## First Part of an Elementary Treatise on Spherical Trigonometry |

### From inside the book

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Page 21

Benjamin Peirce. Solution .

Benjamin Peirce. Solution .

**Let ABC**(**fig**. 2. ) be the triangle ; a the given leg , and A the given angle . First . To find the hypothenuse h ; a is the middle part , and co . h and co . A are the opposite parts . sin . a = sin . h sin ... Page 23

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**Let ABC**(**fig**. 2. ) be the triangle ; a the given leg , and B the given angle . First . To find the hypothenuse h ; co . B is the middle part , and co . h and a are adjacent parts . Hence , by ( 474 ) , cos . B = tang . a cotan . h ... Page 24

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**Let ABC**(**fig**. 2. ) be the triangle , a and b the given legs . First . To find the hypotheneuse h ; co . h is the ... ( fig . 2. ) , a = 1 ° and b 100 ° ; to solve the triangle . Ans . h = 100 ° , = A = 10 1 ' , f B = 90 ° 12 ' . 32 ... Page 26

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**Let ABC**(**figs**.4 . and 5. ) be the given triangle . De- Fig.4 note by a , b , c , the sides respective- B ly opposite to the angles A , B , C. A From either of the vertices let fall the perpendicular BP upon the opposite side AC . Then ... Page 28

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**Let ABC**(**figs**. 4. and 5. ) be the trian- gle ; a and b the given sides , and C the given angle . From B let fall ... ( fig . 4. ) AP - b - PC , or ( fig . 5. ) APPC - b . Thirdly . To find the side c . BPC , co . a is the middle part ...### 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

୦୯ A'BC AC the perpendicular adjacent angles angle are known angles are given angles respectively equal AP and PC ar.co B'OC Corollary cosec cotan Demonstration differs from 90 equal to 90 fall on AC given angle given sides given value greater than 90 h tang half the sum Hence hypothenuse included angle legs are known Lemma less than 90 Let ABC fig let fall logarithm lunary surface means of 496 middle Napier's Rules negative obtuse opposite angle opposite side perpendicular BP perpendicular to OA planes BOC Problem quotient right angle right triangle fig right triangle PBC Scholium second member Secondly side BC sides and angles sides equal Solution of Spherical solve a spherical solve the triangle Spherical Oblique Triangles spherical right triangle spherical triangle ABC SPHERICAL TRIGONOMETRY substituted surface ABC tang.C tangent of half Thirdly trian triangle ABC figs

### 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.