First Part of an Elementary Treatise on Spherical Trigonometry |
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Page 18
EXAMPLES . 1. Given in the spherical right triangle ( fig . 2. ) , h = 145o and A = 23 ° 28 ' ; to solve the triangle . Solution . By ( 558 ) , by ( 559 18 SPHERICAL TRIGONOMETRY . [ cz . II . Ø 11 .
EXAMPLES . 1. Given in the spherical right triangle ( fig . 2. ) , h = 145o and A = 23 ° 28 ' ; to solve the triangle . Solution . By ( 558 ) , by ( 559 18 SPHERICAL TRIGONOMETRY . [ cz . II . Ø 11 .
Page 20
When h and a are both equal to 90 ° , it may be shown , as in ( 563 ) , that the values of B and b are indeterminate , Example . Given , in the spherical right triangle ( fig .
When h and a are both equal to 90 ° , it may be shown , as in ( 563 ) , that the values of B and b are indeterminate , Example . Given , in the spherical right triangle ( fig .
Page 22
EXAMPLE . Given , in the spherical right triangle ( fig . 2. ) , a = 35 ° 44 ' and A = 37 ° 28 ' ; to solve the triangle . Ans . h 73 ° 45 ' , h 106 ° 15 ' , B 72 ' 51 " , or B 1070 9 , 6 69 ° 50 ' , 2 30. Problem .
EXAMPLE . Given , in the spherical right triangle ( fig . 2. ) , a = 35 ° 44 ' and A = 37 ° 28 ' ; to solve the triangle . Ans . h 73 ° 45 ' , h 106 ° 15 ' , B 72 ' 51 " , or B 1070 9 , 6 69 ° 50 ' , 2 30. Problem .
Page 23
B tang . b EXAMPLE . Given , in the spherical right triangle , ( fig . 2. ) , a = 118 ° 54 ' and B = 12 ° 19 ' ; to solve the triangle . Ans . h h = 118 ° 20 ' , A 95 ° 55 ' , 10 ° 49 ' . - 0 = 31. Problem . To solve a spherical right ...
B tang . b EXAMPLE . Given , in the spherical right triangle , ( fig . 2. ) , a = 118 ° 54 ' and B = 12 ° 19 ' ; to solve the triangle . Ans . h h = 118 ° 20 ' , A 95 ° 55 ' , 10 ° 49 ' . - 0 = 31. Problem . To solve a spherical right ...
Page 24
A = cotan . a sin . b . tang . a In the same way ( 592 ) cotan . B = cotan . b sin . a . EXAMPLE . Given , in the spherical right triangle ( fig . 2. ) , a = 1 ° and b = 100 ° ; to solve the triangle . Ans . h = 100 ° , A 1 ° 1 ' .
A = cotan . a sin . b . tang . a In the same way ( 592 ) cotan . B = cotan . b sin . a . EXAMPLE . Given , in the spherical right triangle ( fig . 2. ) , a = 1 ° and b = 100 ° ; to solve the triangle . Ans . h = 100 ° , A 1 ° 1 ' .
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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 angles are given becomes calculated called Corollary corresponding cosec cosine cotan deduced Demonstration denote determined differs divided equal to 90 equation EXAMPLES expressions factor fall on AC Fourthly fractions given angle gives greater than 90 half the sum hemisphere Hence hypothenuse impossible included angle known leads legs Lemma less than 90 Let ABC fig let fall logarithm lunary surface measured middle negative numerator obtained obtuse opposite angle opposite side perpendicular perpendicular BP positive preceding Problem proportion proved quantity quotient reduce result right angle Rules satisfy Scholium second member Secondly sides and angles sides equal signs sine Solution 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 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 1 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
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 51 - 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.
Page 51 - The cosine of half the sum of two angles of a spherical triangle is to the cosine of half their difference as the tangent of half the included side is to the tangent of half the sum of the other two sides.
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 |
| Publisher | J. Munroe, 1836 |
| Original from | the University of Michigan |
| Digitized | 11 Jun 2007 |
| Length | 71 pages |
| Export Citation | BiBTeX EndNote RefMan |