) trace the change in form of the curve, as a changes from 0 to 2A. 7197. (By the Editor)--With two given sides, construct a triangle such that the square of the sine of half the angle contained by the sides may be equal to the squares of the sines of the other two semi-angles. II. Solution by M. Collins, B.A.; J. O'regan; and others. The triangle is of the form in part (3) of Quest. 6287. Thus, if e be the third or required side, Q the centre, r the radius of the inscribed circle, and O the centre of the circumscribed circle, we have r (a + b + c) = ah sin C, which differentiated gives (since r is a maximum, and therefore dr = 0) rdc = ab cos C. dG; also c2 = a2 + As--lab cos C; whence, differentiating, c. d = ab sin C. dC; and from these equations we obtain r = ccot C = 2BD. cot BOD-2OD; hence DE = DE', therefore E'F = 2OG, that is, r'= radius of circle escribed to side c = 2R = diameter of circumscribed circle. abe, '. .._-_.-_-, _., 4A2 Now E = therefore abe(s-c) = 2a2=2a(---) (s-a) (3-e), and as e=2s--a--b, we get 2s3--2 (a + b) s2 + -A(- + b) = 0, an equation which determines -, and therefore the construction of the triangle. It is easy to see that the line DQ, which joins the middle D of the base to the centre Q of the inscribed circle, would pass through the point of intersectien of the three altitudes of the required Aabc when its inscribed circle (or its radius QE) is a maximum. Solution by the Rev. J. L. Kitchin, M.A.; D. Edwardes; and others. We have 6878. (By T. C. Simmons, M.A.)--If A, B are two points in the plane of a given circle; give a geometrical construction for f...

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ... (By Professor Hudson, M.A.)--Prove (1) that, if the curve r = a + b cos fl have a point of inflexion, a must lie between b and lb; and (2) trace the change in form of the curve, as a changes from 0 to 2A. 7197. (By the Editor)--With two given sides, construct a triangle such that the square of the sine of half the angle contained by the sides may be equal to the squares of the sines of the other two semi-angles. II. Solution by M. Collins, B.A.; J. O'regan; and others. The triangle is of the form in part (3) of Quest. 6287. Thus, if e be the third or required side, Q the centre, r the radius of the inscribed circle, and O the centre of the circumscribed circle, we have r (a + b + c) = ah sin C, which differentiated gives (since r is a maximum, and therefore dr = 0) rdc = ab cos C. dG; also c2 = a2 + As--lab cos C; whence, differentiating, c. d = ab sin C. dC; and from these equations we obtain r = ccot C = 2BD. cot BOD-2OD; hence DE = DE', therefore E'F = 2OG, that is, r'= radius of circle escribed to side c = 2R = diameter of circumscribed circle. abe, '. .._-_.-_-, _., 4A2 Now E = therefore abe(s-c) = 2a2=2a(---) (s-a) (3-e), and as e=2s--a--b, we get 2s3--2 (a + b) s2 + -A(- + b) = 0, an equation which determines -, and therefore the construction of the triangle. It is easy to see that the line DQ, which joins the middle D of the base to the centre Q of the inscribed circle, would pass through the point of intersectien of the three altitudes of the required Aabc when its inscribed circle (or its radius QE) is a maximum. Solution by the Rev. J. L. Kitchin, M.A.; D. Edwardes; and others. We have 6878. (By T. C. Simmons, M.A.)--If A, B are two points in the plane of a given circle; give a geometrical construction for f...

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Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

March 2012

Availability

Supplier out of stock. If you add this item to your wish list we will let you know when it becomes available.

First published

March 2012

Authors

Dimensions

246 x 189 x 2mm (L x W x T)

Format

Paperback - Trade

Pages

30

ISBN-13

978-1-130-31404-5

Barcode

9781130314045

Categories

LSN

1-130-31404-9



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