 | Thomas Tate (mathematical master.) - 1848 - 284 pages
...AC x sin A, .'. area triangle ABC = I x AB x AC X sin A; that is, the area of a triangle is equal to half the product of any two sides by the sine of the included angle. Since a parallelogram is double the area of the triangle cut off by a diagonal, it follows that, the... | |
 | John Hind - Trigonometry - 1855 - 540 pages
...Article (76) ; S=±AB.ACsmA: or, the area of a triangle may be arithmetically obtained by multiplying half the product of any two sides by the sine of the angle included between them. It is also evident that the Area of the triangle may always be used as... | |
 | James Edward Oliver - Trigonometry - 1881 - 120 pages
...VERTICES TO THE OPPOSITE SIDES : Let ABC be any triangle, and let к stand for its area, and pa, pb, p0, for the perpendiculars on a, b, c, respectively. CASE...side by the sine of the included angle. For, draw DcJ-Ав; then, v к =£DC. AB = £POC; . [geom. and v pe=&sinA; [I. § 7, note 120] .'. K So, K sinc... | |
 | James Edward Oliver - Trigonometry - 1881 - 140 pages
...perpendiculars on a, 6, c, respectively. CASE 1 . Given two sides and the lncluded angle; for example, 6, c, and A. (1) For the area, multiply half the product...side by the sine of the included angle. For, draw DC _L AB ; then, K = £DC . AB = $Pec » \_geom. and vj?,.= 6sinA; [I. § 7, note 126] .'. K = So, K =... | |
 | James Edward Oliver - Trigonometry - 1889 - 178 pages
...pь, pe, for the ordinates of A, в, c as to a, 6, c. (a) Given two sides and the included angle : For the area, multiply half the product of any two sides by the sine of the included amjle. For the ordinate of either vertex, multiply an adjacent side by the sine of the angle included... | |
 | James Edward Oliver - Trigonometry - 1890 - 186 pages
...pb, pe, for the ordinates of А, в, с as to а, b, с. (a) Given two sides and the included angle : For the area, multiply half the product of any two sides by the sine of the included angle. For the ordinate of either vertex, multiply an adjacent side by the sine of the angle included by this... | |
 | Edward Ira Edgerton, Perry Amherst Carpenter - Algebra - 1924 - 490 pages
...depth of the canon? 169. Areas of Triangles. — It can be proved that the area of any triangle equals half the product of any two sides by the sine of the included angle. Hence in formula form Area = £ ab sin C% Therefore, log area = log a + log b + log sin C — log 2.... | |
 | Richard Courant, Herbert Robbins - Juvenile Nonfiction - 1996 - 596 pages
...elementary means. We recall that the area of a triangle is equal to Jfbase X altitude) and is also given by half the product of any two sides by the sine of the included angle. We then have, in Figure 75, area OCA = \h.CA = \OA-OC sin Z CO A areaOCB = \h.CB = \OB-OC sin £ COB... | |
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Given two sides and the included angle; for example, b, c, and A. (1) For the area, multiply half the product of any two sides by the sine of the included angle. (2) For the perpendicular upon either given side, multiply the adjacent side by the sine... 