Artículos relacionados a A treatise of practical arithmetic
Sinopsis
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. 1819 Excerpt: ...64.46 nat. rad. for 36" 30'. 9400 9555 64.46: 420:: 36.5: 237.8 = BC. 237.8 X 237.8 420 849 )56548.84( 73.8 7 5390 346.3 = AB. fiO 2648 73 2091 697 55700 38 55456 6952 Ex. 2. Suppose angle C = 30s, consequently angle C = 60, and leg AB = 56; required the leg BC, and hyp. AC. The three angles of any plane triangle are together = 180; consequently the two oblique angles will be complements of eacli other. OPERATION. '. 64 X 64 144 72 30 )4096(39 TTi Si--AB 39 676 75 675 72 32 104: 86;:: 33: 27 17' =y/C 62.43= A. Note: All eases in oblique-angled trigonometry triangles having no right angle may be solved by the above theorem, having first drawn a perpendicular from one of the angles to the opposite side or its production, and thus forming two right-angled triangles, either distinct from each other, or one included in the other, according as the perpendicular falls within or without the triangle--observing, that the perpendicular be so drawn, that in one of the right-angled triangles thus formed, two parts, besides the right angle, may be given. In the case when the three sides arc given, the perpendicular may be drawn to the greatest side from its opposite angle, and then the segments of the base, or side on which the perpendicular falls, may be found thus: multiply together the sum and difference of the other two sides, and divide the product by the base; then the quotient will be the difference of the segments, which added to the whole base will give double the greater segment, and this segment substracted from the base will leave the less segment. The application, however, of the foregoing theorem, to the sundry cases in oblique trigonometry; as well as in solving a variety of questions that might be proposed, relative to the calculation of heights and...
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Reseña del editor
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. 1819 Excerpt: ...64.46 nat. rad. for 36" 30'. 9400 9555 64.46: 420:: 36.5: 237.8 = BC. 237.8 X 237.8 420 849 )56548.84( 73.8 7 5390 346.3 = AB. fiO 2648 73 2091 697 55700 38 55456 6952 Ex. 2. Suppose angle C = 30s, consequently angle C = 60, and leg AB = 56; required the leg BC, and hyp. AC. The three angles of any plane triangle are together = 180; consequently the two oblique angles will be complements of eacli other. OPERATION. '. 64 X 64 144 72 30 )4096(39 TTi Si--AB 39 676 75 675 72 32 104: 86;:: 33: 27 17' =y/C 62.43= A. Note: All eases in oblique-angled trigonometry triangles having no right angle may be solved by the above theorem, having first drawn a perpendicular from one of the angles to the opposite side or its production, and thus forming two right-angled triangles, either distinct from each other, or one included in the other, according as the perpendicular falls within or without the triangle--observing, that the perpendicular be so drawn, that in one of the right-angled triangles thus formed, two parts, besides the right angle, may be given. In the case when the three sides arc given, the perpendicular may be drawn to the greatest side from its opposite angle, and then the segments of the base, or side on which the perpendicular falls, may be found thus: multiply together the sum and difference of the other two sides, and divide the product by the base; then the quotient will be the difference of the segments, which added to the whole base will give double the greater segment, and this segment substracted from the base will leave the less segment. The application, however, of the foregoing theorem, to the sundry cases in oblique trigonometry; as well as in solving a variety of questions that might be proposed, relative to the calculation of heights and...
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