Elements of Geometry, with Both Plane and Spherical Trigonometry (Paperback)


This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1813 edition. Excerpt: ...to their intersections. 4. The angle made by two plane sections through the centre of a sphere is the same as the angle which they contain of a third plane section through the centre perpendicular to them. Because the two lines of intersection of the third, with the other two, are perpendicular to the intersection of the two. 153. THEOREM 153, THEOREM 5. Three or more line?, passing through two parallel planes, and meeting at the same point, describe similar triangles or similar polygons on the two parallel pianos. DEM. Suppose the two planes PL and JyS (Fig43.) to be parallel, five lines from the same point R describe on the two planes two similar pentagons ABDFE and ubdfe. For, AB and ab may be considered as the intersections of the same plane RAB with, the two parallel planes PL and KS: therefore AB is parallel to ab. For the same reason, the plane RBD makes DB parallel to db; and the plane RAD makes AD parallel to ad; and the two triangles ABD and abd have their three sides parallel, and are similar (79.); and the angle B is equal to the angle b. A similar demonstration will prove the angles E=e, D=ef, F==/, A = a. Therefore the five angles of the two pentagons described on the two parallel planes, by joining the intersections of the five lines from R, are equal each to each. Besides, the triangles BRD and bfyd are similar (78.); and so are ARB and allb; whence AB: ab:: BR; 6R:: BD: bd. For the same reason, AB: pb:: AE: qe:: EF: ef:: FD: fd. Then, Jthfi two pentagons have their angles equal two and two, and their homologous sides proportional. Therefore, the two pentagons are similar (94.). Tne same demonstration may be applied to any two polygons. 154. COROL. If the two parallel planes PL and KS be produced qn any side, and

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1813 edition. Excerpt: ...to their intersections. 4. The angle made by two plane sections through the centre of a sphere is the same as the angle which they contain of a third plane section through the centre perpendicular to them. Because the two lines of intersection of the third, with the other two, are perpendicular to the intersection of the two. 153. THEOREM 153, THEOREM 5. Three or more line?, passing through two parallel planes, and meeting at the same point, describe similar triangles or similar polygons on the two parallel pianos. DEM. Suppose the two planes PL and JyS (Fig43.) to be parallel, five lines from the same point R describe on the two planes two similar pentagons ABDFE and ubdfe. For, AB and ab may be considered as the intersections of the same plane RAB with, the two parallel planes PL and KS: therefore AB is parallel to ab. For the same reason, the plane RBD makes DB parallel to db; and the plane RAD makes AD parallel to ad; and the two triangles ABD and abd have their three sides parallel, and are similar (79.); and the angle B is equal to the angle b. A similar demonstration will prove the angles E=e, D=ef, F==/, A = a. Therefore the five angles of the two pentagons described on the two parallel planes, by joining the intersections of the five lines from R, are equal each to each. Besides, the triangles BRD and bfyd are similar (78.); and so are ARB and allb; whence AB: ab:: BR; 6R:: BD: bd. For the same reason, AB: pb:: AE: qe:: EF: ef:: FD: fd. Then, Jthfi two pentagons have their angles equal two and two, and their homologous sides proportional. Therefore, the two pentagons are similar (94.). Tne same demonstration may be applied to any two polygons. 154. COROL. If the two parallel planes PL and KS be produced qn any side, and

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Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

September 2013

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

September 2013

Authors

Dimensions

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

Format

Paperback - Trade

Pages

42

ISBN-13

978-1-230-14887-8

Barcode

9781230148878

Categories

LSN

1-230-14887-6



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