## First Part of an Elementary Treatise on Spherical Trigonometry1836 - Spherical trigonometry - 71 pages |

### From inside the book

Results 1-5 of 15

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**substituted**in ( 454 ) and ( 455 ) , give sin . a = cotan . B sin . b cos . b cotan . A sin . a But , by ( 436 ) , sin . a Cos . a sin . b sin . b = cos . a Multiplying ( 458 ) by cos . b and ( 459 ) by cos . a , we have ; sin . a cos ... Page 13

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**substituted**in equation ( 448 ) , sin . a = sin . h sin . A , ( 525 ) h = 90 ° , it follows , from ( 157 ) and ( 524 ) , that = cos , a cos . b , 0 = ( 528 ) and therefore either cos . a or cos . b must be zero ; that is , either a or b ... Page 16

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**substituted**for it in ( 543 ) , and we have sin . ( A 90 ° ) < sin . B ; whence , by ( 507 ) , or or - ―――― B 90 ° < 90 ° ― ――― ( A — 90 ° ) , B90 ° 180 ° - A , < A + B 270 ° . Second Case . When one of the legs is equal to 90 ° , its ... Page 42

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**substituted**in ( 665 ) , gives ( 669 ) Cos . a : cos . c :: cos . PC : cos . b cos . PC + sin . b sin . PC . Dividing the two terms of the last ratio of this proportion by cos . PC , and reducing by ( 10 ) , we have ( 667 ) ( 668 ) ... Page 43

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**substituted**in the numerator of ( 675 ) , gives cos . c— cos . ( a + b ) . 1 + cos . C = sin . a sin . b = COS . C cos . c — cos . a cos . b + sin . a sin . b sin . a sin . b whence Now we have ( 122 ) ( 679 ) cos . ( MN ) cos . ( M + N ) ...### 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.