First Part of an Elementary Treatise on Spherical Trigonometry1836 - Spherical trigonometry - 71 pages |
From inside the book
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Page 10
... Lemma ( 496 ) , be both greater or both less than 90 ° . But when his greater than 90 ° , the first member of ( 492 ) is by ( 496 ) negative ; and therefore one of ( 494 ) the factors of the second member must be positive , while the ...
... Lemma ( 496 ) , be both greater or both less than 90 ° . But when his greater than 90 ° , the first member of ( 492 ) is by ( 496 ) negative ; and therefore one of ( 494 ) the factors of the second member must be positive , while the ...
Page 11
... Lemma . ( 506 ) 10. Lemma . Of angles less than 180 ° , the one which differs the least from 90 ° has the largest sine , tangent , and secant ; and the smallest cosine , cotan- ( 507 ) gent , and cosecant ; no regard being had to the ...
... Lemma . ( 506 ) 10. Lemma . Of angles less than 180 ° , the one which differs the least from 90 ° has the largest sine , tangent , and secant ; and the smallest cosine , cotan- ( 507 ) gent , and cosecant ; no regard being had to the ...
Page 46
... Lemma . The sine , cosine , secant , and cose( 705 ) cant of an angle , greater than 180 ° and less than 270 ° , are negative ; but its tangent and cotangent are positive . Demonstration . Let the excess of the angle above 180 ° be M ...
... Lemma . The sine , cosine , secant , and cose( 705 ) cant of an angle , greater than 180 ° and less than 270 ° , are negative ; but its tangent and cotangent are positive . Demonstration . Let the excess of the angle above 180 ° be M ...
Page 49
... lemmas . 58. Lemma . If we have the equation tang . M x = tang . N y ' we can deduce from it the following equation , x + y y x sin . ( MN ) sin . ( M — N ) Demonstration . tang . M = = sin . M. Cos . M ' ▭▭▭▭▭▭▭▭▭ We have from ...
... lemmas . 58. Lemma . If we have the equation tang . M x = tang . N y ' we can deduce from it the following equation , x + y y x sin . ( MN ) sin . ( M — N ) Demonstration . tang . M = = sin . M. Cos . M ' ▭▭▭▭▭▭▭▭▭ We have from ...
Page 61
... Lemma . The quotient of ( 128 ) , divided by ( 127 ) , is , by ( 10 ) and ( 11 ) , accenting the letters , cos . B Cos . A ' cos . B'cos . A ' - = = sin . ( 841 ) ( A ' + B ' ) sin . ¿ ( A ' — B ' ) cos . ≥ ( A ' + B ' ) cos . § ( A ...
... Lemma . The quotient of ( 128 ) , divided by ( 127 ) , is , by ( 10 ) and ( 11 ) , accenting the letters , cos . B Cos . A ' cos . B'cos . A ' - = = sin . ( 841 ) ( A ' + B ' ) sin . ¿ ( A ' — B ' ) cos . ≥ ( A ' + B ' ) cos . § ( A ...
Other editions - View all
First Part of an Elementary Treatise on Spherical Trigonometry (Classic Reprint) Benjamin Peirce No preview available - 2017 |
First Part of an Elementary Treatise on Spherical Trigonometry Benjamin Peirce No preview available - 2016 |
Common terms and phrases
A'BC acute adjacent angles angle are known becomes calculated called CHAPTER Corollary corresponding cosec cotan deduced Demonstration determined differs divided equal to 90 equation EXAMPLES factor fractions given angle gives greater than 90 half the sum hemisphere Hence hypothenuse impossible included angle legs Lemma less than 90 Let ABC fig let fall logarithm lunary surface measured middle Napier's Rules negative numerator obtained obtuse opposite angle opposite side perpendicular perpendicular BP positive Problem proportion proved quantity quotient reduce result right angle satisfy Scholium second member Secondly sides and angles sides equal signs sine Solution Solution of Spherical solve a spherical solve the triangle spherical right triangle spherical triangle ABC substituted supplements surface ABC tang tangent of half Theorem Thirdly tive trian triangle ABC figs whence
Popular passages
Page 1 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
Page 69 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Page 69 - THEOREM. The surface of a spherical triangle is measured by the excess of the sum of its three angles above two right angles, multiplied by the tri-rectangular triangle.
Page 8 - I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.
Page 8 - II. The sine of the middle part is equal to the product of the cosines of the opposite parts.
Page 30 - Any angle is greater than the difference between 180° and the sum of the other two angles.
Page 63 - The cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference as the cotangent of half the included angle is to the tangent of half the sum of the other two angles. The sine of half the sum of two sides of a spherical...
Page 62 - The sine of half the sum of two sides of a spherical triangle is to the sine of half their difference as the cotangent of half the included angle is to the tangent of half the difference of the other two angles.
Page 71 - ... and the sum of the angles in all the triangles is evidently the same as that of all the angles of the polygon ; hence, the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides less two, into the tri-rectangular triangle.
Bibliographic information
| Title | First Part of an Elementary Treatise on Spherical Trigonometry |
| Author | Benjamin Peirce |
| Published | 1836 |
| Original from | Harvard University |
| Digitized | 2 Mar 2007 |
| Length | 71 pages |
| Export Citation | BiBTeX EndNote RefMan |